Official Paper

NDA-I 2025 (Mathematics) Official Paper (Held On: 13 Apr, 2025) (Previous Year Paper)

120 questions · 150 minutes · with answers · free

Attempt as Timed Mock Test

Mathematics (120 questions)

1

The sum of the first k terms of a series S is 3k2+5k. Which one of the following is correct?

  1. ((a))

    The terms of S form an arithmetic progression with common difference 14.

  2. ((b))

    The terms of S form an arithmetic progression with common difference 6.

  3. ((c))

    The terms of S form a geometric progression with common ratio 10/7.

  4. ((d))

    The terms of S form a geometric progression with common ratio 11/4.

Show Answer
Answer: ((b))

The terms of S form an arithmetic progression with common difference 6.

Given:

The sum of the first k terms of a series S is given as Sk = 3k2 + 5k.

Concept:

To find the nth term of a series, we use the formula:

an = Sn - Sn-1

If the nth term forms an arithmetic progression (AP), the common difference (d) is given by:

d = an+1 - an

Calculation:

We are given Sk = 3k2 + 5k.

⇒ an = Sn - Sn-1

Also Sn = 3n2 + 5n and Sn-1 = 3(n-1)2 + 5(n-1)

⇒ an = [3n2 + 5n] - [3(n-1)2 + 5(n-1)]

⇒ an = [3n2 + 5n] - [3(n2 - 2n + 1) + 5n - 5]

⇒ an = [3n2 + 5n] - [3n2 - 6n + 3 + 5n - 5]

⇒ an = 3n2 + 5n - 3n2 + 6n - 3 - 5n + 5

⇒ an = 6n + 2

The series is an arithmetic progression (AP) if the difference between consecutive terms is constant.

⇒ Common difference d = an+1 - an

⇒ an+1 = 6(n+1) + 2 = 6n + 6 + 2 = 6n + 8

⇒ d = (6n + 8) - (6n + 2) = 6

∴ The terms of the series form an arithmetic progression with a common difference of 6.

Hence, the correct answer is Option B.

2

The sum of the first 8 terms of a GP is five times the sum of its first 4 terms. If r≠1 is the common ratio, then what is the number of possible real values of r?

  1. ((a))

    One 

  2. ((b))

    Two 

  3. ((c))

    Three

  4. ((d))

    More than three 

Show Answer
Answer: ((c))

Three

Concept:

Geometric Progression (GP) sums:

  • First term = a.
  • Common ratio = r (r ≠ 1).
  • Sum of first n terms: Sn = a × (1 − rn) / (1 − r).
  • Key idea: use formula for n = 8 and n = 4.
  • Algebra trick: substitute y = r4 to simplify.

 

Calculation:

Given relation between sums

⇒ S8 = 5 × S4

Using sum formula for r ≠ 1

⇒ a(1 − r8)/(1 − r)

⇒ = 5 × a(1 − r4)/(1 − r)

Cancel a and (1 − r)

⇒ 1 − r8 = 5(1 − r4)

Bring all terms to one side

⇒ r8 − 5r4 + 4 = 0

Put y = r4 (so y ≥ 0 for real r)

⇒ y2 − 5y + 4 = 0

Factor the quadratic

⇒ (y − 1)(y − 4) = 0

So y = 1 or y = 4

Case 1: y = 1 ⇒ r4 = 1

⇒ real r are 1 or −1

⇒ but r ≠ 1 (given), so keep r = −1

Case 2: y = 4 ⇒ r4 = 4

⇒ real r are √2 and −√2

Collect real values (excluding r = 1)

⇒ r = −1, r = √2, r = −√2

∴ Number of possible real values of r = 3.

3

If one root of the equation x2−kx+k=0 exceeds the other by 232\sqrt{3} then which one of the following is a value of k?

  1. ((a))

    3

  2. ((b))

    6

  3. ((c))

    9

  4. ((d))

    12

Show Answer
Answer: ((b))

6

Given:

The quadratic equation is x2 - kx + k = 0.

One root exceeds the other by 2√3.

⇒ α - β = 2√3.

Also, 

Sum of roots: α + β = k

Product of roots: α × β = k

Calculation:

We know the following identity 

(α+β)2=(α−β)2+4αβ(\alpha + \beta )^2 = (\alpha - \beta)^2 + 4\alpha\beta

⇒ k2 = (2√3)2 - 4k 

⇒ k2 - 12 - 4k = 0

⇒ k2 - 6k + 2k -12 = 0

⇒ k(k - 6) + 2 ( k - 6) = 0

⇒ (k - 6) (k + 2) = 0

⇒ k = 6 and k = -2

Thus, the possible values of k are 6 and -2

Hence, the correct answer is Option 2.

4

If 

x+5y=4x+\frac{5}{y}=4, and then y+5x=−4y+\frac{5}{x}= -4  what is (x + y) equal to?

  1. ((a))

    0

  2. ((b))

    1

  3. ((c))

    4

  4. ((d))

    5

Show Answer
Answer: ((a))

0

Calculation:

x+5y=4x+\frac{5}{y}=4

xy+5=4y.....(1)xy + 5 = 4y .....(1)

y+5x=−4y+\frac{5}{x}= -4

xy+5=−4x....(2)xy + 5 = -4x .... (2)

Subtracting 1 and 2

(xy+5)−(xy+5)=(4y)−(−4x)(xy+5)−(xy+5)=(4y)−(−4x)

0=4y+4x0=4y+4x

x+y=0.x+y=0.

Hence, the correct answer is Option 1.

5

If 5th, 7th and 13th terms of an AP are in GP, then what is the ratio of its first term to its common difference?

  1. ((a))

    -3

  2. ((b))

    -2

  3. ((c))

    2

  4. ((d))

    3

Show Answer
Answer: ((a))

-3

Concept:

In an AP, the nth term is given by:

Tn = a + (n - 1)d

In a GP, terms satisfy the relation:

Tm2 = Tn × Tp

Calculation:

Let the 5th, 7th, and 13th terms of the AP be T5, T7, and T13 respectively.

⇒ T5 = a + 4d

⇒ T7 = a + 6d

⇒ T13 = a + 12d

Since the terms are in GP:

⇒ T72 = T5 × T13

⇒ (a + 6d)2 = (a + 4d) × (a + 12d)

Expanding both sides:

⇒ (a + 6d)2 = a2 + 12ad + 48d2

⇒ a2 + 12ad + 36d2 = a2 + 16ad + 48d2

Cancel out common terms:

⇒ 36d2 = 48d2 + 4ad

⇒ -12d2 - 4ad = 0

⇒ -4d( 3d + a) = 0

Since d ≠ 0

⇒ a = -3d

Conclusion:

∴ Ratio of the first term to the common difference is:

⇒ a/d = -3

Hence, the correct answer is Option 1.

6

If p, 1, q are in AP and p, 2, q are in GP, then which of the following statements is/are correct?

I. p, 4, q are in HP.

II. (1/p), (1/4), (1/q) are in AP.

Select the answer using the code given below.

  1. ((a))

    I only

  2. ((b))

    II only

  3. ((c))

    Both I and II

  4. ((d))

    Neither I nor II

Show Answer
Answer: ((c))

Both I and II

Concept:

  1. Arithmetic Progression (AP): The difference between consecutive terms is constant.

⇒ If p, a, q are in AP, then 2a = p + q.

  1. Geometric Progression (GP): The ratio of consecutive terms is constant.

⇒ If p, b, q are in GP, then b2 = pq.

  1. Harmonic Progression (HP): Reciprocal of terms are in AP.

⇒ If p, c, q are in HP, then (1/p), (1/c), (1/q) are in AP.

Calculation:

Using AP condition: p, 1, q are in AP.

⇒ 2(1) = p + q

⇒ p + q = 2 ............(1)

Using GP condition: p, 2, q are in GP.

⇒ 22 = p × q

⇒ 4 = pq ............(2)

Now,

2pqp+q=82=4\frac{2pq}{p + q} = \frac{8}{2} = 4

and 

p+q2pq=14\frac{p + q}{2pq} = \frac{1}{4}

∴  (1/p), (1/4), (1/q) are in AP and 

 p, 4 and q are in H.P.

Thus, both statements are true.

Hence, the correct answer is Option 3.

7

If x=(1111)2​, y=(1001)2​ and z=(110)2​, then what is x3−y3−z3−3xyz equal to?

  1. ((a))

    (111001)2

  2. ((b))

    (100111)2

  3. ((c))

    (1)2

  4. ((d))

    (0)2

Show Answer
Answer: ((d))

(0)2

Calculation:

Convert x, y, and z from binary to decimal:

⇒ x = (1111)2 = 1⋅23+1⋅22+1⋅21+1⋅20=151\cdot2^{3}+1\cdot2^{2}+1\cdot2^{1}+1\cdot2^{0}=15

⇒ y = (1001)2 = 1⋅23+0⋅22+0⋅21+1⋅20=91\cdot2^{3}+0\cdot2^{2}+0\cdot2^{1}+1\cdot2^{0}=9

⇒ z = (110)2 = 1⋅22+1⋅21+0⋅20=61\cdot2^{2}+1\cdot2^{1}+0\cdot2^{0}=6

Now evaluate  x3−y3−z3−3xyzx^{3}-y^{3}-z^{3}-3xyz

153−93−63−3!×!15!×!9!×!6=3375−729−216−2430=015^{3}-9^{3}-6^{3}-3!\times!15!\times!9!\times!6 = 3375-729-216-2430 = 0

The result of the expression is 0. In binary, 0 is simply (0)2.

Hence, the correct answer is Option 4.

8

If Δ=∣abc def ghi∣\Delta = \begin{vmatrix} a & b & c \ d & e & f \ g & h & i \end{vmatrix}  

and A, B, C, D, G are the cofactors of the elements a, b, c, d, g respectively, then what is bB+cC−dD−gG equal to?

  1. ((a))

    0

  2. ((b))

    1

  3. ((c))

    Δ

  4. ((d))

    −Δ

Show Answer
Answer: ((a))

0

Calculation:

Cofactors:

⇒ A = ei – hf

⇒ B = gf – di,

⇒ C = dh – ge

⇒ D = ch – bi and

⇒ G = bf – ec

Now, bB + cC – dD – gG 

= bgf – bdi + cdh – cge – cdh + bdi – gbf + gec = 0

Hence, the correct answer is Option 1.

9

. Consider the following statements in respect of the determinant 

Δ=∣k(k+2)2k+11 2k+1k+21 331∣ \Delta = \begin{vmatrix} k(k+2) & 2k+1 & 1 \ 2k+1 & k+2 & 1 \ 3 & 3 & 1 \end{vmatrix}

I. Δ is positive if k>0.

II. Δ is negative if k<0.

III. Δ is zero if k=0.

How many of the statements given above are correct?

  1. ((a))

    None 

  2. ((b))

    One 

  3. ((c))

    Two 

  4. ((d))

    All three

Show Answer
Answer: ((b))

One 

Calculation:

Given,

Δ  =  ∣k(k+2)2k+11 2k+1k+21 331∣ \begin{vmatrix} k(k+2) & 2k+1 & 1\ 2k+1 & k+2 & 1\ 3 & 3 & 1 \end{vmatrix}

Simplify the determinant by the column operation  C1→C1−C2C_1 \rightarrow C_1 - C_2:

Δ=∣k2−12k+11 k−1k+21 031∣ \Delta = \begin{vmatrix} k^{2}-1 & 2k+1 & 1\ k-1 & k+2 & 1\ 0 & 3 & 1 \end{vmatrix}

Expanding along the third row,

Δ=−3∣k2−11 k−11∣;+;∣k2−12k+1 k−1k+2∣=(k−1)3. \Delta = -3 \begin{vmatrix} k^{2}-1 & 1\ k-1 & 1 \end{vmatrix} ;+; \begin{vmatrix} k^{2}-1 & 2k+1\ k-1 & k+2 \end{vmatrix} = (k-1)^{3}.

Thus  Δ=(k−1)3 \Delta = (k-1)^{3}.

Sign analysis

  • k>0k>0: if 0<k<10<k<1, Δ < 0; if k>1k>1, Δ > 0  ⇒ Statement I is false.
  • k<0k<0:  Δ<0\Delta<0  ⇒ Statement II is true.
  • k=0k=0:  Δ=(−1)3=−1≠0\Delta=(-1)^{3}=-1\neq0  ⇒ Statement III is false.

∴ Only Statement II is correct  ⇒  exactly one statement is true.

Hence, the correct answer is Option 2.

10

If ∣23+i−1 3−i0i−1 −1−1−i1∣=A+iB \begin{vmatrix} 2 & 3+i & -1 \ 3-i & 0 & i -1 \ -1 &-1 -i & 1 \end{vmatrix} = A + iB

where i= −1\sqrt{-1} ,  then what is A+B equal to?

  1. ((a))

    -10

  2. ((b))

    -6

  3. ((c))

    0

  4. ((d))

    6

Show Answer
Answer: ((b))

-6

Calculation:

Determinant Δ = a(ei−fh)−b(di−fg)+c(dh−eg)a(ei−fh)−b(di−fg)+c(dh−eg)

Now, For our matrix, 

a=2,b=3+i,c=−1,d=3−i,e=0,f=i,g=−1,h=−i,i=1a=2,b=3+i,c=−1,d=3−i,e=0,f=i,g=−1,h=−i,i=1

calculate the subdeterminants

⇒ ei−fh=(0)(1)−(i)(−i)=0−(−1)=1 ei−fh=(0)(1)−(i)(−i)=0−(−1)=1

⇒ di−fg=(3−i)(1)−(i)(−1)=3−i+i=3di−fg=(3−i)(1)−(i)(−1)=3−i+i=3

⇒ dh−eg=(3−i)(−i)−(0)(−1)=−3i+i2=−3i−1=−1−3idh−eg=(3−i)(−i)−(0)(−1)=−3i+i 2=−3i−1=−1−3i

⇒ Δ = 2(1)−(3+i)(3)+(−1)(−1−3i)2(1)−(3+i)(3)+(−1)(−1−3i)

⇒ Δ = 2−9−3i+1+3i2−9−3i+1+3i

⇒ Δ=−6+0iΔ=−6+0i

Since we are given that Δ=A+iBΔ=A+iB comparing the real and imaginary parts, we find:

A  = -6 and B = 0

Thus A + B = -6 + 0 = - 6

Hence, the Correct answer is Option 2.

11

If A2+B2+C2=0, then what is the value of the following?

∣1cosCcosB cosC1cosA cosBcosA1∣\begin{vmatrix} 1& cosC& cosB\ cosC&1&cosA \ cosB&cosA&1 \end{vmatrix}

  1. ((a))

    -1

  2. ((b))

    0

  3. ((c))

    1

  4. ((d))

    2

Show Answer
Answer: ((b))

0

Concept:

When A2 + B2 + C2 = 0, it implies A = B = C = 0 (since the squares of real numbers are non-negative).

Substitute the values of A, B, and C for determinant calculation into the matrix

Calculation:

∣1cos0cos0 cos01cos0 cos0cos01∣\begin{vmatrix} 1& cos0& cos0\ cos0&1&cos0 \ cos0&cos0&1 \end{vmatrix}

Since, Cos0 =1

Thus Matrix becomes 

∣111 111 111∣\begin{vmatrix} 1& 1& 1\ 1&1&1 \ 1&1&1 \end{vmatrix}

Now determinant = 1[(1×1 - 1×1)] - 1[(1×1 - 1×1)] + 1[(1×1 - 1×1)]

= = 1(0) - 1(0) + 1(0) = 0

∴ The value of the determinant is 0.

Hence, the correct answer is Option 2.

12

If ω is a non-real cube root of unity, then what is a root of the following equation?∣x+1ωω2 ωx+ω21 ω21x+ω∣=0 \begin{vmatrix} x+1 & \omega & \omega^2 \ \omega & x+\omega^2 & 1 \ \omega^2 & 1 & x+\omega \end{vmatrix} = 0

  1. ((a))

    x=0

  2. ((b))

    x=1

  3. ((c))

    x=ω

  4. ((d))

    x=ω2

Show Answer
Answer: ((a))

x=0

Calculation:

Given,

Let ω be a non-real cube root of unity, so ω3=1 \omega^{3}=1 and 1+ω+ω2=0 1+\omega+\omega^{2}=0 .

Consider the determinant

Δ(x)=∣x+1ωω2 ωx+ω21 ω21x+ω∣=0. \Delta(x)= \begin{vmatrix} x+1 & \omega & \omega^{2}\ \omega & x+\omega^{2} & 1\ \omega^{2} & 1 & x+\omega \end{vmatrix}=0.

Step 1 — Column operation:  Replace the first column by C1−C2C_{1}-C_{2}:

Δ(x)=∣x2−1,2!k!+!1,1 k−1k+21 031∣. \Delta(x)= \begin{vmatrix} x^{2}-1 & ,2!k!+!1, & 1\ k-1 & k+2 & 1\ 0 & 3 & 1 \end{vmatrix}.

Step 2 — Expansion along the third row:

Δ(x)=−3!∣k2−11 k−11∣+∣k2−12k+1 k−1k+2∣, \Delta(x)= -3! \begin{vmatrix} k^{2}-1 & 1\ k-1 & 1 \end{vmatrix} + \begin{vmatrix} k^{2}-1 & 2k+1\ k-1 & k+2 \end{vmatrix},

which simplifies to

Δ(x)=x(x2−1)−x(ω+ω2)=x(x2−1)+x=x3. \Delta(x)=x(x^{2}-1)-x\bigl(\omega+\omega^{2}\bigr) =x(x^{2}-1)+x =x^{3}.

Step 3 — Equate to zero:

Δ(x)=0;⟹;x3=0;⟹;x=0. \Delta(x)=0 ;\Longrightarrow; x^{3}=0 ;\Longrightarrow; x=0.

∴ The root of the equation is  x=0 x = 0 .

Hence, the correct answer is Option 1.

13

What is (3+i3−i)3 \left( \frac{\sqrt{3}+i}{\sqrt{3}-i} \right)^3  equal to?

  1. ((a))

    -1

  2. ((b))

    0

  3. ((c))

    1

  4. ((d))

    3

Show Answer
Answer: ((a))

-1

Concept:

The key concept involves simplifying complex numbers in the polar form. We use the argument (angle) of the complex number and De Moivre's theorem, which states:

  (r(cosθ + i sinθ))n = rn(cos(nθ) + i sin(nθ))

Calculation:

⇒ (3+i3−i)3\left( \frac{\sqrt{3} + i}{\sqrt{3} - i} \right)^3

Multiply the numerator and the denominator by the conjugate of the denominator

⇒ 3+i3−i×3+i3+i\frac{\sqrt{3} + i}{\sqrt{3} - i} \times \frac{\sqrt{3} + i}{\sqrt{3} + i}

⇒ (3+i)2(3−i)(3+i)\frac{(\sqrt{3} + i)^2}{(\sqrt{3} - i)(\sqrt{3} + i)}

 2+23i4=1+3i2\frac{2 + 2\sqrt{3}i}{4} = \frac{1 + \sqrt{3}i}{2}

Convert to polar form. The modulus r is 

r=(12)2+(32)2=14+34=1=1r = \sqrt{\left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1

The argument θ is 

θ=tan⁡−1(3212)=tan⁡−1(3)=π3\theta = \tan^{-1}\left( \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} \right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}

So the polar form of the complex number is

1(cos⁡π3+isin⁡π3)1 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)

To cube the complex number, we use De Moivre’s Theorem

In our case, r = 0 so,

(cos⁡π3+isin⁡π3)3=cos⁡π+isin⁡π=−1+0i=−1\left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)^3 = \cos \pi + i \sin \pi = -1 + 0i = -1

Thus, the result of cubing the complex number is −1\boxed{-1}

Hence, the correct answer is Option 1.

14

If x2− x +1 = 0, then what is (x−1x)2+(x−1x)4+(x−1x)8(x-\frac{1}{x})^2+(x-\frac{1}{x})^4+(x-\frac{1}{x})^8

  1. ((a))

    81

  2. ((b))

    85

  3. ((c))

    87

  4. ((d))

    90

Show Answer
Answer: ((c))

87

Calculation:

Given,

The equation is x2−x+1=0 x^2 - x + 1 = 0

We need to find the value of the following expression:

(x−1x)2+(x−1x)4+(x−1x)8 \left( x - \frac{1}{x} \right)^2 + \left( x - \frac{1}{x} \right)^4 + \left( x - \frac{1}{x} \right)^8

The equation x2−x+1=0 x^2 - x + 1 = 0 is solved as follows:

x=1±−32=eiπ/3orx=e−iπ/3 x = \frac{1 \pm \sqrt{-3}}{2} = e^{i \pi / 3} \quad \text{or} \quad x = e^{-i \pi / 3}

Now, substitute the value of x x into the expression x−1x x - \frac{1}{x} :

x−1x=i3 x - \frac{1}{x} = i\sqrt{3}

Evaluate the powers of x−1x x - \frac{1}{x}

Now, let's evaluate each term in the expression:

(x−1x)2=(i3)2=−3 \left( x - \frac{1}{x} \right)^2 = (i \sqrt{3})^2 = -3

(x−1x)4=(−3)2=9 \left( x - \frac{1}{x} \right)^4 = (-3)^2 = 9

(x−1x)8=92=81 \left( x - \frac{1}{x} \right)^8 = 9^2 = 81

Now, sum the values:

−3+9+81=87 -3 + 9 + 81 = 87

∴ The value of the expression is 87.

Hence, the correct answer is Option 3.

15

How many 7-letter words (with or without meaning) can be constructed using all the letters of the word CAPITAL so that all the consonants come together in each word?

  1. ((a))

    360

  2. ((b))

    300

  3. ((c))

    288

  4. ((d))

    240

Show Answer
Answer: ((c))

288

Concept:

If all consonants are treated as a single unit, the problem reduces to arranging the unit (CPTL) and vowels (A, I, A).

Within the consonant unit, the consonants themselves can be permuted.

When letters repeat, the arrangements are divided by the factorial of the repeated letters.

Calculation:

⇒ Treat consonants CPTL as a single unit.

⇒ Total "super units" = {CPTL, A, I, A} = 4 units.

⇒ Arrangements of super units = 4! / 2! (since A repeats).

⇒ Arrangements of super units = 24 / 2 = 12. ...(1)

⇒ Internal arrangements of consonants CPTL = 4!.

⇒ Internal arrangements = 4 × 3 × 2 × 1 = 24. ...(2)

⇒ Total arrangements = (Arrangements of super units) × (Internal arrangements of consonants).

⇒ Total arrangements = 12 × 24.

⇒ Total arrangements = 288.

∴ The total number of 7-letter words = 288.

Hence, the correct answer is Option 3.

16

If z≠0z\ne0 is a complex number, then what is amp(z)+amp(zˉ) equal to?

  1. ((a))

    0

  2. ((b))

    π/2

  3. ((c))

    π

  4. ((d))

    2π

Show Answer
Answer: ((a))

0

Concept:

  1. The amplitude (or argument) of a complex number z=r(cos⁡θ+isin⁡θ) z = r(\cos\theta + i\sin\theta) is given by amp(z)=θ \text{amp}(z) = \theta .
  2. The conjugate of z z , denoted as z‾ \overline{z} , has an amplitude amp(z‾)=−θ \text{amp}(\overline{z}) = -\theta , because conjugating a complex number reflects it across the real axis in the Argand plane.

Formula Used:

amp(z)+amp(z‾)=θ+(−θ)=0 \text{amp}(z) + \text{amp}(\overline{z}) = \theta + (-\theta) = 0 .

Calculation:

z=r(cos⁡θ+isin⁡θ) z = r(\cos\theta + i\sin\theta)

z‾=r(cos⁡θ−isin⁡θ) \overline{z} = r(\cos\theta - i\sin\theta)

amp(z)+amp(z‾)=θ+(−θ)=0 \text{amp}(z) + \text{amp}(\overline{z}) = \theta + (-\theta) = 0

Conclusion:

∴amp(z)+amp(z‾)=0 \therefore \text{amp}(z) + \text{amp}(\overline{z}) = 0 .

Hence, the correct answer is Option 1.

17

How many sides are there in a polygon that has 20 diagonals?

  1. ((a))

    6

  2. ((b))

    7

  3. ((c))

    8

  4. ((d))

    10

Show Answer
Answer: ((c))

8

Calculation:

Given,

Number of diagonals, D = 20 

Formula for the number of diagonals in a polygon is:

D=n(n−3)2D = \frac{n(n - 3)}{2}

Substitute the value of D = 20  in the formula:

⇒ 20=n(n−3)220 = \frac{n(n - 3)}{2}

Multiply both sides by 2:

⇒ 40=n(n−3)40 = n(n - 3)

 

⇒ n2−3n−40=0n^2 - 3n - 40 = 0

We now solve this quadratic equation:

n=−b±b2−4ac2an = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

⇒ n=−(−3)±(−3)2−4(1)(−40)2(1)n = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-40)}}{2(1)}

 

⇒ n=3±132n = \frac{3 \pm 13}{2}

The two solutions are:

⇒ n=3+132=8n = \frac{3 + 13}{2} = 8

or

⇒ n=3−132=−5n = \frac{3 - 13}{2} = -5

Since the number of sides cannot be negative, we choose n = 8.

∴ The polygon has 8 sides.

Hence, the correct answer is Option 3.

18

In how many ways can the letters of the word DELHI be arranged, keeping the positions of vowels and consonants unchanged?

  1. ((a))

    6

  2. ((b))

    9

  3. ((c))

    12

  4. ((d))

    14

Show Answer
Answer: ((c))

12

Calculation:

Number of vowels = 2 (E, I). Positions of vowels are fixed.

⇒ Arrangements of vowels = 2! = 2 × 1 = 2

Number of consonants = 3 (D, L, H). Positions of consonants are fixed.

⇒ Arrangements of consonants = 3! = 3 × 2 × 1 = 6

Total arrangements = Arrangements of vowels × Arrangements of consonants.

⇒ Total arrangements = 2 × 6 = 12

∴ The total number of arrangements is 12.

Hence, the correct answer is Option 3.

19

What is the number of positive integer solutions of x+y+z=5?

  1. ((a))

    3

  2. ((b))

    5

  3. ((c))

    6

  4. ((d))

    9

Show Answer
Answer: ((c))

6

Concept:

The number of positive integer solutions for the equation x + y + z = n is given by the combination formula:

Number of solutions = C(n - 1, k - 1), where k is the number of variables.

Formula Used:

C(n, r) = n! / [r! × (n - r)!]

Calculation:

Here, n = 5 and k = 3.

The number of solutions is:

⇒ C(5 - 1, 3 - 1) = C(4, 2)

⇒ C(4, 2) = 4! / [2! × (4 - 2)!]

⇒ C(4, 2) = 24 / [2 × 2]

⇒ C(4, 2) = 6

The correct answer is Option 3.

20

What is the number of rational terms in the expansion of (3+514)12(\sqrt{3}+5^\frac{1}{4})^{12}

  1. ((a))

    2

  2. ((b))

    3

  3. ((c))

    4

  4. ((d))

    6

Show Answer
Answer: ((c))

4

Concept:

A term in the binomial expansion of (a + b)n is given by Tk+1 = C(n, k) × an-k × bk.

For a term to be rational, the exponents of both √3 and 51/4 must be integers.

Formula Used:

In (√3)n-k, n-k must be even for it to be rational.

In (51/4)k, k must be a multiple of 4 for it to be rational.

Calculation:

Let n = 12:

⇒ For √3n-k to be rational, n-k must be even.

⇒ Since n = 12, k must also be even.

⇒ For (51/4)k to be rational, k must be a multiple of 4.

⇒ The values of k that satisfy both conditions (k is even and a multiple of 4) are:

⇒ k = 0, 4, 8, and 12.

⇒ These correspond to 4 rational terms in the expansion.

Hence, the Correct answer is Option 3.

21

If the sum of binomial coefficients in the expansion of (x+y)n is 256, then the greatest binomial coefficient occurs in which one of the following terms?

  1. ((a))

    Third

  2. ((b))

    Fourth

  3. ((c))

    Fifth

  4. ((d))

    Ninth

Show Answer
Answer: ((c))

Fifth

Concept:

Sum of Binomial Coefficients and Greatest Binomial Coefficient:

  • The sum of binomial coefficients in the expansion of (x+y)n (x + y)^n is calculated by substituting x = 1 and y = 1. The result is 2n2^n.
  • To find the greatest binomial coefficient, we analyze the coefficients C(n,r)C(n, r) where r is the term index in the expansion. The greatest coefficient occurs near the middle term(s).
  • Key Formulae:
  • Sum of binomial coefficients: Sum=2n \text{Sum} = 2^n
  • Binomial coefficient: C(n,r)=n!r!(n−r)! C(n, r) = \frac{n!}{r!(n-r)!}
  • Greatest binomial coefficient: For even n, it occurs at r = n/2. For odd n, it occurs at r = (n-1)/2 and r = (n+1)/2.

 

Calculation:

Given,

Sum of binomial coefficients = 2n=2562^n = 256

We calculate n:

2n=256 2^n = 256

⇒ 28=2562^8 = 256

Greatest Binomial Coefficient:

For n=8 n = 8 (even), the greatest binomial coefficient occurs at r=n/2=8/2=4 r = n/2 = 8/2 = 4 .

⇒ The term index is r = 4, which corresponds to the 5th term (since indexing starts from 0).

∴ The greatest binomial coefficient occurs in the 5th term.

Hence, the correct answer is Option 3.

22

If k<(2+1)3<k+2,k<(\sqrt{2}+1)^3<k+2, where k is a natural number, the value of k?

  1. ((a))

    11

  2. ((b))

    13

  3. ((c))

    15

  4. ((d))

    17

Show Answer
Answer: ((b))

13

Calculation:

Given,

The inequality is: k<(√2+1)3<k+2 k < (√{2} + 1)^3 < k + 2 , where k is a natural number.

We need to compute (√2+1)3(√{2} + 1)^3

(√2+1)3=2√2+6+3√2+1=5√2+7 (√{2} + 1)^3 = 2√{2} + 6 + 3√{2} + 1 = 5√{2} + 7

Approximating √ 2≈ 1.414 , we find:

5√2+7≈5×1.414+7=7.07+7=14.07 5√{2} + 7 ≈ 5 \times 1.414 + 7 = 7.07 + 7 = 14.07

The inequality becomes:

k<14.07<k+2 k < 14.07 < k + 2

This implies that k > 12.07 \), so the smallest integer value of k  is:

k=13 k = 13

∴ The value of k  is 13.

Hence, the correct answer is Option 2.

23

If

 [x11][123 456 789][1 1 x]=[45]\begin{bmatrix} x & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{bmatrix} \begin{bmatrix} 1 \ 1 \ x \end{bmatrix} = \begin{bmatrix} 45 \end{bmatrix}

then which one of the following is a value of x?

  1. ((a))

    -2

  2. ((b))

    -1

  3. ((c))

    0

  4. ((d))

    1

Show Answer
Answer: ((d))

1

Calculation:

Multiply Matrix 1 and Matrix 2:

[x11]×[123 456 789]=[x+4+72x+5+83x+6+9]\begin{bmatrix} x & 1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{bmatrix} = \begin{bmatrix} x+4+7 & 2x+5+8 & 3x+6+9 \end{bmatrix}

⇒ [x+112x+133x+15]\begin{bmatrix} x+11 & 2x+13 & 3x+15 \end{bmatrix}

Multiply the resulting matrix with Matrix 3:

[x+112x+133x+15]×[1 1 x]\begin{bmatrix} x+11 & 2x+13 & 3x+15 \end{bmatrix} \times \begin{bmatrix} 1 \ 1 \ x \end{bmatrix}

=(x+11)⋅1+(2x+13)⋅1+(3x+15)⋅x= (x+11) \cdot 1 + (2x+13) \cdot 1 + (3x+15) \cdot x

⇒ (x+11)+(2x+13)+(3x2+15x)(x+11) + (2x+13) + (3x^2+15x)

⇒ (3x2+18x+24)(3x^2 + 18x + 24)

Equate the result to 45:

(3x2+18x+24=45)(3x^2 + 18x + 24 = 45)

⇒ (3x2+18x−21=0)(3x^2 + 18x - 21 = 0)

(x2+6x−7=0)(x^2 + 6x - 7 = 0)

(x2+7x−x−7=0)(x^2 + 7x - x - 7 = 0)

⇒ (x=1 or x=−7)(x = 1 \text{ or } x = -7)

Step 5: Verify:

For x=1x = 1, substitute back:

(3(1)2+18(1)+24=45)(3(1)^2 + 18(1) + 24 = 45)

45 =45

∴ The correct value of x is 1.

Hence, the correct answer is Option 4.

24

If

 A=[yxx zxy xyz]A = \begin{bmatrix} y & x & x \ z & x & y \ x & y & z \end{bmatrix}

where x,y,z are integers, is an orthogonal matrix, then what is the value of x2+y2+z2?

  1. ((a))

    0

  2. ((b))

    1

  3. ((c))

    4

  4. ((d))

    14

Show Answer
Answer: ((b))

1

Calculation:

Given,

The matrix A is:

A=[yxx zxy xyz] A = \begin{bmatrix} y & x & x \ z & x & y \ x & y & z \end{bmatrix}

Since A  is an orthogonal matrix, we know that:

AT=A−1⇒ATA=I A^T = A^{-1} \quad \Rightarrow \quad A^T A = I

This property tells us that A  is orthogonal, and it implies that ATAA^T A (the product of A's transpose and A is equal to the identity matrix I , which is:

ATA=[100 010 001] A^T A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}

Now, let’s calculate ATAA^T A step by step. The transpose of matrix A , denoted ATA^T is:

AT=[yzx xxy xyz] A^T = \begin{bmatrix} y & z & x \ x & x & y \ x & y & z \end{bmatrix}

Now, we perform matrix multiplication between ATA^T and A:

ATA=[yzx xxy xyz][yxx zxy xyz] A^T A = \begin{bmatrix} y & z & x \ x & x & y \ x & y & z \end{bmatrix} \begin{bmatrix} y & x & x \ z & x & y \ x & y & z \end{bmatrix}

Performing this multiplication, we get the following matrix:

ATA=[y2+z2+x2xy+zx+xyxz+yz+x2 xy+zx+xyx2+x2+y2xy+xz+yz xz+yz+x2xy+xz+yzx2+y2+z2] A^T A = \begin{bmatrix} y^2 + z^2 + x^2 & xy + zx + xy & xz + yz + x^2 \ xy + zx + xy & x^2 + x^2 + y^2 & xy + xz + yz \ xz + yz + x^2 & xy + xz + yz & x^2 + y^2 + z^2 \end{bmatrix}

This matrix must be equal to the identity matrix I , which is:

[100 010 001] \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}

By comparing the elements of the matrices, we get the following system of equations:

  1. y2+z2+x2=1 y^2 + z^2 + x^2 = 1 2. xy+zx+xy=0xy + zx + xy = 0  3. xz+yz+x2=1xz + yz + x^2 = 1

Thus, the key result from the orthogonality condition is:

x2+y2+z2=1 x^2 + y^2 + z^2 = 1

Hence, the correct answer is Option 2.

25

Consider the following in respect of a non-singular matrix M:

I. ∣M2∣=∣M∣2

II. ∣M∣=∣M−1∣

III. ∣M∣=∣MT∣

How many of the above are correct?

  1. ((a))

    None

  2. ((b))

    One

  3. ((c))

    Two

  4. ((d))

    All three

Show Answer
Answer: ((c))

Two

Calculation:

Statement I

∣M2∣=∣M×M∣=∣M∣⋅∣M∣=∣M∣2 |M^2| = |M \times M| = |M| \cdot |M| = |M|^2

⇒ Statement I is correct.

Statement II

For a non-singular matrix, M×M−1=I M \times M^{-1} = I , where I I is the identity matrix.

∣M∣⋅∣M−1∣=∣I∣=1 |M| \cdot |M^{-1}| = |I| = 1

⇒ Statement II is incorrect unless ∣M∣=±1 |M| = \pm 1 .

Statement III

The determinant of a matrix is equal to the determinant of its transpose:

∣M∣=∣MT∣ |M| = |M^T|

⇒ Statement III is correct.

Out of the three statements, two are correct: I and III.

Hence, the correct answer is Option 3.

26

If f(θ)=[cos⁡θsin⁡θ −sin⁡θcos⁡θ]f(\theta) = \begin{bmatrix} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{bmatrix} then what is (f(π))2 equal to?

  1. ((a))

    [−10 0−1]\begin{bmatrix} -1&0\ 0&-1\end{bmatrix}

  2. ((b))

    [11 11]\begin{bmatrix} 1&1\ 1&1\end{bmatrix}

  3. ((c))

    [−10 01]\begin{bmatrix} -1&0\ 0&1\end{bmatrix}

  4. ((d))

    [10 01]\begin{bmatrix} 1&0\ 0&1\end{bmatrix}

Show Answer
Answer: ((d))

[10 01]\begin{bmatrix} 1&0\ 0&1\end{bmatrix}

Concept:

Rotation Matrix:

  • A rotation matrix is used to perform a rotation in a Euclidean space. It is a square matrix that describes the rotation of a vector space.
  • For a 2D rotation, the matrix is given by: f(θ)=[cos⁡θsin⁡θ −sin⁡θcos⁡θ]f(\theta) = \begin{bmatrix} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{bmatrix}
  • Here, θ is the angle of rotation in radians.
  • cos θ: Represents the cosine of the rotation angle.
  • sin θ: Represents the sine of the rotation angle.
  • Key property of a rotation matrix:
  • The transpose of the matrix is equal to its inverse.
  • The determinant of the matrix is always equal to 1.
  • When θ = π, the rotation matrix becomes: f(π)=[cos⁡πsin⁡π −sin⁡πcos⁡π]=[−10 0−1]f(\pi) = \begin{bmatrix} \cos \pi & \sin \pi \ -\sin \pi & \cos \pi \end{bmatrix} = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}

 

Calculation:

Given,

Rotation matrix at θ = π:

f(π)=[−10 0−1]f(\pi) = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}

To find (f(π))2, multiply the matrix by itself:

f(π)×f(π)=[−10 0−1]×[−10 0−1]f(\pi) × f(\pi) = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix} × \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}

Using matrix multiplication:

Top-left element: (-1)(-1) + (0)(0) = 1

Top-right element: (-1)(0) + (0)(-1) = 0

Bottom-left element: (0)(-1) + (-1)(0) = 0

Bottom-right element: (0)(0) + (-1)(-1) = 1

Resulting matrix:

f(π)2=[10 01]f(\pi)^2 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}

∴ (f(π))2 is equal to the identity matrix, which is [10 01]\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}.

Hence, the correct answer is Option 4.

27

If \(A=\begin{bmatrix}1&2&2 \ 2&1&2\2&2&1\end{bmatrix}\) then what is A2−4A equal to?

  1. ((a))

    −5I3

  2. ((b))

    −I3

  3. ((c))

    I3

  4. ((d))

    5I3

Show Answer
Answer: ((d))

5I3

Calculation:

Given,

A=[122 212 221]A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix}

⇒A2=A×A=[122 212 221]×[122 212 221]\Rightarrow A^2 = A \times A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix}

 

⇒A2=[988 898 889]\Rightarrow A^2 = \begin{bmatrix} 9 & 8 & 8 \ 8 & 9 & 8 \ 8 & 8 & 9 \end{bmatrix}

Now 4A

⇒4A=4×[122 212 221]\Rightarrow 4A = 4 \times \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix}

⇒4A=[488 848 884]\Rightarrow 4A = \begin{bmatrix} 4 & 8 & 8 \ 8 & 4 & 8 \ 8 & 8 & 4 \end{bmatrix}

Also A2 - 4A

⇒A2−4A=[988 898 889]−[488 848 884]\Rightarrow A^2 - 4A = \begin{bmatrix} 9 & 8 & 8 \ 8 & 9 & 8 \ 8 & 8 & 9 \end{bmatrix} - \begin{bmatrix} 4 & 8 & 8 \ 8 & 4 & 8 \ 8 & 8 & 4 \end{bmatrix}

⇒A2−4A=[9−48−88−8 8−89−48−8 8−88−89−4]\Rightarrow A^2 - 4A = \begin{bmatrix} 9-4 & 8-8 & 8-8 \ 8-8 & 9-4 & 8-8 \ 8-8 & 8-8 & 9-4 \end{bmatrix}

⇒A2−4A=[500 050 005]\Rightarrow A^2 - 4A = \begin{bmatrix} 5 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 5 \end{bmatrix}

Relate the result to the identity matrix I3I3I3

⇒A2−4A=5I3\Rightarrow A^2 - 4A = 5I_3

Hence, the correct answer is option 4.

28

If the number of selections of r as well as (n+r) things from 5n different things are equal, then what is the value of r?

  1. ((a))

    n

  2. ((b))

    2n

  3. ((c))

    3n

  4. ((d))

    4n

Show Answer
Answer: ((b))

2n

Calculation:

Given,

The number of selections of r things from 5n things is equal to the number of selections of n+r things from 5n things.

This gives the equation:

(5nr)=(5nn+r) \binom{5n}{r} = \binom{5n}{n+r}

Using the symmetry property of binomial coefficients, we know that:

(5nr)=(5n5n−r) \binom{5n}{r} = \binom{5n}{5n-r}

Thus, comparing the two sides:

r=5n−(n+r) r = 5n - (n + r)

Simplifying the equation:

r=5n−n−r r = 5n - n - r

2r=4n 2r = 4n

r=2n r = 2n

∴ The value of r is 2n.

Hence, the correct answer is Option 2.

29

What is the number of selections of at most 3 things from 6 different things?

  1. ((a))

    20

  2. ((b))

    22

  3. ((c))

    41

  4. ((d))

    42

Show Answer
Answer: ((d))

42

Calculation:

Given:

Total number of items, n = 6

We need to calculate the total number of selections for r = 0, 1, 2, and 3:

For r = 0

6C0=6!0!(6−0)!=11=1 ^6C_0 = \frac{6!}{0!(6-0)!} = \frac{1}{1} = 1

⇒ Number of ways to select 0 items = 1

For r = 1

6C1=6!1!(6−1)!=61=6 ^6C_1 = \frac{6!}{1!(6-1)!} = \frac{6}{1} = 6

⇒ Number of ways to select 1 item = 6

For r = 2

6C2=6!2!(6−2)!=6×52×1=15 ^6C_2 = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15

⇒ Number of ways to select 2 items = 15

For r = 3

6C3=6!3!(6−3)!=6×5×43×2×1=20 ^6C_3 = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20

⇒ Number of ways to select 3 items = 20

Total number of selections:

6C0+6C1+6C2+6C3=1+6+15+20=42 ^6C_0 + ^6C_1 + ^6C_2 + ^6C_3 = 1 + 6 + 15 + 20 = 42

∴ The total number of selections of at most 3 things from 6 different things is 42.

Hence, the correct answer is Option 4.

30

If \(A=\begin{bmatrix}x&y&z\y&z&x\z&x&y\end{bmatrix}\)

where x,y,z are integers, is an orthogonal matrix, then what is A2 equal to?

  1. ((a))

    Null matrix 

  2. ((b))

    Identity matrix

  3. ((c))

    A

  4. ((d))

    -A

Show Answer
Answer: ((b))

Identity matrix

Calculation:

Given,

Matrix A is defined as:

A=[xyz yzx zxy]A = \begin{bmatrix} x & y & z \ y & z & x \ z & x & y \end{bmatrix}

It is mentioned that A is an orthogonal matrix. Therefore,

ATA=IA^T A = I

Now, we calculate A2:

A2=A⋅AA^2 = A \cdot A

A2=[xyz yzx zxy]⋅[xyz yzx zxy]A^2 = \begin{bmatrix} x & y & z \ y & z & x \ z & x & y \end{bmatrix} \cdot \begin{bmatrix} x & y & z \ y & z & x \ z & x & y \end{bmatrix}

Since A is orthogonal, ATA=IA^T A = I, and hence:

A2=IA^2 = I

∴ A2 = Identity matrix (I).

Hence, the Correct answer is Option 2.

Consider the following for the three (03) items that follow: Let p=sin35∘, q=sin25∘ and r=sin(−95∘).

31

What is (p+q+r) equal to?

  1. ((a))

    -1

  2. ((b))

    0

  3. ((c))

    2sin5∘

  4. ((d))

    2cos5∘

Show Answer
Answer: ((b))

0

Calculation:

Given,

Letp=sin⁡35∘p = \sin 35^\circ ,q=sin⁡25∘q = \sin 25^\circ and r=sin⁡(−95∘)r = \sin(-95^\circ)

We need to find the value of p + q + r .

First, we know that:

sin⁡(−95∘)=−sin⁡(95∘) \sin(-95^\circ) = -\sin(95^\circ) .

Thus, the expression p + q + r becomes:

p+q+r=sin⁡35∘+sin⁡25∘−sin⁡95∘ p + q + r = \sin 35^\circ + \sin 25^\circ - \sin 95^\circ .

Using the identity for the sum of two sine functions:

sin⁡A+sin⁡B=2sin⁡(A+B2)cos⁡(A−B2) \sin A + \sin B = 2 \sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) .

Substitute A=35∘A = 35^\circ  and B=25∘B = 25^\circ

sin⁡35∘+sin⁡25∘=2sin⁡30∘cos⁡5∘ \sin 35^\circ + \sin 25^\circ = 2 \sin 30^\circ \cos 5^\circ .

Since sin⁡30∘=12\sin 30^\circ = \frac{1}{2} , we get:

sin⁡35∘+sin⁡25∘=cos⁡5∘ \sin 35^\circ + \sin 25^\circ = \cos 5^\circ .

Now, we know that:

sin⁡95∘=cos⁡5∘ \sin 95^\circ = \cos 5^\circ .

Therefore:

p+q+r=cos⁡5∘−cos⁡5∘=0 p + q + r = \cos 5^\circ - \cos 5^\circ = 0 .

∴ The value of p + q + r  is 0.

Hence, the correct answer is Option 0.

32

What is (pq+qr+rp) equal to?

  1. ((a))

    -3/4

  2. ((b))

    0

  3. ((c))

    1/4

  4. ((d))

    3/4

Show Answer
Answer: ((a))

-3/4

Calculation:

Given,

The expression to simplify is pq + qr + rp, where:

p=sin⁡(35∘)p = \sin(35^\circ)

q=sin⁡(25∘)q = \sin(25^\circ)

r=sin⁡(−95∘)=−cos⁡(5∘)r = \sin(-95^\circ) = -\cos(5^\circ)

Thus, we need to evaluate:

pq+qr+rp=sin⁡(35∘)sin⁡(25∘)+sin⁡(25∘)(−cos⁡(5∘))+(−cos⁡(5∘))sin⁡(35∘) pq + qr + rp = \sin(35^\circ) \sin(25^\circ) + \sin(25^\circ) (-\cos(5^\circ)) + (-\cos(5^\circ)) \sin(35^\circ)

Using trigonometric identities, we simplify the expression:

pq+qr+rp=−sin⁡2(25∘)−sin⁡(95∘)sin⁡(35∘) pq + qr + rp = -\sin^2(25^\circ) - \sin(95^\circ) \sin(35^\circ)

Using the sum-to-product identity and known values for sine and cosine:

=12[−2sin⁡(25∘)−2sin⁡(95∘)⋅sin⁡(35∘)] = \frac{1}{2} [-2 \sin(25^\circ) - 2 \sin(95^\circ) \cdot \sin(35^\circ)]

=12[−1+cos⁡(50∘)+cos⁡(130∘)−cos⁡(60∘)] = \frac{1}{2} [-1 + \cos(50^\circ) + \cos(130^\circ) - \cos(60^\circ)]

=12[−1+cos⁡(50∘)−cos⁡(50∘)−12] = \frac{1}{2} [-1 + \cos(50^\circ) - \cos(50^\circ) - \frac{1}{2}]

=12[−1−12]=−34 = \frac{1}{2} \left[ -1 - \frac{1}{2} \right] = -\frac{3}{4}

∴ The value of pq + qr + rp  is −34. -\frac{3}{4} .

Hence, the correct answer is Option 1.

33

What is (p2+q2+r2) equal to?

  1. ((a))

    1/2

  2. ((b))

    1

  3. ((c))

    3/2

  4. ((d))

    2

Show Answer
Answer: ((c))

3/2

Calculation:

Given,

The expression to simplify is p2+q2+r2p^2 + q^2 + r^2, where:

p2+q2+r2p^2 + q^2 + r^2

q=sin⁡(25∘)q = \sin(25^\circ)

r=sin⁡(−95∘)=−cos⁡(5∘)r = \sin(-95^\circ) = -\cos(5^\circ)

We are given that p + q + r = 0 . Using the identity for squaring a sum:

(p+q+r)2=p2+q2+r2+2(pq+qr+rp) (p + q + r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp)

Since p + q + r = 0, we substitute this into the identity:

0=p2+q2+r2+2(pq+qr+rp) 0 = p^2 + q^2 + r^2 + 2(pq + qr + rp)

From previous calculations, we know that:

pq+qr+rp=−34 pq + qr + rp = -\frac{3}{4} .

Substitute this value into the equation:

0=p2+q2+r2+2(−34) 0 = p^2 + q^2 + r^2 + 2\left( -\frac{3}{4} \right)

Simplifying:

0=p2+q2+r2−32 0 = p^2 + q^2 + r^2 - \frac{3}{2}

p2+q2+r2=32 p^2 + q^2 + r^2 = \frac{3}{2} .

Hence, the correct answer is Option 3.

Consider the following for the two (02) items that follow: Let p=∣sinα−sin(α−90∘)∣.

34

What is the minimum value of p?

  1. ((a))

    0

  2. ((b))

    1/2

  3. ((c))

    121\over\sqrt{2}

  4. ((d))

    1

Show Answer
Answer: ((a))

0

Calculation:

Given,

We are given the expression:

p=∣sin⁡α−sin⁡(α−90∘)∣ p = | \sin \alpha - \sin(\alpha - 90^\circ) |

Using the identity for sine of a difference:

sin⁡(α−90∘)=−cos⁡α \sin(\alpha - 90^\circ) = -\cos \alpha

Substituting this identity into the expression for p :

p=∣sin⁡α+cos⁡α∣ p = | \sin \alpha + \cos \alpha |

The minimum value of ∣sin⁡α+cos⁡α∣ | \sin \alpha + \cos \alpha |  occurs when sin⁡α=−cos⁡α\sin \alpha = -\cos \alpha , which happens whenα=135∘\alpha = 135^\circ . At this point:

sin⁡135∘=12,cos⁡135∘=−12 \sin 135^\circ = \frac{1}{\sqrt{2}}, \cos 135^\circ = -\frac{1}{\sqrt{2}}

Therefore, at α=135∘\alpha = 135^\circ , the sum of sin⁡α\sin \alpha and cos⁡α\cos \alpha  is 0, so:

p=0 p = 0

∴ The minimum value of p  is 0.

Hence, the correct answer is option (a) 0.

Consider the following for the two (02) items that follow: Let p=∣sinα−sin(α−90∘)∣.

35

What is the maximum value of p?

  1. ((a))

    1

  2. ((b))

    2\sqrt{2}

  3. ((c))

    3\sqrt{3}

  4. ((d))

    2

Show Answer
Answer: ((b))

2\sqrt{2}

Calculation:

Given,

The expression for p  is:

p=∣sin⁡α−sin⁡(α−90∘)∣ p = |\sin \alpha - \sin(\alpha - 90^\circ)|

Using the identity for sin⁡(α−90∘)\sin(\alpha - 90^\circ)

sin⁡(α−90∘)=cos⁡α \sin(\alpha - 90^\circ) = \cos \alpha

Substituting this into the expression for p :

p=∣sin⁡α−cos⁡α∣ p = |\sin \alpha - \cos \alpha|

The expression  ∣sin⁡α−cos⁡α∣|\sin \alpha - \cos \alpha|  achieves its maximum value when the difference between sin⁡α\sin \alpha and cos⁡α\cos \alpha is largest.

We can rewrite sin⁡α−cos⁡α\sin \alpha - \cos \alpha as:

sin⁡α−cos⁡α=2(sin⁡(α−45∘)) \sin \alpha - \cos \alpha = \sqrt{2} \left( \sin \left( \alpha - 45^\circ \right) \right)

Since sin⁡(α−45∘)\sin(\alpha - 45^\circ) has a maximum value of 1, the maximum value of 2(sin⁡(α−45∘))\sqrt{2} \left( \sin(\alpha - 45^\circ) \right) is:

2 \sqrt{2}

 

Hence, the correct answer is Option 2

Consider the following for the three (03) items that follow: The sides of a triangle ABC are AB=3cm, BC=5cm and CA=7cm.

36

Consider the following statements:

I. The triangle is obtuse-angled triangle.

II. The sum of acute angles of the triangle is also acute.

Which of the statements given above is/are correct?

  1. ((a))

    I only

  2. ((b))

    II only

  3. ((c))

    Both I and II

  4. ((d))

    Neither I nor II

Show Answer
Answer: ((c))

Both I and II

Calculation:

Given,

The sides of the triangle are:

a = 5 cm

b = 7 cm

c = 3 cm

Statement I: The triangle is obtuse-angled.

Using the Cosine Rule:

cos⁡B=a2+c2−b22ac \cos B = \frac{a^2 + c^2 - b^2}{2ac}

Substitute the values for a, b = 7 and c

cos⁡B=52+32−722×5×3=25+9−4930=−1530=−12 \cos B = \frac{5^2 + 3^2 - 7^2}{2 \times 5 \times 3} = \frac{25 + 9 - 49}{30} = \frac{-15}{30} = -\frac{1}{2}

Now, calculate angle B:

B=cos⁡−1(−12)=120∘ B = \cos^{-1}\left(-\frac{1}{2}\right) = 120^\circ

Thus, angle B  is 120∘120^\circ , which confirms that the triangle is an obtuse-angled triangle, 

Statement II: The sum of the acute angles of the triangle is also acute.

Since the sum of the angles in a triangle is 180∘180^\circ , the sum of the acute angles is:

A+C=180∘−120∘=60∘ A + C = 180^\circ - 120^\circ = 60^\circ

This confirms that the sum of the acute angles is acute,

∴ Both statements are correct.

Hence, the correct answer is Option 3.

37

What is ∠B equal to?

  1. ((a))

    60∘

  2. ((b))

    105∘

  3. ((c))

    120∘

  4. ((d))

    150∘

Show Answer
Answer: ((c))

120∘

Calculation:

Given,

The sides of the triangle are: a = 5 Cm,  b = 7 cm,  c = 3 cm.

 

Apply the Cosine Rule to find angle B:

Using the Cosine Rule:

cos⁡B=a2+c2−b22ac \cos B = \frac{a^2 + c^2 - b^2}{2ac}

Substitute the values of a, b and c

cos⁡B=52+32−722×5×3=25+9−4930=−1530=−12 \cos B = \frac{5^2 + 3^2 - 7^2}{2 \times 5 \times 3} = \frac{25 + 9 - 49}{30} = \frac{-15}{30} = -\frac{1}{2}

Now, calculate angle B :

B=cos⁡−1(−12)=120∘ B = \cos^{-1}\left(-\frac{1}{2}\right) = 120^\circ

Thus, angle B  is 120∘120^\circ

Hence, the correct answer is Option 3.

38

What is the area of the triangle?

  1. ((a))

    153/415\sqrt{3}/4 square cm

  2. ((b))

    153/215\sqrt{3}/2 square cm

  3. ((c))

    15315\sqrt{3} square cm

  4. ((d))

    30330\sqrt{3} square cm

Show Answer
Answer: ((a))

153/415\sqrt{3}/4 square cm

Calculation:

Given,

The sides of the triangle are: a = 5 cm, c = 3 cm, and the angle B = 120∘120^\circ between sides a and c.

The area of the triangle is given by the formula:

Area=12×a×c×sin⁡B \text{Area} = \frac{1}{2} \times a \times c \times \sin B

Area=12×5×3×sin⁡(120∘) \text{Area} = \frac{1}{2} \times 5 \times 3 \times \sin(120^\circ)

We know thatsin⁡(120∘)=32\sin(120^\circ) = \frac{\sqrt{3}}{2} , so:

Area=12×5×3×32 \text{Area} = \frac{1}{2} \times 5 \times 3 \times \frac{\sqrt{3}}{2}

Area=12×15×32=15×34 \text{Area} = \frac{1}{2} \times 15 \times \frac{\sqrt{3}}{2} = \frac{15 \times \sqrt{3}}{4}

∴ The area of the triangle is 1534\frac{15\sqrt{3}}{4}  square cm.

Hence, the Correct answer is Option 1.

Consider the following for the two (02) items that follow:

The top (M) of a tower is observed from three points P, Q and R lying in a horizontal straight line which passes directly along the foot (N) of the tower. The angles of elevations of M from P, Q and R are 30°, 45° and 60° respectively. Let PQ = a and QR = b

39

What is PN equal to?

  1. ((a))

    (3−32)a \left(\frac{3-\sqrt{3}}{2}\right)a

  2. ((b))

    (3+32)a \left(\frac{3+\sqrt{3}}{2}\right)a

  3. ((c))

    (3−34)a \left(\frac{3-\sqrt{3}}{4}\right)a

  4. ((d))

    (3+34)a \left(\frac{3+\sqrt{3}}{4}\right)a

Show Answer
Answer: ((b))

(3+32)a \left(\frac{3+\sqrt{3}}{2}\right)a

Calculation:

The angles of elevation are:

 

From point P,θ=30∘\theta = 30^\circ ; from point Q, θ=45∘\theta = 45^\circ ; and from point R, θ=60∘\theta = 60^\circ

Using the tangent formula for each angle:

tan⁡(60∘)=hx  ⟹  h=3x \tan(60^\circ) = \frac{h}{x} \implies h = \sqrt{3}x

tan⁡(45∘)=hQN  ⟹  h=QN \tan(45^\circ) = \frac{h}{QN} \implies h = QN

tan⁡(30∘)=hPN  ⟹  PN=h3 \tan(30^\circ) = \frac{h}{PN} \implies PN = h{\sqrt{3}}

from the figure PN = h + a 

h+a=h3h+a = h\sqrt3

h=a3−1  ⟹  a3+12h = \frac{a}{\sqrt3 -1} \implies \frac{a \sqrt3 +1}{2}

= (3+32)a(\frac{3 + \sqrt3}{2}) a

Hence, the correct answer is Option 2.

40

What is MN equal to?

  1. ((a))

    (3+32)b \left(\frac{3+\sqrt{3}}{2}\right)b

  2. ((b))

    (3−32)b \left(\frac{3-\sqrt{3}}{2}\right)b

  3. ((c))

    (3−34)b \left(\frac{3-\sqrt{3}}{4}\right)b

  4. ((d))

    (3+34)b \left(\frac{3+\sqrt{3}}{4}\right)b

Show Answer
Answer: ((a))

(3+32)b \left(\frac{3+\sqrt{3}}{2}\right)b

Calculation

 

tan⁡(60∘)=hx⇒h=3x \tan(60^\circ) = \frac{h}{x} \Rightarrow h = \sqrt{3}x

tan⁡(45∘)=hb+x⇒h=b+x \tan(45^\circ) = \frac{h}{b + x} \Rightarrow h = b + x

h=b+x⇒3x=b+x h = b + x \Rightarrow \sqrt{3}x = b + x

x=b(3+1)2 x = \frac{b(\sqrt{3} + 1)}{2}

h=MN=3x=(3+32)b h = MN = \sqrt{3}x = \left( \frac{3 + \sqrt{3}}{2} \right) b

Hence, the correct answer is Option 1.

Consider the following for the three (03) items that follow:

Let p = tan 2α   - tanα and q = cotα - cot 2α

41

What is (p/q) equal to?

  1. ((a))

    −tanα⋅tan2α

  2. ((b))

    −cotα⋅cot2α

  3. ((c))

    tanα⋅tan2α

  4. ((d))

    cotα⋅cot2α

Show Answer
Answer: ((c))

tanα⋅tan2α

Explanation:

We are given:

p=tan⁡(2α)−tan⁡(α) p = \tan(2\alpha) - \tan(\alpha)

q=cot⁡(α)−cot⁡(2α) q = \cot(\alpha) - \cot(2\alpha)

We rewrite q  in terms of tangent since cot⁡(θ)=1tan⁡(θ)\cot(\theta) = \frac{1}{\tan(\theta)}

q=1tan⁡(α)−1tan⁡(2α) q = \frac{1}{\tan(\alpha)} - \frac{1}{\tan(2\alpha)}

Now we compute cot⁡(θ)=1tan⁡(θ)\cot(\theta) = \frac{1}{\tan(\theta)}

pq=tan⁡(2α)−tan⁡(α)1tan⁡(α)−1tan⁡(2α) \frac{p}{q} = \frac{\tan(2\alpha) - \tan(\alpha)}{\frac{1}{\tan(\alpha)} - \frac{1}{\tan(2\alpha)}}

Next, we find a common denominator for q :

1tan⁡(α)−1tan⁡(2α)=tan⁡(2α)−tan⁡(α)tan⁡(α)⋅tan⁡(2α) \frac{1}{\tan(\alpha)} - \frac{1}{\tan(2\alpha)} = \frac{\tan(2\alpha) - \tan(\alpha)}{\tan(\alpha) \cdot \tan(2\alpha)}

Substituting this back into the formula for pq\frac{p}{q}

pq=tan⁡(2α)−tan⁡(α)tan⁡(2α)−tan⁡(α)tan⁡(α)⋅tan⁡(2α) \frac{p}{q} = \frac{\tan(2\alpha) - \tan(\alpha)}{\frac{\tan(2\alpha) - \tan(\alpha)}{\tan(\alpha) \cdot \tan(2\alpha)}}

Simplifying, we get:

pq=tan⁡(α)⋅tan⁡(2α) \frac{p}{q} = \tan(\alpha) \cdot \tan(2\alpha)

∴ The correct answer is Option (c): tan⁡(α)⋅tan⁡(2α) \tan(\alpha) \cdot \tan(2\alpha)

42

What is (p+q) equal to?

  1. ((a))

    sec4α

  2. ((b))

    cosec4α

  3. ((c))

    2sec4α

  4. ((d))

    2cosec4α

Show Answer
Answer: ((d))

2cosec4α

Calculation: 

We are given:

p=tan⁡(2α)−tan⁡(α) p = \tan(2\alpha) - \tan(\alpha)

q=cot⁡(α)−cot⁡(2α) q = \cot(\alpha) - \cot(2\alpha)

We need to find p + q.

p+q=(sin⁡2(2α)−cos⁡2(2α)sin⁡(2α)⋅cos⁡(2α))+(cos⁡2(α)−sin⁡2(α)sin⁡(α)⋅cos⁡(α)) p + q = \left( \frac{\sin^2(2\alpha) - \cos^2(2\alpha)}{\sin(2\alpha) \cdot \cos(2\alpha)} \right) + \left( \frac{\cos^2(\alpha) - \sin^2(\alpha)}{\sin(\alpha) \cdot \cos(\alpha)} \right)

Simplifying both terms:

=−2cos⁡(4α)sin⁡(4α)+2cos⁡(2α)sin⁡(2α) = \frac{-2\cos(4\alpha)}{\sin(4\alpha)} + \frac{2\cos(2\alpha)}{\sin(2\alpha)}

Now, factorizing and simplifying further:

=2(sin⁡(4α)⋅cos⁡(2α)−cos⁡(4α)⋅sin⁡(2α))sin⁡(4α)⋅sin⁡(2α) = \frac{2(\sin(4\alpha) \cdot \cos(2\alpha) - \cos(4\alpha) \cdot \sin(2\alpha))}{\sin(4\alpha) \cdot \sin(2\alpha)}

Recognizing the sine identity, we get:

=2sin⁡(4α−2α)sin⁡(4α)⋅sin⁡(2α) = \frac{2 \sin(4\alpha - 2\alpha)}{\sin(4\alpha) \cdot \sin(2\alpha)}

Finally, simplifying this:

=2sin⁡(2α)sin⁡(4α)⋅sin⁡(2α) = \frac{2 \sin(2\alpha)}{\sin(4\alpha) \cdot \sin(2\alpha)}

The final result is:

p+q=2cosec(4α) p + q = 2 cosec (4\alpha)

∴ The correct answer is Option (4)

43

What is tan2α equal to?

  1. ((a))

    (pq)/(p+q)

  2. ((b))

    (p+2q)/p

  3. ((c))

    p/(p+2q)

  4. ((d))

    p/(2p+q)

Show Answer
Answer: ((c))

p/(p+2q)

Calculation: 

We are given:

p=tan⁡(2α)−tan⁡(α) p = \tan(2\alpha) - \tan(\alpha)

q=cot⁡(α)−cot⁡(2α) q = \cot(\alpha) - \cot(2\alpha)

We need to find tan⁡2(α)\tan^2(\alpha) in terms of p  and q .

We start by simplifying the expression pp+2q\frac{p}{p + 2q}

pp+2q=11+2qp \frac{p}{p + 2q} = \frac{1}{1 + \frac{2q}{p}}

This simplifies further as:

11+2tan⁡(α)⋅tan⁡(2α) \frac{1}{1 + \frac{2}{\tan(\alpha) \cdot \tan(2\alpha)}}

Now, simplify the fraction inside the denominator:

=11+2tan⁡(α)⋅tan⁡(2α)=11+12⋅tan⁡(α)⋅tan⁡(2α) = \frac{1}{1 + \frac{2}{\tan(\alpha) \cdot \tan(2\alpha)}} = \frac{1}{1 + \frac{1}{2} \cdot \tan(\alpha) \cdot \tan(2\alpha)}

Now, expand both terms using trigonometric identities:

=sin⁡(α)⋅sin⁡(2α)sin⁡(α)⋅cos⁡(α)+cos⁡(α)⋅cos⁡(2α) = \frac{\sin(\alpha) \cdot \sin(2\alpha)}{\sin(\alpha) \cdot \cos(\alpha) + \cos(\alpha) \cdot \cos(2\alpha)}

Simplifying this expression gives:

=tan⁡2(α) = \tan^2(\alpha)

∴ The correct answer is Option (c):

Consider the following for the two (02) items that follow:

Let 2sinα + cosα = 2 where 0 < α < 90°

44

What is tanα equal to?

  1. ((a))

    1/2

  2. ((b))

    1

  3. ((c))

    3/4

  4. ((d))

    2

Show Answer
Answer: ((c))

3/4

Calculation: 

We are given:

2sin⁡(α)+cos⁡(α)=2 2\sin(\alpha) + \cos(\alpha) = 2

 

cos⁡(α)=2(1−sin⁡(α)) \cos(\alpha) = 2(1 - \sin(\alpha))

cos⁡2(α)=4(1−sin⁡(α))2 \cos^2(\alpha) = 4(1 - \sin(\alpha))^2

Use the identity cos⁡2(α)=1−sin⁡2(α)\cos^2(\alpha) = 1 - \sin^2(\alpha)

1−sin⁡2(α)=4(1−2sin⁡(α)+sin⁡2(α)) 1 - \sin^2(\alpha) = 4(1 - 2\sin(\alpha) + \sin^2(\alpha))

1−sin⁡2(α)=4−8sin⁡(α)+4sin⁡2(α) 1 - \sin^2(\alpha) = 4 - 8\sin(\alpha) + 4\sin^2(\alpha)

Rearrange the terms to form a quadratic equation:

5sin⁡2(α)−8sin⁡(α)+3=0 5\sin^2(\alpha) - 8\sin(\alpha) + 3 = 0

Using the quadratic formula:

sin⁡(α)=−(−8)±(−8)2−4(5)(3)2(5) \sin(\alpha) = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(5)(3)}}{2(5)}

sin⁡(α)=8±64−6010 \sin(\alpha) = \frac{8 \pm \sqrt{64 - 60}}{10}

sin⁡(α)=8±210 \sin(\alpha) = \frac{8 \pm 2}{10}

sin⁡(α)=1 \sin(\alpha) = 1 or sin⁡(α)=35 \sin(\alpha) = \frac{3}{5}

Since 0∘<α<90∘0^\circ < \alpha < 90^\circ , we select sin⁡(α)=35\sin(\alpha) = \frac{3}{5}

cos⁡2(α)=1−sin⁡2(α)=1−(35)2=1−925=1625 \cos^2(\alpha) = 1 - \sin^2(\alpha) = 1 - \left( \frac{3}{5} \right)^2 = 1 - \frac{9}{25} = \frac{16}{25}

cos⁡(α)=45 \cos(\alpha) = \frac{4}{5}

tan⁡(α)=sin⁡(α)cos⁡(α)=3545=34 \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}

∴ The correct answer is Option (c):

45

What is 2sin2α+cos2α equal to?

  1. ((a))

    11/10

  2. ((b))

    11/5

  3. ((c))

    12/5

  4. ((d))

    13/5

Show Answer
Answer: ((b))

11/5

Calculation: 

We know:

sin⁡(α)=35andcos⁡(α)=45 \sin(\alpha) = \frac{3}{5} \quad \text{and} \quad \cos(\alpha) = \frac{4}{5}

 

sin⁡(2α)=2sin⁡(α)cos⁡(α) \sin(2\alpha) = 2\sin(\alpha)\cos(\alpha)

cos⁡(2α)=1−2sin⁡2(α) \cos(2\alpha) = 1 - 2\sin^2(\alpha)

sin⁡(2α)=2×35×45=2425 \sin(2\alpha) = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25}

cos⁡(2α)=1−2×(35)2=1−2×925=725 \cos(2\alpha) = 1 - 2 \times \left( \frac{3}{5} \right)^2 = 1 - 2 \times \frac{9}{25} = \frac{7}{25}

2sin⁡(2α)+cos⁡(2α)=2×2425+725 2\sin(2\alpha) + \cos(2\alpha) = 2 \times \frac{24}{25} + \frac{7}{25}

Simplifying:

2sin⁡(2α)+cos⁡(2α)=4825+725=5525=115 2\sin(2\alpha) + \cos(2\alpha) = \frac{48}{25} + \frac{7}{25} = \frac{55}{25} = \frac{11}{5}

∴ The correct answer is Option (2):

Consider the following for the two (02) items that follow:

In a triangle ABC, two sides BC and CA are in the ratio 2:1 and their opposite corresponding angles are in the ratio 3: 1.

46

One of the angles of the triangle is

  1. ((a))

    15∘

  2. ((b))

    30∘

  3. ((c))

    45∘

  4. ((d))

    75∘

Show Answer
Answer: ((b))

30∘

Calculation:

 

We are given the equation for the ratio of sides using the Sine Rule:

asin⁡(3x)=bsin⁡(x) \frac{a}{\sin(3x)} = \frac{b}{\sin(x)}

 

ab=sin⁡(3x)sin⁡(x) \frac{a}{b} = \frac{\sin(3x)}{\sin(x)}

Step 3: Use the identity for sin(3x), which is sin⁡(3x)=3sin⁡(x)−4sin⁡3(x)\sin(3x) = 3\sin(x) - 4\sin^3(x) , and substitute it into the equation:

2=3sin⁡(x)−4sin⁡3(x)sin⁡(x) 2 = \frac{3\sin(x) - 4\sin^3(x)}{\sin(x)}

2sin⁡(x)=3sin⁡(x)−4sin⁡3(x) 2\sin(x) = 3\sin(x) - 4\sin^3(x)

−sin⁡(x)+4sin⁡3(x)=0 -\sin(x) + 4\sin^3(x) = 0

sin⁡(x)(4sin⁡2(x)−1)=0 \sin(x)(4\sin^2(x) - 1) = 0

We have two possible solutions for this equation:

sin⁡(x)=0 \sin(x) = 0 , which gives x = 0∘0^\circ (not valid in this case).

4sin⁡2(x)−1=0 4\sin^2(x) - 1 = 0 , which simplifies to:

sin⁡2(x)=14⇒sin⁡(x)=12 \sin^2(x) = \frac{1}{4} \quad \Rightarrow \quad \sin(x) = \frac{1}{2}

x=30∘ x = 30^\circ

∴ The correct answer is Option (2)

47

Consider the following statements:

I. The triangle is right-angled.

II. One of the sides of the triangle is 3 times the other.

III. The angles A, C and B of the triangle are in AP.

Which of the statements given above is/are correct?

  1. ((a))

    I only

  2. ((b))

    II and III only

  3. ((c))

    I and III only

  4. ((d))

    I, II and III

Show Answer
Answer: ((c))

I and III only

Explanation:

We are given the triangle with angles:

∠B=x=30∘ \angle B = x = 30^\circ

∠A=3x=90∘ \angle A = 3x = 90^\circ

∠C=180∘−120∘=60∘\angle C = 180^\circ - 120^\circ = 60^\circ

Step 1: Check if the sum of the angles is 180°:

∠A+∠B+∠C=90∘+30∘+60∘=180∘ \angle A + \angle B + \angle C = 90^\circ + 30^\circ + 60^\circ = 180^\circ

This confirms that the angles satisfy the angle sum property of a triangle.

Statement I. The triangle is right-angled.

Since∠A=90∘ \angle A = 90^\circ , the triangle is right-angled.

Statement III: III. The angles A, C and B of the triangle are in AP.

The angles 30∘60∘,and90∘30^\circ 60^\circ , \text and 90^\circ are in Arithmetic Progression because:

60∘−30∘=30∘and90∘−60∘=30∘ 60^\circ - 30^\circ = 30^\circ \quad \text{and} \quad 90^\circ - 60^\circ = 30^\circ

This confirms that the angles are in AP.

Statement II is not correct because there is no mention of a side being 3 times the other.

∴ The correct answer is Option (I) and (III) are correct.

Hence, the correct answer is Option 3.

48

A man at M, standing 100 m away from the base (P) of a chimney of height 50 m, observes the angle of elevation of the highest point (Q) of the smoke to be 45∘. The highest point of the chimney is at R. Further P, R and Q are in a straight line and the straight line is perpendicular to PM. What is the angle RMQ equal to?

  1. ((a))

    tan−1(12\frac{1}{2})

  2. ((b))

    tan−1(13\frac{1}{3}​)

  3. ((c))

    tan−1(23\frac{2}{3}​)

  4. ((d))

    tan−1(34\frac{3}{4}​)

Show Answer
Answer: ((b))

tan−1(13\frac{1}{3}​)

Given:

  • The height of the chimney (PR) is given as 50 meters.
  • The man is 100 meters away from the base of the chimney (point P).
  • The angle of elevation from the man (at point M) to the highest point of the smoke (point Q) is 45∘.
  • There is a straight line from point P through point R to point Q, perpendicular to PM.

Calculation:

For the right-angled triangle △ MPQ the tangent of the angle of elevation at point M is

⇒ tan 45∘ = PQ/PM = PQ100\frac{PQ}{100} 

Since tan 45∘ = 1, we have PQ = 100 m.

Finding the angle ∠RMQ:

∠RMQ = ∠PMQ - ∠PMR

In △ PMR, tan ∠PMR = PR/PM = 50/100 = 1/2.

Using the formula tan(A - B) = (tan A - tan B) / (1 + tan A ⋅ tan B):

⇒ tan ∠RMQ = (tan 45∘ - tan ∠PMR) / (1 + tan 45∘ ⋅ tan ∠PMR)

⇒ tan ∠RMQ = (1 - 1/2) / (1 + 1 ⋅ 1/2) = (1/2) / (3/2) = 1/3

⇒ ∠RMQ = tan-113\frac{1}{3}

Hence, the Correct answer is Option 2.

49

If k is a root of x2−4x+1=0, then what is tan−1k+tan−11k\frac{1}{k}​ equal to?

  1. ((a))

    −π/2

  2. ((b))

    0

  3. ((c))

    π/4

  4. ((d))

    π/2

Show Answer
Answer: ((d))

π/2

Calculation:

We are given that k is a root of x2−4x+1=0

Solving the quadratic equation:

x=−(−4)±(−4)2−4(1)(1)2(1)=4±16−42=4±122=4±232x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(1)}}{2(1)} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2}

⇒ x=2±3x = 2 \pm \sqrt{3}

Thus, the two possible values of k are

⇒ k=2+3ork=2−3k = 2 + \sqrt{3} \quad \text{or} \quad k = 2 - \sqrt{3}

We need to find  tan⁡−1(k)+tan⁡−1(1k)\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{k}\right)

Using the identity for the sum of inverse tangents

tan⁡−1(a)+tan⁡−1(b)=tan⁡−1(a+b1−ab)\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right)

⇒ tan⁡−1(k)+tan⁡−1(1k)=tan⁡−1(k+1k1−k⋅1k)\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{k}\right) = \tan^{-1}\left(\frac{k + \frac{1}{k}}{1 - k \cdot \frac{1}{k}}\right)

=tan⁡−1(k+1k1−1)=tan⁡−1(k+1k0)= \tan^{-1}\left(\frac{k + \frac{1}{k}}{1 - 1}\right) = \tan^{-1} \left(\frac{k + \frac{1}{k}}{0}\right)

This expression results in an undefined value, but we know from properties of the inverse tangent that

⇒ tan⁡−1(k)+tan⁡−1(1k)=π2\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{k}\right) = \frac{\pi}{2}

Hence, the correct answer is Option 4.

50

If tan−1k+tan−1 12\frac{1}{2}​=π4​\frac{π}{4} ​ , then what is the value of k?

  1. ((a))

    1

  2. ((b))

    1/2

  3. ((c))

    1/3

  4. ((d))

    1/4

Show Answer
Answer: ((c))

1/3

Calculation:

Given,

tan⁡−1(k)+tan⁡−1(12)=π4 \tan^{-1}(k) + \tan^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{4}

Use the addition formula for inverse tangents:

tan⁡−1(a)+tan⁡−1(b)=tan⁡−1(a+b1−ab) \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right)

Using this for tan⁡−1(k)+tan⁡−1(12)\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{2}\right) , we get:

tan⁡−1(k+121−k⋅12)=π4 \tan^{-1}\left(\frac{k + \frac{1}{2}}{1 - k \cdot \frac{1}{2}}\right) = \frac{\pi}{4}

Since tan⁡(π4)=1\tan\left(\frac{\pi}{4}\right) = 1 , we have:

k+121−k2=1 \frac{k + \frac{1}{2}}{1 - \frac{k}{2}} = 1

k+12=1−k2 k + \frac{1}{2} = 1 - \frac{k}{2}

2k+1=2−k 2k + 1 = 2 - k

2k+k=2−1 2k + k = 2 - 1

3k=1 3k = 1

k=13 k = \frac{1}{3}

Hence, the correct answer is Option 3.

51

Under what condition will the lines m2x+ny−1=0 and n2x−my+2=0 be perpendicular?

  1. ((a))

    mn−1=0

  2. ((b))

    mn + 1 = 0

  3. ((c))

    m + n = 0

  4. ((d))

    m - n = 0

Show Answer
Answer: ((a))

mn−1=0

Calculation:

Given,

Line 1: m2x+ny−1=0m^2x + ny - 1 = 0

Line 2: n2x−my+2=0n^2x - my + 2 = 0

Slope of Line 1, m1=−m2nm_1 = -\dfrac{m^2}{n}

Slope of Line 2, m2=n2mm_2 = \dfrac{n^2}{m}

For perpendicular lines, m1×m2=−1m_1 \times m_2 = -1.

−m2n×−n2m=−1\frac{-m^2}{n} \times \frac{-n^2}{m} = -1

mn -1 = 0

Hence, the correct answer is Option 1.

52

If p and q are real numbers between 0 and 1 such that the points (p,1), (1,q) and (0,0) form an equilateral triangle, then what is (p+q) equal to?

  1. ((a))

    2\sqrt{2}

  2. ((b))

    2−1\sqrt{2}-1

  3. ((c))

    2−32-\sqrt{3}

  4. ((d))

    4−234-2\sqrt{3}

Show Answer
Answer: ((d))

4−234-2\sqrt{3}

Calculation:

Given points A(0,0), B(p,1), and C(1,q), which form an equilateral triangle. We are told (0 < p,q < 1).

Compute the squared lengths of the sides:

AB2=(p−0)2+(1−0)2=p2+1AB^{2} = (p - 0)^{2} + (1 - 0)^{2} = p^{2} + 1

AC2=(1−0)2+(q−0)2=1+q2AC^{2} = (1 - 0)^{2} + (q - 0)^{2} = 1 + q^{2}

BC2=(1−p)2+(,q−1,)2=(1−p)2+(q−1)2=2,(1−p)2BC^{2} = (1 - p)^{2} + (,q - 1,)^{2} = (1 - p)^{2} + (q - 1)^{2} = 2,(1 - p)^{2}

Because the triangle is equilateral, all three squared lengths are equal:

⇒ AB2=AC2AB^{2} = AC^{2}

⇒ p2+1;=;1+q2;⟹;p2=q2;⟹;p=qp^{2} + 1 ;=; 1 + q^{2} ;\Longrightarrow; p^{2} = q^{2} ;\Longrightarrow; p = q

both p and q are positive in (0,1)

Let p = q = t Then

⇒ AB2=t2+1AB^{2} = t^{2} + 1

⇒ BC2=2,(1−t)2BC^{2} = 2,(1 - t)^{2}

Equate AB2 and BC2:

⇒ t2+1;=;2,(1−t)2;=;2(1−2t+t2);=;2−4t+2t2.t^{2} + 1 ;=; 2,(1 - t)^{2} ;=; 2\bigl(1 - 2t + t^{2}\bigr) ;=; 2 - 4t + 2t^{2}.

Simplify:

⇒ t2+1=2−4t+2t2;⟹;0=2−4t+2t2−(t2+1)=t2−4t+1.t^{2} + 1 = 2 - 4t + 2t^{2};\Longrightarrow;0 = 2 - 4t + 2t^{2} - (t^{2} + 1) = t^{2} - 4t + 1.

So t satisfies:

⇒ t2−4t+1=0⟹t=4±√16−42=4±2√32=2±√3.t^{2} - 4t + 1 = 0 \quad\Longrightarrow\quad t = \frac{4 \pm √{16 - 4}}{2} = \frac{4 \pm 2√{3}}{2} = 2 \pm √{3}.

Since 0 < t < 1, we take t = 2 - √3 (note 2 + √3> 1) is not allowed. Hence,

⇒ p=q=2−√3.p = q = 2 - √{3}.

Therefore:

 p+q;=;(2−√3)+(2−√3);=;4−2√3.p + q ;=; (2 - √{3}) + (2 - √{3}) ;=; 4 - 2√{3}.

Hence, the correct answer is Option 4.

53

The vertices of a triangle are A(1, 1),B(0, 0) and C(2, 0). The angular bisectors of the triangle meet at P. What are the coordinates of P?

  1. ((a))

    (1,2−1)(1, \sqrt{2} −1)

  2. ((b))

    (1,3−1)(1, \sqrt{3} −1)

  3. ((c))

    '(1,1/2)

  4. ((d))

    (1/2,2−1)(1/2, \sqrt{2}-1)

Show Answer
Answer: ((a))

(1,2−1)(1, \sqrt{2} −1)

Calculation:

Given points A(1,1), B(0,0), and C(2,0). The angle bisectors meet at P (the incenter).

Compute the side lengths:

a=BC=(2−0)2+(0−0)2=2a = BC = \sqrt{(2 - 0)^{2} + (0 - 0)^{2}} = 2

b=CA=(1−2)2+(1−0)2=2b = CA = \sqrt{(1 - 2)^{2} + (1 - 0)^{2}} = \sqrt{2}

c=AB=(1−0)2+(1−0)2=2c = AB = \sqrt{(1 - 0)^{2} + (1 - 0)^{2}} = \sqrt{2}

Therefore, a+b+c=2+2+2=2+22.a + b + c = 2 + \sqrt{2} + \sqrt{2} = 2 + 2\sqrt{2}.

By the incenter formula,

XP;=;a,xA+b,xB+c,xCa+b+c;=;,2⋅1+2⋅0+2⋅2,,2+22,;=;2+222+22;=;1.X_{P} ;=; \frac{a,x_{A} + b,x_{B} + c,x_{C}}{a + b + c} ;=; \frac{,2\cdot 1 + \sqrt{2}\cdot 0 + \sqrt{2}\cdot 2,}{,2 + 2\sqrt{2},} ;=; \frac{2 + 2\sqrt{2}}{2 + 2\sqrt{2}} ;=; 1.

YP;=;a,yA+b,yB+c,yCa+b+c;=;,2⋅1+2⋅0+2⋅0,,2+22,;=;22+22;=;1,1+2,;=;2;−;1.Y_{P} ;=; \frac{a,y_{A} + b,y_{B} + c,y_{C}}{a + b + c} ;=; \frac{,2\cdot 1 + \sqrt{2}\cdot 0 + \sqrt{2}\cdot 0,}{,2 + 2\sqrt{2},} ;=; \frac{2}{2 + 2\sqrt{2}} ;=; \frac{1}{,1 + \sqrt{2},} ;=; \sqrt{2} ;-; 1.

∴ The incenter is P=(1,;2−1)P = \bigl(1,;\sqrt{2} - 1\bigr) .

Hence, the correct answer is Option 1.

54

Let A(3, -1) and B(1, 1) be the end points of line segment AB. Let P be the middle point of the line segment AB. Let Q be the point situated at a distance 2\sqrt{2} units from P on the perpendicular bisector line of AB. What are the possible coordinates of Q?

  1. ((a))

    (2,1)

  2. ((b))

    (3,1)

  3. ((c))

    (2,2)

  4. ((d))

    (1,3)

Show Answer
Answer: ((b))

(3,1)

Calculation:

Given points A(3, −1) and B(1, 1). Let P be the midpoint of AB, and Q be a point on the perpendicular bisector of AB that lies √2 units from P.

Compute the midpoint P of AB:

P=(3+12,,−1+12)=(2,,0)P = \Bigl(\tfrac{3 + 1}{2},,\tfrac{-1 + 1}{2}\Bigr) = (2,,0)

Compute the slope of AB:

mAB=,1−(−1),,1−3,=2−2=−1m_{AB} = \frac{,1 - (-1),}{,1 - 3,} = \frac{2}{-2} = -1

Therefore, the equation of line AB is:

y−1=−1,(x−1);⟹;x+y−2=0y - 1 = -1,(x - 1);\Longrightarrow;x + y - 2 = 0

The perpendicular bisector of AB must pass through P(2, 0) and have slope perpendicular to −1 (i.e. slope +1):

y−0=1,(x−2);⟹;y=x−2y - 0 = 1,(x - 2);\Longrightarrow;y = x - 2

Any point Q on this bisector satisfies y = x - 2. Write Q = (x, x − 2).

We require the distance PQ = √2. Since P(2, 0),

Distance2=(x−2)2+((,x−2,)−0)2=2\text{Distance}^2 = (x - 2)^2 + \bigl((,x - 2,) - 0\bigr)^2 = 2

⇒ (x−2)2+(x−2)2=2;⟹;2,(x−2)2=2;⟹;(x−2)2=1(x - 2)^2 + (x - 2)^2 = 2;\Longrightarrow;2,(x - 2)^2 = 2;\Longrightarrow;(x - 2)^2 = 1

Thus,

⇒ x−2=±1;⟹;x=3 or x=1x - 2 = \pm 1;\Longrightarrow;x = 3\text{ or }x = 1

If x = 3, then y = 3 - 2 = 1. So one solution is Q(3, 1).

If x = 1, then y = 1 - 2 = -1. So the other solution is Q(1, −1).

Hence, the correct answer is Option 2.

55

ABC is an equilateral triangle and AD is the altitude on BC. If the coordinates of A are (1,2) and that of D are (−2,6), then what is the equation of BC?

  1. ((a))

    3x+4y−18=0

  2. ((b))

    4x+3y−1=0

  3. ((c))

    4x−3y+26=0

  4. ((d))

    3x−4y+30=0

Show Answer
Answer: ((d))

3x−4y+30=0

Calculation:

Given vertices A(1, 2) and D(−2, 6), where AD is the altitude from A onto BC in an equilateral triangle ABC.

Compute the slope of AD:

mAD=,yD−yA,,xD−xA,=,6−2,,−2−1,=4−3=−43.m_{AD} = \frac{,y_D - y_A,}{,x_D - x_A,} = \frac{,6 - 2,}{,-2 - 1,} = \frac{4}{-3} = -\tfrac{4}{3}.

Because AD ⟂ BC, the slope of BC, mBCm_{BC}, satisfies

mAD⋅mBC=−1;⟹;(−43),mBC=−1;⟹;mBC=−1,−43,=34.m_{AD} \cdot m_{BC} = -1 ;\Longrightarrow; \Bigl(-\tfrac{4}{3}\Bigr) ,m_{BC} = -1 ;\Longrightarrow; m_{BC} = \frac{-1}{, -\tfrac{4}{3},} = \tfrac{3}{4}.

Equation of the line BC 

y−6=34,(x+2).y - 6 = \tfrac{3}{4},(x + 2).

4(y−6)=3(x+2);⟹;4y−24=3x+6.4(y - 6) = 3(x + 2);\Longrightarrow;4y - 24 = 3x + 6.

4y−24−3x−6=04y - 24 - 3x - 6 = 0

;⟹;−3x+4y−30=0;⟹;3x−4y+30=0.;\Longrightarrow; -3x + 4y - 30 = 0 ;\Longrightarrow; 3x - 4y + 30 = 0.

Hence, the correct answer is Option 4.

56

What is the equation of the circle whose diameter is 10 cm and the equations of two of its diameters are x+y=0 and x−y=0?

  1. ((a))

    x2+y2=1

  2. ((b))

    x2+y2=25

  3. ((c))

    x2+y2=100

  4. ((d))

    x2+y2−2x−2y−23=0

Show Answer
Answer: ((b))

x2+y2=25

Calculation:

Given two diameters of the circle are along the lines x + y = 0 and x - y = 0. The center of the circle is at their intersection.

We have, 

x−y=0;⟹;x=y.x - y = 0 ;\Longrightarrow; x = y.

Substitute into x + y = 0:

x+x=0;⟹;2x=0;⟹;x=0,;y=0.x + x = 0 ;\Longrightarrow; 2x = 0 ;\Longrightarrow; x = 0,; y = 0.

Thus, the center is (0, 0).

Also radius,

r=102=5.r = \frac{10}{2} = 5.

the equation of the circle with center (0,0) and radius 5 is

x2+y2=52=25.x^{2} + y^{2} = 5^{2} = 25.

Hence, the correct answer is Option 2.

57

A square is inscribed in a circle x 2 + y 2 + 2x + 2y + 1 = 0 and its sides are parallel to coordinate axes. Which one of the following is a vertex of the square?

  1. ((a))

    (-2,2)

  2. ((b))

    (-2,-2)

  3. ((c))

    (−1+12,−1−12)(-1+\frac{1}{\sqrt{2}},-1-\frac{1}{\sqrt{2}})

  4. ((d))

    None of the above

Show Answer
Answer: ((c))

(−1+12,−1−12)(-1+\frac{1}{\sqrt{2}},-1-\frac{1}{\sqrt{2}})

Calculation:

Given,

The circle’s equation is

x2+y2+2x+2y+1=0x^{2} + y^{2} + 2x + 2y + 1 = 0

Rewrite by completing squares:

(x+1)2+(y+1)2=1(x + 1)^{2} + (y + 1)^{2} = 1

∴ center = (-1, -1), radius r = 1

For a square inscribed in this circle with sides parallel to the axes, its diagonal equals the circle’s diameter = 2. If the square has side length a, then

a2=2;⟹;a=2.a\sqrt{2} = 2 ;\Longrightarrow; a = \sqrt{2}.

Each vertex lies a half‐side =a2=12= \tfrac{a}{2} = \tfrac{1}{\sqrt{2}} from the center along both axes. Since the center is ( -1, -1), the four vertices are

(−1±12,;−1±12).\displaystyle \bigl(-1 \pm \tfrac{1}{\sqrt{2}},; -1 \pm \tfrac{1}{\sqrt{2}}\bigr).

One of these vertices is

(−1+12,;−1−12).\bigl(-1 + \tfrac{1}{\sqrt{2}},; -1 - \tfrac{1}{\sqrt{2}}\bigr).

Hence, the correct answer is Option 3

58

A tangent to the parabola y2 = 4x is inclined at an angle 45° deg with the positive direction of x-axis. What is the point of contact of the tangent and the parabola?

  1. ((a))

    (1, 1)

  2. ((b))

    (2,22)(2,2\sqrt{2})

  3. ((c))

    (12,12)(\frac{1}{2},\frac{1}{\sqrt{2}})

  4. ((d))

    (1,2)

Show Answer
Answer: ((d))

(1,2)

Calculation:

Given the parabola

y2 = 4x 

and a tangent to this parabola that is inclined at an angle of 45°with the positive x-axis, Hence, its slope is 

m;=;tan⁡(45∘);=;1.m ;=; \tan(45^\circ) ;=; 1.

A standard parametric form for y2 = 4x is 

(x(t),,y(t));=;(t2,,2t),\bigl(x(t),,y(t)\bigr) ;=;\bigl(t^{2},,2t\bigr),

since  y2=4x  ⟹  (2t)2=4,t2y^{2} = 4x \implies (2t)^{2} = 4,t^{2}

The slope of the tangent at (t2,,2t)\bigl(t^{2},,2t\bigr)

Differentiate y2 = 4x 

2y,dydx;=;42y,\frac{dy}{dx} ;=; 4

;⟹;dydx;=;4,2y,;=;2,y,.;\Longrightarrow; \frac{dy}{dx} ;=; \frac{4}{,2y,} ;=; \frac{2}{,y,}.

At the point (x,y)=(t2,,2t)\bigl(x,y\bigr) = \bigl(t^{2},,2t\bigr)  one has y=2ty = 2t

dydx∣,(t2,,2t)\left.\frac{dy}{dx}\right|_{,(t^{2},,2t)}  =2,2t,=1t.= \frac{2}{,2t,} = \frac{1}{t}.

We require this slope to equal 1.

1t;=;1;⟹;t=1.\frac{1}{t} ;=; 1 ;\Longrightarrow; t = 1.

Now point of contact

Substitute t = 1 in (x(t),,y(t))=(t2,,2t)\bigl(x(t),,y(t)\bigr) = \bigl(t^{2},,2t\bigr). 

x(1)=12=1,x(1) = 1^{2} = 1,

y(1)=2⋅1=2.y(1) = 2 \cdot 1 = 2.

Thus the point of contact of the tangent of slope 1 is (1, 2)

Hence, the correct answer is Option 4.

59

. What is the distance between the two foci of the hyperbola 25x2 - 75y 2= 225 ?

  1. ((a))

    232\sqrt{3}  units

  2. ((b))

    434 \sqrt{3}  units

  3. ((c))

    6\sqrt{6}  units

  4. ((d))

    262\sqrt{6}  units

Show Answer
Answer: ((b))

434 \sqrt{3}  units

Calculation:

Given,

Hyperbola equation: 25x2−75y2=22525x^{2} - 75y^{2} = 225

Divide both sides by 225 to obtain standard form:

25x2225−75y2225=1;⟹;x29−y23=1\frac{25x^{2}}{225} - \frac{75y^{2}}{225} = 1 ;\Longrightarrow; \frac{x^{2}}{9} - \frac{y^{2}}{3} = 1

Thus, a2=9a^{2} = 9 and b2=3b^{2} = 3.

Compute cc from c2=a2+b2c^{2} = a^{2} + b^{2}:

c2=9+3=12;⟹;c=12=23.c^{2} = 9 + 3 = 12 ;\Longrightarrow; c = \sqrt{12} = 2\sqrt{3}.

The foci are at (±c,,0)(\pm c,,0), so the distance between them is 2c2c:

2c=2×23=43.2c = 2 \times 2\sqrt{3} = 4\sqrt{3}.

∴ The distance between the two foci is 434\sqrt{3} units.

Hence, the correct answer is Option 2.

60

If any point on an ellipse is (3sinα\alpha, 5cosα\alpha), then what is the eccentricity of the ellipse?

  1. ((a))

    4/3

  2. ((b))

    4/5

  3. ((c))

    3/4

  4. ((d))

    1/2

Show Answer
Answer: ((b))

4/5

Calculation:

Given any point on the ellipse is 3sinα, 5cosα. In the standard parametric form of an ellipse,

x=a,sin⁡α,y=b,cos⁡αx = a,\sinα,\quad y = b,\cosα

we identify

a=3,b=5a = 3,\quad b = 5

Since (b > a), the semi-major axis is (b = 5) and the semi-minor axis is (a = 3). The eccentricity e of an ellipse is

e=,1−(semi-minor)2(semi-major)2,=,1−a2b2,e = \sqrt{,1 - \frac{(\text{semi-minor})^{2}}{(\text{semi-major})^{2}},} = \sqrt{,1 - \frac{a^{2}}{b^{2}},}

Substitute (a = 3) and (b = 5):

e=,1−3252,=,1−925,=1625=45e = \sqrt{,1 - \frac{3^{2}}{5^{2}},} = \sqrt{,1 - \frac{9}{25},} = \sqrt{\frac{16}{25}} = \frac{4}{5}

Hence, the correct answer is Option 2.

61

If a line in 3 dimensions makes angles α, β and γ with the positive directions of the coordinate axes, then what is cos(α+β)cos(α−β) equal to?

  1. ((a))

    cos2γ

  2. ((b))

    −cos2γ

  3. ((c))

    sin2γ

  4. ((d))

    −sin2γ

Show Answer
Answer: ((b))

−cos2γ

Calculation:

We know that cos⁡2α+cos⁡2β+cos⁡2γ=1 \cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1 ... (i)

And cos⁡(α+β)⋅cos⁡(α−β)=cos⁡2α−sin⁡2β \cos(\alpha + \beta) \cdot \cos(\alpha - \beta) = \cos^{2} \alpha - \sin^{2} \beta

Substitute and simplify:

=cos⁡2α+cos⁡2β−1=1−cos⁡2γ⇒−cos⁡2γ = \cos^{2} \alpha + \cos^{2} \beta - 1 = 1 - \cos^{2} \gamma \quad \Rightarrow \quad - \cos^{2} \gamma

Hence, the correct answer is Option 2.

62

A(1,2,−1), B(2,5,−2) and C(4,4,−3) are three vertices of a rectangle. What is the area of the rectangle?

  1. ((a))

    8 square units

  2. ((b))

    9 square units

  3. ((c))

    66\sqrt{66}  square units

  4. ((d))

    68\sqrt{68}  square units

Show Answer
Answer: ((c))

66\sqrt{66}  square units

Calculation:

Given,

Points A(1, 2, -1), B(2, 5, -2), and C(4, 4, -3) are three vertices of the rectangle.

We need to find the area of the rectangle formed by the vectors AB AB and BC BC .

Length of vector AB AB :

Length of AB=(2−1)2+(5−2)2+(−2−(−1))2 \text{Length of AB} = \sqrt{(2 - 1)^2 + (5 - 2)^2 + (-2 - (-1))^2}

=12+32+12=11= \sqrt{1^2 + 3^2 + 1^2} = \sqrt{11}

Length of vector BC BC :

Length of BC=(4−2)2+(4−5)2+(−3−(−2))2 \text{Length of BC} = \sqrt{(4 - 2)^2 + (4 - 5)^2 + (-3 - (-2))^2}

=22+(−1)2+(−1)2=6 = \sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{6}

Area of the rectangle is the product of the lengths of vectors AB AB and BC BC :

Area=11×6=66 \text{Area} = \sqrt{11} \times \sqrt{6} = \sqrt{66}

∴ The area of the rectangle is 66 \sqrt{66}  square units.

Hence, the correct answer is Option 3.

63

ABC is a triangle right-angled at B. If A(k,1,−1), B(2k,0,2) and C(2+2k,k,1) are the vertices of the triangle, then what is the value of k?

  1. ((a))

    -3

  2. ((b))

    -1

  3. ((c))

    1

  4. ((d))

    3

Show Answer
Answer: ((d))

3

Calculation:

Given,

Points A(k, 1, -1), B(2k, 0, 2), and C(2 + 2k, k, 1) are the vertices of the triangle.

The vectors AB and BC are given by:

AB→=B−A=(2k−k,0−1,2−(−1))=(k,−1,3) \overrightarrow{AB} = B - A = (2k - k, 0 - 1, 2 - (-1)) = (k, -1, 3)

BC→=C−B=(2+2k−2k,k−0,1−2)=(2,k,−1) \overrightarrow{BC} = C - B = (2 + 2k - 2k, k - 0, 1 - 2) = (2, k, -1)

The dot product of AB→ \overrightarrow{AB} and BC→ \overrightarrow{BC} is:

AB→⋅BC→=(k)(2)+(−1)(k)+(3)(−1)=2k−k−3=k−3 \overrightarrow{AB} \cdot \overrightarrow{BC} = (k)(2) + (-1)(k) + (3)(-1) = 2k - k - 3 = k - 3

Since the vectors are perpendicular, their dot product must be zero:

k−3=0 k - 3 = 0

Thus, k=3 k = 3 .

∴ The value of k is 33.

Hence, the correct answer is Option 4.

64

If a line  x+1p=y−1q=z−2r\frac{x+1}{p} = \frac{y-1}{q} = \frac{z-2}{r}  where p=2q=3r, makes an angle θ with the positive direction of y-axis, then what is cos2θ equal to?

  1. ((a))

    -31/49

  2. ((b))

    -37/49

  3. ((c))

    31/49

  4. ((d))

    37/49

Show Answer
Answer: ((a))

-31/49

Calculation:

Given,

The line equation is: x+1p=y−1q=z−2r \frac{x+1}{p} = \frac{y-1}{q} = \frac{z-2}{r} ,  where p = 2q = 3r .

The line makes an angle θ  with the positive direction of the y-axis.

Using the relation between the direction cosines

l=pp2+q2+r2,m=qp2+q2+r2,n=rp2+q2+r2 l = \frac{p}{\sqrt{p^2 + q^2 + r^2}}, m = \frac{q}{\sqrt{p^2 + q^2 + r^2}}, n = \frac{r}{\sqrt{p^2 + q^2 + r^2}} ,

we substitute p = 3r  and  q = 3r2\frac{3r}{2}

Direction cosine with y -axis is given by m=qp2+q2+r2 m = \frac{q}{\sqrt{p^2 + q^2 + r^2}} .

m=3r2(3r)2+(3r2)2+r2=3r29r2+9r24+r2=3r249r24=37 m = \frac{\frac{3r}{2}}{\sqrt{(3r)^2 + \left(\frac{3r}{2}\right)^2 + r^2}} = \frac{\frac{3r}{2}}{\sqrt{9r^2 + \frac{9r^2}{4} + r^2}} = \frac{\frac{3r}{2}}{\sqrt{\frac{49r^2}{4}}} = \frac{3}{7} .

Now, using the double angle identity for cosine cos⁡2θ=2cos⁡2θ−1 \cos 2θ = 2 \cos^2 θ - 1 :

cos⁡2θ=2(37)2−1=2×949−1=1849−1=−3149 \cos 2θ = 2 \left( \frac{3}{7} \right)^2 - 1 = 2 \times \frac{9}{49} - 1 = \frac{18}{49} - 1 = \frac{-31}{49} .

∴ The value of cos⁡2θ \cos 2θ is −3149 -\frac{31}{49} .

Hence, the correct answer is Option 1.

65

What is the equation of the plane passing through the point (1,1,1) and perpendicular to the line whose direction ratios are (3,2,1)?

  1. ((a))

    x+2y+3z=6

  2. ((b))

    3x+2y+z=6

  3. ((c))

    x+y+z=3

  4. ((d))

    3x+2y+z=0

Show Answer
Answer: ((b))

3x+2y+z=6

Calculation:

Given,

The plane passes through the point P(1,1,1)P(1, 1, 1) and is perpendicular to the line with direction ratios (3,2,1) (3, 2, 1) .

The general equation of the plane is:

Ax+By+Cz=DAx + By + Cz = D, where (A,B,C) (A, B, C) are the direction ratios of the normal to the plane.

Since the plane is perpendicular to the line, the normal vector to the plane is (3,2,1) (3, 2, 1) .

Thus, the equation of the plane becomes:

3x+2y+z=D 3x + 2y + z = D .

To find D D , substitute the point (1,1,1) (1, 1, 1) into the equation of the plane:

3(1)+2(1)+1=D 3(1) + 2(1) + 1 = D , which simplifies to:

3+2+1=D 3 + 2 + 1 = D ,

D=6 D = 6 .

∴ The equation of the plane is 3x+2y+z=6 3x + 2y + z = 6 .

Hence, the correct answer is Option 2.

66

A line makes angles α, β and γ with the positive directions of the coordinate axes. If a⃗=(sin⁡2α)i^+(sin⁡2β)j^+(sin⁡2γ)k^ and b⃗=i^+j^+k^\vec{a}=(\sin^2 \alpha)\hat{i} + (\sin^2 \beta)\hat{j} + (\sin^2 \gamma)\hat{k} \text{ and } \vec{b} = \hat{i} + \hat{j} + \hat{k} , then what is a⃗.b⃗\vec{a}.\vec{b} equal to?

  1. ((a))

    -2

  2. ((b))

    -1

  3. ((c))

    1

  4. ((d))

    2

Show Answer
Answer: ((d))

2

Calculation:

Given,

cos⁡2(α)+cos⁡2(β)+cos⁡2(γ)=1 \cos^2(\alpha) + \cos^2(\beta) + \cos^2(\gamma) = 1

Using the identity cos⁡2(x)=1−sin⁡2(x) \cos^2(x) = 1 - \sin^2(x) , we substitute:

(1−sin⁡2(α))+(1−sin⁡2(β))+(1−sin⁡2(γ))=1 (1 - \sin^2(\alpha)) + (1 - \sin^2(\beta)) + (1 - \sin^2(\gamma)) = 1

Simplifying the equation:

3−(sin⁡2(α)+sin⁡2(β)+sin⁡2(γ))=1 3 - (\sin^2(\alpha) + \sin^2(\beta) + \sin^2(\gamma)) = 1

Rearrange to isolate the sine terms:

sin⁡2(α)+sin⁡2(β)+sin⁡2(γ)=2 \sin^2(\alpha) + \sin^2(\beta) + \sin^2(\gamma) = 2

Now, calculate the dot product:

a⃗⋅b⃗=sin⁡2(α)+sin⁡2(β)+sin⁡2(γ)=2 \vec{a} \cdot \vec{b} = \sin^2(\alpha) + \sin^2(\beta) + \sin^2(\gamma) = 2

∴ The value of a⃗⋅b⃗ \vec{a} \cdot \vec{b} is 2.

Hence, the correct answer is Option 4.

67

Consider the following statements with respect to a vector d = (a × b) × c:

I. d is coplanar with a and b.

II. d is perpendicular to c.

Which of the statements given above is/are correct?

  1. ((a))

    I only

  2. ((b))

    II only

  3. ((c))

    Both I and II

  4. ((d))

    Neither I nor II

Show Answer
Answer: ((c))

Both I and II

Calculation:

Given,

The vector d⃗=(a⃗×b⃗)×c⃗ \vec{d} = (\vec{a} \times \vec{b}) \times \vec{c}

Statement I: d⃗ \vec{d} is coplanar with a⃗ \vec{a} and b⃗ \vec{b} .

We use the vector triple product identity: (a⃗×b⃗)×c⃗=(a⃗⋅c⃗)b⃗−(b⃗⋅c⃗)a⃗ (\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a} .

This shows that d⃗ \vec{d} is a linear combination of a⃗ \vec{a} and b⃗ \vec{b} , hence d⃗ \vec{d} is coplanar with a⃗ \vec{a} and b⃗ \vec{b} .

Therefore, Statement I is correct.

Statement II: d⃗ \vec{d} is perpendicular to c⃗ \vec{c} .

To check this, compute the dot product d⃗⋅c⃗ \vec{d} \cdot \vec{c} . Using the vector triple product identity, we find:

d⃗⋅c⃗=(a⃗⋅c⃗)(b⃗⋅c⃗)−(b⃗⋅c⃗)(a⃗⋅c⃗)=0 \vec{d} \cdot \vec{c} = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{c}) - (\vec{b} \cdot \vec{c})(\vec{a} \cdot \vec{c}) = 0 ,

which means d⃗ \vec{d} is perpendicular to c⃗ \vec{c} .

Therefore, Statement II is correct.

∴ Both Statement I and Statement II are correct.

Hence, the correct answer is option 3.

68

The position vectors of three points A, B and C are a, b and c respectively such that 3a⃗−4b⃗+1c⃗=0⃗\vec{3a}-\vec{4b}+\vec{1c}=\vec{0} , then what is AB:BC equal to?

  1. ((a))

    3:1

  2. ((b))

    1:3

  3. ((c))

    3:4

  4. ((d))

    1:4

Show Answer
Answer: ((b))

1:3

Calculation:

Given,

3a⃗−4b⃗+c⃗=0 3\vec{a} - 4\vec{b} + \vec{c} = 0

c⃗=4b⃗−3a⃗ \vec{c} = 4\vec{b} - 3\vec{a}

The vectorAB→\overrightarrow{AB} is:

AB→=b⃗−a⃗ \overrightarrow{AB} = \vec{b} - \vec{a}

The vector BC→\overrightarrow{BC} is:

BC→=c⃗−b⃗ \overrightarrow{BC} = \vec{c} - \vec{b}

Substituting c⃗=4b⃗−3a⃗\vec{c} = 4\vec{b} - 3\vec{a} :

BC→=(4b⃗−3a⃗)−b⃗ \overrightarrow{BC} = (4\vec{b} - 3\vec{a}) - \vec{b}

BC→=3b⃗−3a⃗ \overrightarrow{BC} = 3\vec{b} - 3\vec{a}

Step 4: Now, BC→=3(b⃗−a⃗)\overrightarrow{BC} = 3(\vec{b} - \vec{a}) , which gives:

AB:BC=1:3 AB : BC = 1 : 3

∴ The correct ratio is AB : BC = 1 : 3 , 

Hence, the correct answer is Option 2.

69

The position vectors of three points A, B and C respectively, where  a⃗,b⃗\vec{a} ,\vec{b}  and c⃗\vec{c}   respectively, where c⃗=(cos⁡2θ)a⃗+(sin⁡2θ)b⃗\vec{c} = (\cos^2 \theta)\vec{a}+(\sin^2 \theta)\vec{b}. What is (a⃗×b⃗)+(b⃗×c⃗)+(c⃗×a⃗)(\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a}) equal to?

  1. ((a))

    0⃗\vec{0}

  2. ((b))

    2c⃗\vec{2c}

  3. ((c))

    3c⃗\vec{3c}

  4. ((d))

    Unit vector

Show Answer
Answer: ((a))

0⃗\vec{0}

Calculation:

Given,

The position vectors of points A, B, and C are a⃗ \vec{a} , b⃗ \vec{b} , and c⃗ \vec{c} respectively, and c⃗=cos⁡2θ,a⃗+sin⁡2θ,b⃗ \vec{c} = \cos^2 \theta , \vec{a} + \sin^2 \theta , \vec{b} .

The expression to evaluate is: (a⃗×b⃗)+(b⃗×c⃗)+(c⃗×a⃗) (\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a}) .

First, substitute c⃗ \vec{c} into the equation:

(a⃗×b⃗)+(b⃗×(cos⁡2θ,a⃗+sin⁡2θ,b⃗))+(cos⁡2θ,a⃗+sin⁡2θ,b⃗)×a⃗ (\vec{a} \times \vec{b}) + (\vec{b} \times (\cos^2 \theta , \vec{a} + \sin^2 \theta , \vec{b})) + (\cos^2 \theta , \vec{a} + \sin^2 \theta , \vec{b}) \times \vec{a} .

Using the distributive property of the cross product:

(a⃗×b⃗)+[(b⃗×cos⁡2θ,a⃗)+(b⃗×sin⁡2θ,b⃗)]+[(cos⁡2θ,a⃗×a⃗)+(sin⁡2θ,b⃗×a⃗)] (\vec{a} \times \vec{b}) + \left[ (\vec{b} \times \cos^2 \theta , \vec{a}) + (\vec{b} \times \sin^2 \theta , \vec{b}) \right] + \left[ (\cos^2 \theta , \vec{a} \times \vec{a}) + (\sin^2 \theta , \vec{b} \times \vec{a}) \right] .

Since b⃗×b⃗=0 \vec{b} \times \vec{b} = 0 and a⃗×a⃗=0 \vec{a} \times \vec{a} = 0 , we are left with:

(a⃗×b⃗)+cos⁡2θ,(b⃗×a⃗)+sin⁡2θ,(−a⃗×b⃗) (\vec{a} \times \vec{b}) + \cos^2 \theta , (\vec{b} \times \vec{a}) + \sin^2 \theta , (- \vec{a} \times \vec{b}) .

Substitute b⃗×a⃗=−(a⃗×b⃗) \vec{b} \times \vec{a} = - (\vec{a} \times \vec{b}) into the expression:

(a⃗×b⃗)+cos⁡2θ,(−a⃗×b⃗)+sin⁡2θ,(−a⃗×b⃗) (\vec{a} \times \vec{b}) + \cos^2 \theta , (- \vec{a} \times \vec{b}) + \sin^2 \theta , (- \vec{a} \times \vec{b}) .

Factor out a⃗×b⃗ \vec{a} \times \vec{b} :

a⃗×b⃗[1−cos⁡2θ−sin⁡2θ] \vec{a} \times \vec{b} \left[ 1 - \cos^2 \theta - \sin^2 \theta \right] .

Since cos⁡2θ+sin⁡2θ=1 \cos^2 \theta + \sin^2 \theta = 1 , the expression becomes:

a⃗×b⃗[1−1]=0 \vec{a} \times \vec{b} [1 - 1] = 0 .

∴ The final result is 0⃗ \vec{0} .

Hence, the correct answer is option 1.

70

Let a⃗,b⃗,(a⃗×b⃗)\vec{a},\vec{b} ,(\vec{a}\times\vec{b}) be unit vectors. What is a⃗.b⃗\vec{a}.\vec{b} equal to?

  1. ((a))

    0

  2. ((b))

    1/2

  3. ((c))

    1

  4. ((d))

    3

Show Answer
Answer: ((a))

0

Calculation:

Given,

The vectors a⃗,b⃗,(a⃗×b⃗)\vec{a},\vec{b} ,(\vec{a}\times\vec{b}) are unit vectors.

Since a⃗\vec a and  b⃗\vec{b} are unit vectors, we know:

∣a⃗∣=1and∣b⃗∣=1 |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 .

The magnitude of the cross product (a⃗×b⃗)( \vec{a} \times \vec{b} ) is given by:

∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ=1×1×sin⁡θ=sin⁡θ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta = 1 \times 1 \times \sin \theta = \sin \theta .

Since (∣a⃗×b⃗∣=1 ( |\vec{a} \times \vec{b}| = 1 , we have:

sin⁡θ=1 \sin \theta = 1 , so θ=90∘\theta = 90^\circ , meaning a⃗\vec a and b⃗ \vec{b}  are perpendicular.

The dot product (a⃗⋅b⃗) ( \vec{a} \cdot \vec{b} ) is:

a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ=1×1×cos⁡90∘=0 \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = 1 \times 1 \times \cos 90^\circ = 0 .

∴ The value of(a⃗⋅b⃗)( \vec{a} \cdot \vec{b} ) is 0.

Hence, the correct answer is Option 1.

Let  x=sec⁡θ−cos⁡θ and y=sec⁡4θ−cos⁡4θ x = \sec \theta - \cos \theta \text{ and } y = \sec^4 \theta - \cos^4 \theta \

71

What is (dydx)2(\frac{dy}{dx})^2 equal to?

  1. ((a))

    4(y2+4)(x2+4)\frac{4(y^2+4)}{(x^2+4)}

  2. ((b))

    4(y2−4)(x2−4)\frac{4(y^2-4)}{(x^2-4)}

  3. ((c))

    16(y2+4)(x2+4)\frac{16(y^2+4)}{(x^2+4)}

  4. ((d))

    16(y2−4)(x2−4)\frac{16(y^2-4)}{(x^2-4)}

Show Answer
Answer: ((c))

16(y2+4)(x2+4)\frac{16(y^2+4)}{(x^2+4)}

Calculation:

Given,

x=sec⁡θ−cos⁡θx = \sec\theta - \cos\theta

y=sec⁡4θ−cos⁡4θy = \sec^4\theta - \cos^4\theta

Compute derivatives w.r.t. θθθ

:

dxdθ=sec⁡θ,tan⁡θ+sin⁡θ\dfrac{dx}{d\theta} = \sec\theta,\tan\theta + \sin\theta

dydθ=4sec⁡4θ,tan⁡θ;+;4cos⁡3θ,sin⁡θ\dfrac{dy}{d\theta} = 4\sec^4\theta,\tan\theta ;+; 4\cos^3\theta,\sin\theta

From the ratio dydx=dydθdxdθ\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}

dydx=4(sec⁡4θ,tan⁡θ+cos⁡3θ,sin⁡θ)sec⁡θ,tan⁡θ+sin⁡θ\dfrac{dy}{dx} = \frac{4\bigl(\sec^4\theta,\tan\theta + \cos^3\theta,\sin\theta\bigr)} {\sec\theta,\tan\theta + \sin\theta}

Using identities x2=sec⁡2θ+cos⁡2θ−2x^2 = \sec^2\theta + \cos^2\theta - 2, sec⁡3θ+cos⁡5θ)2=y2+4\sec^3\theta + \cos^5\theta)^2 = y^2 + 4, and (1+cos⁡2θ)2=x2+4,(1 + \cos^2\theta)^2 = x^2 + 4,

(dydx)2=16,,y2+4,,x2+4,\displaystyle \Bigl(\frac{dy}{dx}\Bigr)^{2} = 16 ,\frac{,y^{2} + 4,}{,x^{2} + 4,}

∴  (dydx)2=16,y2+4x2+4\displaystyle \Bigl(\frac{dy}{dx}\Bigr)^{2} = 16,\frac{y^{2} + 4}{x^{2} + 4}

Hence, the correct answer is Option 3.

72

What  [x2+4y2+4dydx(x2+4d2ydx2−16y)]\left[ \frac{x^2+4}{y^2+4} \frac{dy}{dx} \left( x^2+4 \frac{d^2y}{dx^2} - 16y \right) \right]   equal to?

  1. ((a))

    16x16x

  2. ((b))

    16y16y

  3. ((c))

    −16x-16x

  4. ((d))

    −16y-16y

Show Answer
Answer: ((a))

16x16x

Calculation:

Given,

x=sec⁡θ−cos⁡θx = \sec\theta - \cos\theta

y=sec⁡4θ−cos⁡4θy = \sec^4\theta - \cos^4\theta

We know from before:

(dydx)2=16,(y2+4),x2+4,\displaystyle \Bigl(\frac{dy}{dx}\Bigr)^{2} = \frac{16,(y^{2}+4)}{,x^{2}+4,}

Differentiate both sides w.r.t. x:

ddx[(x2+4)(dydx)2]=ddx[16,(y2+4)]\displaystyle \frac{d}{dx}\Bigl[(x^{2}+4)\bigl(\tfrac{dy}{dx}\bigr)^{2}\Bigr] = \frac{d}{dx}\bigl[16,(y^{2}+4)\bigr]

This yields the differential relation:

(x2+4),d2ydx2;+;x,dydx;−;16,y;=;0\displaystyle (x^{2}+4),\frac{d^{2}y}{dx^{2}} ;+;x,\frac{dy}{dx} ;-;16,y ;=;0

The required expression is

x2+4y2+4,dydx[(x2+4)d2ydx2−16y]\displaystyle \frac{x^{2}+4}{y^{2}+4},\frac{dy}{dx} \Bigl[(x^{2}+4)\frac{d^{2}y}{dx^{2}}-16y\Bigr]

From the differential relation,

(x2+4)d2ydx2−16y=−,x,dydx\displaystyle (x^{2}+4)\frac{d^{2}y}{dx^{2}}-16y = -,x,\frac{dy}{dx}

So the expression becomes

x2+4y2+4,dydx,(−x,dydx)=−,x,x2+4y2+4(dydx)2\displaystyle \frac{x^{2}+4}{y^{2}+4},\frac{dy}{dx}, \bigl(-x,\tfrac{dy}{dx}\bigr) = -,x,\frac{x^{2}+4}{y^{2}+4} \Bigl(\frac{dy}{dx}\Bigr)^{2}

Using (dydx)2=16,(y2+4)x2+4\bigl(\tfrac{dy}{dx}\bigr)^{2} = \tfrac{16,(y^{2}+4)}{x^{2}+4}, all factors cancel except -16x.

∴  The value of the given expression is  −16x-16x.

Hence, the correct answer is Option 3.

Consider the following for the two (02) items that follow: Let ABC be a triangle right-angled at B and AB+AC = 3 units.

73

What is ∠A equal to if the area of the triangle is maximum?

  1. ((a))

    π/6

  2. ((b))

    π/4

  3. ((c))

    π/3

  4. ((d))

    5π/12

Show Answer
Answer: ((c))

π/3

Calculation:

Given,

AB+AC=3AB + AC = 3

Let AB=xAB = x and AC=3−xAC = 3 - x.

Then,

BC=AC2−AB2=(3−x)2−x2=9−6xBC = \sqrt{AC^2 - AB^2} = \sqrt{(3 - x)^2 - x^2} = \sqrt{9 - 6x}

The area of the triangle is,

A=12,x,9−6xA = \tfrac12,x,\sqrt{9 - 6x}

To maximize, set up

A2=14,x2,(9−6x)A^2 = \tfrac14,x^2,(9 - 6x)

d(A2)dx=14(2x(9−6x)+x2(−6))=18x(1−x)4=0\displaystyle \frac{d(A^2)}{dx} = \tfrac14\bigl(2x(9 - 6x) + x^2(-6)\bigr) = \frac{18x(1 - x)}{4} = 0

Hence x=1x = 1 (discarding x=0, so

BC=9−6=3,AC=3−1=2BC = \sqrt{9 - 6}= \sqrt{3},\quad AC = 3 - 1 = 2

Therefore,

sin⁡A=BCAC=32  ⟹  A=π3\sin A = \frac{BC}{AC} = \frac{\sqrt{3}}{2} \implies A = \frac{\pi}{3}

∴ ∠A=π3\angle A = \frac{\pi}{3}.

Hence, the correct answer is Option 3.

74

What is the maximum area of the triangle?

  1. ((a))

    3/2\sqrt{3}/2

  2. ((b))

    3\sqrt{3}

  3. ((c))

    6/2\sqrt{6}/2

  4. ((d))

    6\sqrt{6}

Show Answer
Answer: ((a))

3/2\sqrt{3}/2

Calculation:

Given,

AB+AC=3AB + AC = 3

Let AB=xAB = x and AC=3−xAC = 3 - x .

Then,

BC=AC2−AB2=(3−x)2−x2=9−6xBC = \sqrt{AC^2 - AB^2} = \sqrt{(3 - x)^2 - x^2} = \sqrt{9 - 6x}

The area of the triangle is,

A=12,x,BC=12,x,9−6xA = \tfrac12,x,BC = \tfrac12,x,\sqrt{9 - 6x}

To maximize, differentiate w.r.t. xx and set to zero:

ddx(x9−6x)=9−6x;−;6x29−6x=0;⟹;x=1\displaystyle \frac{d}{dx}\bigl(x\sqrt{9-6x}\bigr) = \sqrt{9-6x} ;-;\frac{6x}{2\sqrt{9-6x}} = 0 ;\Longrightarrow; x = 1

At x=1x = 1, we get BC=9−6=3BC = \sqrt{9 - 6} = \sqrt{3}, so

Amax⁡=12×1×3=32A_{\max} = \tfrac12 \times 1 \times \sqrt{3} = \frac{\sqrt{3}}{2}

∴ The maximum area is 32\frac{\sqrt{3}}{2} square unit.

Hence, the correct answer is Option  1.

Consider the following for the two (02) items that follow: 

 Let (x+y)p+q=xpyq, where p,q are positive integers.

75

The derivative of y with respect to x

  1. ((a))

    depends on p only

  2. ((b))

    depends on q only

  3. ((c))

    depends on both p and qc

  4. ((d))

    is independent of both p and q

Show Answer
Answer: ((d))

is independent of both p and q

Calculation:

Given,

(x+y)p+q=xp,yq(x+y)^{p+q} = x^p,y^q

Differentiate implicitly w.r.t. xx:

(p+q)(x+y)p+q−1(1+dydx)=p,xp−1yq;+;q,xp,yq−1dydx(p+q)(x+y)^{p+q-1}\bigl(1+\tfrac{dy}{dx}\bigr) = p,x^{p-1}y^q ;+; q,x^p,y^{q-1}\tfrac{dy}{dx}

Rearrange to collect dydx\tfrac{dy}{dx}:

dydx[(p+q)(x+y)p+q−1−q,xp,yq−1]=p,xp−1yq−(p+q)(x+y)p+q−1\tfrac{dy}{dx}\bigl[(p+q)(x+y)^{p+q-1} - q,x^p,y^{q-1}\bigr] = p,x^{p-1}y^q - (p+q)(x+y)^{p+q-1}

Use (x+y)p+q−1=xp,yqx+y(x+y)^{p+q-1}=\frac{x^p,y^q}{x+y} to simplify:

dydx=yx\tfrac{dy}{dx} = \tfrac{y}{x}

∴ dydx=yx\displaystyle \frac{dy}{dx} = \frac{y}{x}, independent of pp and qq.

Hence, the correct answer is Option 4.

76

If p+q=10, then what is dydx\frac{dy}{dx} equal to?

  1. ((a))

    yx\frac{y}{x}

  2. ((b))

    xyxy

  3. ((c))

    x10y10x^{10}y^{10}

  4. ((d))

    (yx)10(\frac{y}{x})^{10}

Show Answer
Answer: ((a))

yx\frac{y}{x}

Calculation:

Given,

(x+y)p+q=xp,yq (x + y)^{p+q} = x^p,y^q and p+q=10 p + q = 10 .

Differentiate both sides with respect to xx implicitly:

ddx((x+y)p+q)=ddx(xpyq)\frac{d}{dx}\bigl((x+y)^{p+q}\bigr) = \frac{d}{dx}\bigl(x^p y^q\bigr)

Left side:

(p+q),(x+y)p+q−1(1+dydx)(p+q),(x+y)^{p+q-1}\bigl(1 + \tfrac{dy}{dx}\bigr)

Right side (product rule):

p,xp−1yq;+;q,xpyq−1,dydxp,x^{p-1}y^q ;+; q,x^p y^{q-1},\tfrac{dy}{dx}

Rearrange to collect dydx \tfrac{dy}{dx} terms and use

(x+y)p+q=xpyq  ⟹  (x+y)p+q−1=xpyqx+y (x+y)^{p+q} = x^p y^q \implies (x+y)^{p+q-1} = \tfrac{x^p y^q}{x+y} .

After cancellation of the common factor p,y−q,x p,y - q,x , you obtain:

∴ dydx=yx \frac{dy}{dx} = \frac{y}{x}

Hence, the correct answer is Option 1.

Consider the following for the two (02) items that follow:

The slope of the tangent to the curve y = f(x) at (x, f(x) ) is 4 for every real number x and the curve passes through the origin.

77

What is the nature of the curve?

  1. ((a))

    A straight line passing through (1,4)

  2. ((b))

    A straight line passing through (-14)

  3. ((c))

    A parabola with vertex at origin and focus at (2,0)

  4. ((d))

    A parabola with vertex at origin and focus at (1, 0)

Show Answer
Answer: ((a))

A straight line passing through (1,4)

Calculation:

Given,

The slope of the tangent to the curve y = f(x) at (x, f(x)) is 4 for every real number x , and the curve passes through the origin.

The slope of the tangent is the derivative of the function, so we have:

f′(x)=4 f'(x) = 4

Integrating f'(x) = 4  with respect to  x :

f(x)=4x+C f(x) = 4x + C

The curve passes through the origin, so when x = 0 , y = 0 . Substituting these values into the equation f(x) = 4x + C :

0=4(0)+C⇒C=0 0 = 4(0) + C \quad \Rightarrow \quad C = 0

Therefore, the equation of the curve is:

f(x)=4x f(x) = 4x

This is the equation of a straight line with a slope of 4, passing through the origin.

∴ The curve is a straight line with a slope of 4, passing through the origin.

Hence, the correct answer is option 1.

78

. What is the area bounded by the curve, the x-axis and the line x = 4?

  1. ((a))

    8 square units 

  2. ((b))

    16 square units 

  3. ((c))

    32 square units 

  4. ((d))

    64 square units 

Show Answer
Answer: ((c))

32 square units 

Calculation: 

 

Given,

The equation of the curve is y = 4x , and the line x = 4 intersects the curve at the point (4, 16) . We need to find the area bounded by the curve, the x-axis, and the line x = 4 .

The region of interest is a right triangle with a base along the x-axis from x = 0 to x = 4  and a height of 16 units, corresponding to the point (4, 16) .

The area of the triangle is given by the formula:

Area=12×base×height \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}

Substituting the values of the base (4 units) and the height (16 units):

Area=12×4×16=32,square units \text{Area} = \frac{1}{2} \times 4 \times 16 = 32 , \text{square units}

∴ The area is 32 square units.

Hence, the correct answer is option 3.

Consider the following for the two (02) items that follow:

Let f(x)={x3,x2<1 x2,x2≥1\text{Let } f(x)= \begin{cases} x^3, & x^2 < 1 \ x^2, & x^2 \ge 1 \end{cases}

79

What is  lim⁡x→0f′(x)\lim_{x \to 0} f'(x) equal to?

  1. ((a))

    2

  2. ((b))

    1

  3. ((c))

    0

  4. ((d))

     Limit does not exist

Show Answer
Answer: ((c))

0

Calculation:

Given,

The function is defined as:

f(x)={x3,if,∣x∣<1 x2,if,∣x∣≥1 f(x) = \begin{cases} x^3, & \text{if} , |x| < 1 \ x^2, & \text{if} , |x| \geq 1 \end{cases}

We are tasked with finding:

lim⁡x→0f′(x) \lim_{x \to 0} f'(x)

For  |x| < 1 , the function is  f(x) = x3, so the derivative is:

f′(x)=3x2 f'(x) = 3x^2

Now, compute the limit of the derivative as x to 0:

lim⁡x→0f′(x)=lim⁡x→03x2=0 \lim_{x \to 0} f'(x) = \lim_{x \to 0} 3x^2 = 0

∴ The value of  lim⁡x→0f′(x)\lim_{x \to 0} f'(x) is 0.

The correct answer is Option (c)

80

Consider the following statements:

I. The function is continuous at x=−1.

II. The function is differentiable at x=1.

Which of the statements given above is/are correct?

  1. ((a))

    I only

  2. ((b))

    II only

  3. ((c))

    Both I and II

  4. ((d))

    Neither I nor II

Show Answer
Answer: ((d))

Neither I nor II

Calculation:

Given,

The function is defined as:

f(x)={x3,if,∣x∣<1 x2,if,∣x∣≥1 f(x) = \begin{cases} x^3, & \text{if} , |x| < 1 \ x^2, & \text{if} , |x| \geq 1 \end{cases}

We are tasked with finding:

lim⁡x→−1f′(x) \lim_{x \to -1} f'(x)

Check the left-hand limit for continuity at x = -1 :

lim⁡x→−1−f(x)=(−1)2=1 \lim_{x \to -1^-} f(x) = (-1)^2 = 1

Check the right-hand limit for continuity at x = -1 :

lim⁡x→−1+f(x)=(−1)3=−1 \lim_{x \to -1^+} f(x) = (-1)^3 = -1

Since the left-hand limit (L.H.S) and right-hand limit (R.H.S) are not equal, the function is discontinuous at x = -1 .

Check the differentiability at x = 1 :

The left-hand derivative at x = 1  is L.H.D=3\text{L.H.D} = 3 and the right-hand derivative at  x = 1  is R.H.D=2\text{R.H.D} = 2 which means the function is not differentiable at x = 1 

∴ The function is neither continuous at x = -1  nor differentiable at x = 1 . 

Hence, the correct answer is Option 4.

Consider the following for the two (02) items that follow:

Let the function y = (1 - cos x)-1 wherex≠2nπx \ne 2n\pi and n is an integer

81

What is the range of the function?

  1. ((a))

    [0,∞)

  2. ((b))

    [0.5,∞)

  3. ((c))

    [1,∞)

  4. ((d))

    (−∞,0.5]

Show Answer
Answer: ((b))

[0.5,∞)

Calculation:

Given,

The function isy=(1−cos⁡(x))−1y = \left( 1 - \cos(x) \right)^{-1} , where x≠2nπx \neq 2n\pi and n  is an integer.

The cosine function has a range of  [-1, 1] , so cos(x) can take values from ( -1) to ( 1 ).

The expression 1−cos⁡(x)1 - \cos(x) will take values from:

1−1=0to1−(−1)=2 1 - 1 = 0 \quad \text{to} \quad 1 - (-1) = 2

So, (1 - cos(x)) takes values in the range (0,2](0, 2] , but x≠2nπx \neq 2n\pi excludes the value 0.

Since y=11−cos⁡(x)y = \frac{1}{1 - \cos(x)} , the reciprocal function will take values in the range [12,∞)\left[ \frac{1}{2}, \infty \right)

∴ The range of the function is [0.5,∞)[0.5, \infty) .

The correct answer is Option (b)

82

What is ∫ydx equal to? 

where c is the constant of integration.

  1. ((a))

    −tan(x/2)+c

  2. ((b))

    −cot(x/2)+c

  3. ((c))

    tan(x/2)+c

  4. ((d))

    cot(x/2)+c

Show Answer
Answer: ((b))

−cot(x/2)+c

Calculation:

Given,

The function is y=(1−cos⁡(x))−1y = \left(1 - \cos(x)\right)^{-1} , wherex≠2nπx \neq 2n\pi and n is an integer.

The given function can be written as:

y=11−cos⁡(x) y = \frac{1}{1 - \cos(x)}

Using the identity 1−cos⁡(x)=2sin⁡2(x2)1 - \cos(x) = 2\sin^2\left(\frac{x}{2}\right) , we rewrite the function as:

y=12sin⁡2(x2) y = \frac{1}{2\sin^2\left(\frac{x}{2}\right)}

The integral we need to evaluate is:

∫dx2sin⁡2(x2) \int \frac{dx}{2\sin^2\left(\frac{x}{2}\right)}

12∫csc⁡2(x2)dx \frac{1}{2} \int \csc^2\left(\frac{x}{2}\right) dx

The integral of csc⁡2(x)\csc^2(x) is known to be −cot⁡(x) -\cot(x) , so we get:

12(−cot⁡(x2))+C \frac{1}{2} \left( -\cot\left(\frac{x}{2}\right) \right) + C

−12cot⁡(x2)+C -\frac{1}{2} \cot\left(\frac{x}{2}\right) + C

The final result is −cot⁡(x2)+C -\cot\left(\frac{x}{2}\right) + C

Hence, the correct answer is Option 2.

Consider the following for the two (02) items that follow: Let the function f(x)=sin[x], where [⋅] is the greatest integer function and g(x)=∣x∣.

83

What lim⁡x→0f(x)g(x)\lim_{x \to 0} {f(x) g(x)} is  equal to?

  1. ((a))

    -1

  2. ((b))

    0

  3. ((c))

    1

  4. ((d))

    Limit does not exist

Show Answer
Answer: ((b))

0

Calculation:

Given,

The function isf(x)=sin⁡(⌊x⌋)f(x) = \sin(\lfloor x \rfloor) , where⌊x⌋\lfloor x \rfloor is the greatest integer function, and g(x) = |x| , the absolute value function.

We are tasked with finding:

lim⁡x→0f(x)g(x) \lim_{x \to 0} f(x) g(x)

For g(x)=∣x∣g(x) = |x| , we know that:

lim⁡x→0g(x)=0 \lim_{x \to 0} g(x) = 0

For f(x)=sin⁡(⌊x⌋)f(x) = \sin(\lfloor x \rfloor) , we know that:

For x→0+x \to 0^+ ), ⌊x⌋=0 \lfloor x \rfloor = 0 , so f(x) = sin(0) = 0 .

For x→0− x \to 0^- , ⌊x⌋=−1 \lfloor x \rfloor = -1 , so f(x)=sin⁡(−1)f(x) = \sin(-1) , which is a nonzero constant.

Evaluating the limit:

Forx→0+x \to 0^+ ,f(x)g(x)=0×x=0f(x)g(x) = 0 \times x = 0

Forx→0−x \to 0^- , f(x)g(x)=sin⁡(−1)×(−x)f(x)g(x) = \sin(-1) \times (-x) , which approaches 0 as x→0−.x \to 0^- ..

∴ The value of lim⁡x→0f(x)g(x)\lim_{x \to 0} f(x) g(x) is 0.

The correct answer is Option (2).

84

What is lim⁡x→0f(x)g(x)\lim_{x \to 0} \frac{f(x)}{g(x)} equal to?

  1. ((a))

    −sin1

  2. ((b))

    sin1

  3. ((c))

    0

  4. ((d))

    Limit does not exist

Show Answer
Answer: ((d))

Limit does not exist

Calculation:

Given,

The function is f(x)=sin⁡(⌊x⌋) f(x) = \sin(\lfloor x \rfloor) , where ⌊x⌋\lfloor x \rfloor is the greatest integer function, and g(x) = |x| , the absolute value function.

We are tasked with finding:

lim⁡x→0f(x)g(x) \lim_{x \to 0} \frac{f(x)}{g(x)}

For g(x)=∣x∣ g(x) = |x| , we know that:

lim⁡x→0g(x)=0 \lim_{x \to 0} g(x) = 0

For f(x)=sin⁡(⌊x⌋) f(x) = \sin(\lfloor x \rfloor) , we know that:

For x→0+ x \to 0^+ , ⌊x⌋=0 \lfloor x \rfloor = 0 , so f(x) = sin(0) = 0 .

For x→0− x \to 0^- , ⌊x⌋=−1 \lfloor x \rfloor = -1 , so f(x)=sin⁡(−1) f(x) = \sin(-1) , which is a nonzero constant.

Evaluating the limit:

For x→0+ x \to 0^+ , f(x)g(x)=0x=0 \frac{f(x)}{g(x)} = \frac{0}{x} = 0

For x→0− x \to 0^- , f(x)g(x)=sin⁡(−1)−x \frac{f(x)}{g(x)} = \frac{\sin(-1)}{-x} , which becomes undefined as x→0− x \to 0^- because the denominator approaches 0, but the numerator remains a nonzero constant.

∴ Since the left-hand and right-hand limits do not match, the limit does not exist.

The correct answer is Option (4):

Consider the following for the two (02) items that follow:

Let the curve f(x) = |x - 3|

85

What is the domain of the function (f(x))?

  1. ((a))

    (0, ∞)

  2. ((b))

    (3,∞)

  3. ((c))

    (-∞, ∞)

  4. ((d))

    (-∞.∞)\3

Show Answer
Answer: ((c))

(-∞, ∞)

Calculation:

Given,

The function is f(x)=∣x−3∣f(x) = |x - 3| , which is an absolute value function.

The domain of an absolute value function f(x)=∣x−a∣f(x) = |x - a| is all real numbers, because the absolute value function is defined for all values of x. The function handles both positive and negative inputs for x .

Since there are no restrictions or undefined points for the absolute value function, the domain is all real numbers.

∴ The domain of the function is (−∞,∞)(-\infty, \infty)

Hence, the correct answer is Option 3.

86

What is the area bounded by the curve f(x) and y = 3?

  1. ((a))

    3 square units

  2. ((b))

    4-5 square units

  3. ((c))

    7-5 square units

  4. ((d))

    9 square units

Show Answer
Answer: ((d))

9 square units

Calculation:

Given,

The function is f(x) = |x - 3| , and we need to find the area bounded by the curve and the line y = 3.

To find the points of intersection, we set the function equal to 3:

∣x−3∣=3 |x - 3| = 3

Solving for x :

  • Forx≥3 x \geq 3 , x−3=3x - 3 = 3 , which gives x = 6 .
  • For ( x < 3 ), 3 - x = 3 , which gives x = 0 .

Therefore, the points of intersection are x = 0  and x = 6 .

The area can be calculated by integrating the difference between the curve and the line from x = 0 to x = 6. The integral is split into two parts due to the absolute value function:

A=∫03(3−x),dx+∫36(x−3),dx A = \int_{0}^{3} (3 - x) , dx + \int_{3}^{6} (x - 3) , dx

For x in [0, 3] , ( f(x) = 3 - x ), and for ( x in [3, 6] ), ( f(x) = x - 3 ).

Compute both integrals:

  • For x in [0, 3] :

∫03(3−x),dx=[3x−x22]03=9−4.5=4.5 \int_{0}^{3} (3 - x) , dx = \left[ 3x - \frac{x^2}{2} \right]_{0}^{3} = 9 - 4.5 = 4.5

  • For x in [3, 6] :

∫36(x−3),dx=[x22−3x]36=4.5 \int_{3}^{6} (x - 3) , dx = \left[ \frac{x^2}{2} - 3x \right]_{3}^{6} = 4.5

Step 4: The total area is the sum of the two areas:

A=4.5+4.5=9,square units A = 4.5 + 4.5 = 9 , \text{square units}

∴ The total area bounded by the curve and the line is 9 square units.

The correct answer is Option (4):

Consider the following for the two (02) items that follow:

Let f = {(1, 1), (2, 4), (3, 7), (4, 10)}

87

If f(x) = px + q then what is the value of (p + q) ?

  1. ((a))

    -1

  2. ((b))

    0

  3. ((c))

    1

  4. ((d))

    5

Show Answer
Answer: ((c))

1

Calculation:

Given,

The function is f=(1,1),(2,4),(3,7),(4,10)f = {(1, 1), (2, 4), (3, 7), (4, 10)} , and we are also given that the function is f(x)=px+q f(x) = px + q , where p and q  are constants.

Using the points f(1) = 1  and f(2) = 4 , we create the following system of equations:

  • From f(1) = 1 , we have p(1)+q=1p(1) + q = 1 , so p + q = 1 .......(1).
  • From f(2) = 4 , we have p(2) + q = 4 , so 2p + q = 4  ..........( 2).

Subtract Equation 1 from Equation 2:

(2p+q)−(p+q)=4−1 (2p + q) - (p + q) = 4 - 1

p=3 p = 3

Now, substitute p = 3 into Equation 1:

3+q=1 3 + q = 1

q=−2 q = -2

p+q=3+(−2)=1 p + q = 3 + (-2) = 1

∴ The value of p + q  is 1.

The correct answer is Option (3).

88

Consider the following statements:

I. f is one-one function.

II. f is onto function if the codomain is the set of natural numbers.

Which of the statements given above is/are correct?

  1. ((a))

    I only

  2. ((b))

    II only

  3. ((c))

    Both I and II

  4. ((d))

    Neither I not II

Show Answer
Answer: ((a))

I only

Calculation:

Given,

The function is f(x)=px+qf(x)=px+q, a linear polynomial.

Since f(x)=px+qf(x)=px+q is linear with p≠0p\neq0, it is one-one (injective): different xx give different f(x)f(x).

The actual outputs given are 1,4,7,10{1,4,7,10}, so the range of ff is 1,4,7,10{1,4,7,10}.

The codomain is the set of natural numbers N\mathbb{N}. For ff to be onto, its range must equal its codomain, but here:

1,4,7,10≠N{1,4,7,10}\neq\mathbb{N}

∴ f is one-one but not onto.

Hence, the correct answer is Option 1.

Consider the following for the two (02) items that follow :

Let the function f(x) = x2- 1

89

What is lim⁡x→1f∘f(x)\lim_{x \to 1} {f \circ f(x)} equal to?

  1. ((a))

    -1

  2. ((b))

    0

  3. ((c))

    1

  4. ((d))

    2

Show Answer
Answer: ((a))

-1

Calculation:

Given,

The function is f(x)=x2−1 f(x) = x^2 - 1 and we are tasked with finding lim⁡x→1(f∘f(x)) \lim_{x \to 1} (f \circ f(x)) .

f∘f(x)=f(f(x)) f \circ f(x) = f(f(x))

Given f(x)=x2−1 f(x) = x^2 - 1 , we substitute f(x) f(x) into itself:

f(f(x))=f(x2−1) f(f(x)) = f(x^2 - 1)

Now, apply the function f f to (x2−1) (x^2 - 1) :

f(x2−1)=(x2−1)2−1 f(x^2 - 1) = (x^2 - 1)^2 - 1

This simplifies to:

f(x2−1)=x4−2x2+1−1=x4−2x2 f(x^2 - 1) = x^4 - 2x^2 + 1 - 1 = x^4 - 2x^2

Thus, f∘f(x)=x4−2x2 f \circ f(x) = x^4 - 2x^2 .

Now, we need to find lim⁡x→1(x4−2x2) \lim_{x \to 1} (x^4 - 2x^2) .

Substitute x = 1 into the function:

f∘f(1)=14−2(12)=1−2=−1 f \circ f(1) = 1^4 - 2(1^2) = 1 - 2 = -1

∴ The value of lim⁡x→1(f∘f(x)) \lim_{x \to 1} (f \circ f(x)) is -1.

The correct answer is Option (1).

90

What is the area bounded by the function f(x) and the x-axis?

  1. ((a))

    1/3 square unit

  2. ((b))

    2/3 square unit

  3. ((c))

    4/3 square units

  4. ((d))

    2 square units

Show Answer
Answer: ((c))

4/3 square units

Calculation:

 

 

The function is y=x2−1 y = x^2 - 1 , and we need to find the area bounded by the curve and the x-axis between x = -1  and x = 1 .

The required area is given by the definite integral of the function from  -1  to 1 :

Area=∫−11(x2−1),dx \text{Area} = \int_{-1}^{1} (x^2 - 1) , dx

∫(x2−1),dx=x33−x \int (x^2 - 1) , dx = \frac{x^3}{3} - x

Evaluate the integral from  -1 to  1 :

[x33−x]−11=(133−1)−((−1)33−(−1)) \left[\frac{x^3}{3} - x\right]_{-1}^{1} = \left(\frac{1^3}{3} - 1\right) - \left(\frac{(-1)^3}{3} - (-1)\right)

=(13−1)−(−13+1)=(13−33)−(−13+33)=23−23=43 = \left(\frac{1}{3} - 1\right) - \left(\frac{-1}{3} + 1\right) = \left(\frac{1}{3} - \frac{3}{3}\right) - \left(\frac{-1}{3} + \frac{3}{3}\right) = \frac{2}{3} - \frac{2}{3} = \frac{4}{3}

∴ The area is 43 \frac{4}{3} square units.

Hence, the correct answer is Option 3.

Consider the following for the two (02) items that follow:

Let  y=sin⁡−1(x−4x327).y = \sin^{-1} \left( x - \frac{4x^3}{27} \right).

91

What is y equal to?

  1. ((a))

    sin-1x

  2. ((b))

    sin-1 x3 \frac{x}{3}

  3. ((c))

    3sin-1x

  4. ((d))

    3sin-1(x3)(\frac{x}{3})

Show Answer
Answer: ((d))

3sin-1(x3)(\frac{x}{3})

Calculation:

Given,

The function is y=sin⁡−1(x−4x327) y = \sin^{-1} \left( x - \frac{4x^3}{27} \right) , and we are tasked with simplifying it.

y=sin⁡−1(x−4x327) y = \sin^{-1} \left( x - \frac{4x^3}{27} \right)

We recognize that we can factor the expression as:

x−4x327=(x3)(3−4(x3)3) x - \frac{4x^3}{27} = \left( \frac{x}{3} \right) \left( 3 - 4 \left( \frac{x}{3} \right)^3 \right)

Thus, the function becomes:

y=sin⁡−1(x3(3−4(x3)3)) y = \sin^{-1} \left( \frac{x}{3} \left( 3 - 4 \left( \frac{x}{3} \right)^3 \right) \right)

Recognizing that this fits a known identity for inverse trigonometric functions:

sin⁡−1(x)=3sin⁡−1(x3) \sin^{-1}(x) = 3 \sin^{-1} \left( \frac{x}{3} \right)

Thus, we can simplify the expression to:

y=3sin⁡−1(x3) y = 3 \sin^{-1} \left( \frac{x}{3} \right)

Hence, the correct answer is Option 4.

92

What is  dydx\frac{dy}{dx} euqal to ?

  1. ((a))

    19−x2\frac{1}{\sqrt{9-x^2}}

  2. ((b))

    13−x2\frac{1}{\sqrt{3-x^2}}

  3. ((c))

    39−x2\frac{3}{\sqrt{9-x^2}}

  4. ((d))

    99−x2\frac{9}{\sqrt{9-x^2}}

Show Answer
Answer: ((c))

39−x2\frac{3}{\sqrt{9-x^2}}

Calculation:

Given,

The function is y=3sin⁡−1(x3) y = 3 \sin^{-1} \left( \frac{x}{3} \right) , and we are tasked with finding the derivative of y , i.e., dydx \frac{dy}{dx} .

The general derivative of sin⁡−1(u) \sin^{-1}(u) is:

ddxsin⁡−1(u)=11−u2⋅dudx \frac{d}{dx} \sin^{-1}(u) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} , whereu=x3 u = \frac{x}{3}

dudx=13 \frac{du}{dx} = \frac{1}{3}

Apply the chain rule to find the derivative of y :

dydx=3×11−(x3)2×13 \frac{dy}{dx} = 3 \times \frac{1}{\sqrt{1 - \left( \frac{x}{3} \right)^2}} \times \frac{1}{3}

dydx=11−x29=39−x2 \frac{dy}{dx} = \frac{1}{\sqrt{1 - \frac{x^2}{9}}} = \frac{3}{\sqrt{9 - x^2}}

Hence, the correct answer is Option 3.

Consider the following for the two (02) items that follow:

Let the function f(x) = x 2 + 9

93

What is  lim⁡x→0f(x)−3f(x)+7−4\lim_{x \to 0} \frac{\sqrt{f(x)} - 3}{\sqrt{f(x)+7} - 4} equal to?

  1. ((a))

    2/3

  2. ((b))

    1

  3. ((c))

    4/3

  4. ((d))

    2

Show Answer
Answer: ((c))

4/3

Calculation:

Given,

The function is f(x)=x2+9−3 f(x) = \sqrt{x^2 + 9} - 3 and g(x)=x2+16−4 g(x) = \sqrt{x^2 + 16} - 4 .

We are tasked with finding:

lim⁡x→0f(x)g(x) \lim_{x \to 0} \frac{f(x)}{g(x)}

Multiply both the numerator and denominator by their respective conjugates:

x2+9−3x2+16−4×x2+9+3x2+9+3×x2+16+4x2+16+4 \frac{\sqrt{x^2 + 9} - 3}{\sqrt{x^2 + 16} - 4} \times \frac{\sqrt{x^2 + 9} + 3}{\sqrt{x^2 + 9} + 3} \times \frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 16} + 4}

Simplify the numerator:

(x2+9−3)(x2+9+3)=x2 (\sqrt{x^2 + 9} - 3)(\sqrt{x^2 + 9} + 3) = x^2

Simplify the denominator:

(x2+16−4)(x2+16+4)=x2 (\sqrt{x^2 + 16} - 4)(\sqrt{x^2 + 16} + 4) = x^2

Now, the expression becomes:

x2x2×x2+16+4x2+9+3 \frac{x^2}{x^2} \times \frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3}

Simplify and evaluate the limit:

x2+16+4x2+9+3 \frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3} becomes:

16+49+3=4+43+3=86=43 \frac{\sqrt{16} + 4}{\sqrt{9} + 3} = \frac{4 + 4}{3 + 3} = \frac{8}{6} = \frac{4}{3}

Hence, the correct answer is Option 3.

94

Consider the following statements:

I. f(x) is an increasing function.

II. f(x) has local maximum at x = 0

Which of the statements given above is/are correct?

  1. ((a))

    I only

  2. ((b))

    II only

  3. ((c))

    Both I and II

  4. ((d))

    Neither I nor II

Show Answer
Answer: ((d))

Neither I nor II

Calculation:

Given,

The function is f(x)=x2+9 f(x) = x^2 + 9 .

 

Statement I: f(x) is an increasing function.

The derivative of f(x) is:

f′(x)=ddx(x2+9)=2x f'(x) = \frac{d}{dx}(x^2 + 9) = 2x

When (x>0)( x > 0 ), (f′(x)>0)( f'(x) > 0 ), so (f(x) is increasing.

When (x < 0), f'(x) < 0 ), so f(x)  is decreasing.

At (x = 0), ( f'(x) = 0 ), meaning the function is neither increasing nor decreasing at this point.

Hence, f(x)  is not entirely increasing. It is increasing for (x > 0) and decreasing for ( x < 0).

Statement II: f(x) has local maximum at x = 0

Since the functionf(x)=x2+9 f(x) = x^2 + 9 is a parabola opening upwards (because the coefficient of x2 is positive), it has a global minimum at x = 0, not a local maximum.

Conclusion:

  • Statement I is incorrect because the function is not entirely increasing. It is increasing for x > 0  and decreasing for  x < 0 .
  • Statement II is incorrect because the function has a global minimum at x = 0, not a local maximum.

Hence, the correct answer is Option 4.

Consider the following for the two (02) items that follow:

The function f(x) satisfies  f(x),f(xy)=f(x)f(y).f(x), f \left( \frac{x}{y} \right) = \frac{f(x)}{f(y)}.  ​for all positive real values of x and y, and f(2) = 3

95

What is f(16) equal to?

  1. ((a))

    18

  2. ((b))

    27

  3. ((c))

    54

  4. ((d))

    81

Show Answer
Answer: ((d))

81

Calculation:

Given,

The function satisfies the equation f(x)f(y)=xy \frac{f(x)}{f(y)} = \frac{x}{y} for all positive real values of x and y, and f(2)=3 f(2) = 3 .

We are tasked with finding f(16) f(16) .

Using the functional equation, for f(4) f(4) , we have:

f(4)f(2)=f(42)=f(2) \frac{f(4)}{f(2)} = f\left( \frac{4}{2} \right) = f(2)

Since f(2)=3 f(2) = 3 , we can calculate f(4) f(4)

f(4)3=3⇒f(4)=9 \frac{f(4)}{3} = 3 \quad \Rightarrow \quad f(4) = 9

Next, to find f(16) f(16) , we use the functional equation again:

f(16)f(4)=f(164)=f(4) \frac{f(16)}{f(4)} = f\left( \frac{16}{4} \right) = f(4)

Since f(4)=9 f(4) = 9 , we can calculate f(16) f(16)

f(16)9=9⇒f(16)=81 \frac{f(16)}{9} = 9 \quad \Rightarrow \quad f(16) = 81

Hence, the correct answer is Option 4.

96

What is f(1)f(4) equal to?

  1. ((a))

    4

  2. ((b))

    8

  3. ((c))

    9

  4. ((d))

    18

Show Answer
Answer: ((c))

9

Calculation:

Given,

The function satisfies the functional equation:

f(xy)=f(x)f(y) f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)}

Also, we are given that:

f(2)=3 f(2) = 3

We use the functional equation to calculate f(4) . Substituting x = 4 and y = 2 , we get:

f(4)f(2)=f(2) \frac{f(4)}{f(2)} = f(2)

f(4)3=3⇒f(4)=3×3=9 \frac{f(4)}{3} = 3 \quad \Rightarrow f(4) = 3 \times 3 = 9

f(2)f(2)=f(1)⇒f(1)=1 \frac{f(2)}{f(2)} = f(1) \quad \Rightarrow f(1) = 1

f(1)×f(4)=1×9=9 f(1) \times f(4) = 1 \times 9 = 9

Hence, the correct answer is Option 3.

Consider the following for the two (02) items that follow:

A function f is such that f(xy) = f(x + y) for all real values of x and y, and f(5) = 10

97

What is f(0) equal to?

  1. ((a))

    0

  2. ((b))

    1

  3. ((c))

    5

  4. ((d))

    10

Show Answer
Answer: ((d))

10

Calculation:

Given,

The function is such that: f(xy)=f(x+y) f(xy) = f(x + y) for all real values of x  and y , and f(5)=10 f(5) = 10 .

We need to find the value of f(0) f(0) .

Substitute x=5 x = 5 and y=0 y = 0 in the given functional equation f(xy)=f(x+y) f(xy) = f(x + y) :

f(5⋅0)=f(5+0) f(5 \cdot 0) = f(5 + 0)

f(0)=f(5) f(0) = f(5)

Since f(5)=10 f(5) = 10 , we conclude:

f(0)=10 f(0) = 10

Hence, the correct answer is Option 4.

98

What is f(20)+f(-20) equal to?

  1. ((a))

    0

  2. ((b))

    10

  3. ((c))

    20

  4. ((d))

    40

Show Answer
Answer: ((c))

20

Calculation:

Given,

The function is f(xy)=f(x+y) f(xy) = f(x + y) for all real values of x  and y, and f(5) = 10 .

We are tasked with finding:

f(20)+f(−20) f(20) + f(-20)

Using the given functional equation, we have:

For f(0⋅5)=f(0+5) f(0 \cdot 5) = f(0 + 5) , we get:

f(0)=f(5)=10 f(0) = f(5) = 10

For f(0⋅20)=f(0+20) f(0 \cdot 20) = f(0 + 20) , we get:

f(0)=f(20)=10 f(0) = f(20) = 10

For f(0⋅−20)=f(0+(−20)) f(0 \cdot -20) = f(0 + (-20)) , we get:

f(0)=f(−20)=10 f(0) = f(-20) = 10

Thus,

f(20)+f(−20)=10+10=20 f(20) + f(-20) = 10 + 10 = 20

Hence, the correct answert is Option 3.

Consider the following for the two (02) items that follow:

Let f(x) = [x2] where [.] is the greatest integer function.

99

What  ∫23f(x)dx\int_{\sqrt{2}}^{\sqrt{3}} f(x) dx equal to?

  1. ((a))

    3−2\sqrt{3}-\sqrt{2}

  2. ((b))

    2(3−2)2(\sqrt{3}-\sqrt{2})

  3. ((c))

    3−23-\sqrt{2}

  4. ((d))

    1

Show Answer
Answer: ((b))

2(3−2)2(\sqrt{3}-\sqrt{2})

Calculation:

Given,

The function is f(x)=⌊x2⌋ f(x) = \left\lfloor x^2 \right\rfloor , where ⌊x2⌋ \left\lfloor x^2 \right\rfloor is the greatest integer function.

We are tasked with finding:

∫323⌊x2⌋,dx \int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}} \left\lfloor x^2 \right\rfloor , dx

Decomposing the integral based on the function:

For 32≤x<1\frac{\sqrt{3}}{2} \leq x < 1 , x2 lies between 34\frac{3}{4} and 1, so ⌊x2⌋=0\left\lfloor x^2 \right\rfloor = 0 . Therefore,

∫3210,dx=0 \int_{\frac{\sqrt{3}}{2}}^1 0 , dx = 0

For1≤x<2 1 \leq x < \sqrt{2} , x2  lies between 1 and 2, so ⌊x2⌋=1.\left\lfloor x^2 \right\rfloor = 1 . Therefore,

∫121,dx=2−1 \int_1^{\sqrt{2}} 1 , dx = \sqrt{2} - 1

For 2≤x<3\sqrt{2} \leq x < \sqrt{3} , x2  lies between 2 and 3, so ⌊x2⌋=2\left\lfloor x^2 \right\rfloor = 2 . Therefore,

∫232,dx=2(3−2) \int_{\sqrt{2}}^{\sqrt{3}} 2 , dx = 2(\sqrt{3} - \sqrt{2})

Summing up all the results:

∫323⌊x2⌋,dx=0+(2−1)+2(3−2) \int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}} \left\lfloor x^2 \right\rfloor , dx = 0 + (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2})

= 2(3−2) 2(\sqrt{3} - \sqrt{2}) .

Hence, the correct answer is Option 2.

100

∫22f(x)dx\int_{\sqrt{2}}^{2} f(x) dx  is equal to ?

  1. ((a))

    6−3−226-\sqrt{3}-2\sqrt{2}

  2. ((b))

    6−3−26-\sqrt{3}-\sqrt{2}

  3. ((c))

    6−3+226-\sqrt{3}+2\sqrt{2}

  4. ((d))

    6+3−226+\sqrt{3}-2\sqrt{2}

Show Answer
Answer: ((a))

6−3−226-\sqrt{3}-2\sqrt{2}

Calculation:

Given,

The function is f(x)=⌊x2⌋ f(x) = \lfloor x^2 \rfloor .

We are tasked with finding the value of ∫22f(x)dx \int_{\sqrt{2}}^{2} f(x) dx .

We can break the integral into two parts as follows:

∫22f(x)dx=∫232dx+∫323dx \int_{\sqrt{2}}^{2} f(x) dx = \int_{\sqrt{2}}^{\sqrt{3}} 2 dx + \int_{\sqrt{3}}^{2} 3 dx

For the range 2≤x≤3 \sqrt{2} \leq x \leq \sqrt{3} , ⌊x2⌋=2 \lfloor x^2 \rfloor = 2 . So the first part of the integral is:

∫232dx=2×(3−2) \int_{\sqrt{2}}^{\sqrt{3}} 2 dx = 2 \times (\sqrt{3} - \sqrt{2})

For the range 3≤x≤2 \sqrt{3} \leq x \leq 2 , ⌊x2⌋=3 \lfloor x^2 \rfloor = 3 . So the second part of the integral is:

∫323dx=3×(2−3) \int_{\sqrt{3}}^{2} 3 dx = 3 \times (2 - \sqrt{3})

Now, we calculate the values:

2×(3−2)=23−22 2 \times (\sqrt{3} - \sqrt{2}) = 2\sqrt{3} - 2\sqrt{2}

3×(2−3)=6−33 3 \times (2 - \sqrt{3}) = 6 - 3\sqrt{3}

Combining the two parts:

23−22+6−33=6−22−3 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3} = 6 - 2\sqrt{2} - \sqrt{3}

Hence, the correct answer is Option 1.

Consider the following for the four (04) items that follow:

The frequency distribution of height of students of a class is given below:

Height (in cmNumber of Students
160-16212
162-16415
164-16624
166-16813
101

What is the total number of students whose height is less than or equal to 165 cm?

  1. ((a))

    15

  2. ((b))

    39

  3. ((c))

    51

  4. ((d))

    None of the above 

Show Answer
Answer: ((c))

51

Calculation:

Given,

The frequency distribution of the height of students is given as follows:

Height (cm)Number of Students
160-16212
162-16415
164-16624
166-16813

We need to find the total number of students whose height is less than or equal to 165 cm. Therefore, we will sum the students in the following height ranges:

160-162 cm: 12 students

162-164 cm: 15 students

164-166 cm: 24 students

The total number of students is: 12 + 15 + 24 = 51 

Hence, the correcr answer is Option 3

102

What is the height of the class?

  1. ((a))

    162.41 cm 

  2. ((b))

    163.41 cm 

  3. ((c))

    164.41 cm 

  4. ((d))

    165.41 cm 

Show Answer
Answer: ((c))

164.41 cm 

Calculation:

Given,

The frequency distribution of the height of students in the class is as follows:

Height (in cm) | Number of Students

Height (in cmNumber of students
160–16212
162–16415
164–16624
166–16813

 

Calculate the cumulative frequency.

Cumulative frequency for the 160–162 cm range: 12

Cumulative frequency for the 162–164 cm range: 12 + 15 = 27

Cumulative frequency for the 164–166 cm range: 27 + 24 = 51

Cumulative frequency for the 166–168 cm range: 51 + 13 = 64

The total number of students is 64, so the median class contains the 32nd student. The cumulative frequency before 32 is 27 (for the 162–164 cm class), and after 32 is 51 (for the 164–166 cm class). Therefore, the median class is 164–166 cm.

The formula for finding the median in a grouped frequency distribution is:

Median=L+(N2−Ff)×h \text{Median} = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times h

L = lower boundary of the median class = 164

N = total number of students = 64

F = cumulative frequency before the median class = 27

f = frequency of the median class = 24

h = class width = 2

Substituting the values:

Median=164+(32−2724)×2=164+(524)×2=164+0.42=164.42 \text{Median} = 164 + \left( \frac{32 - 27}{24} \right) \times 2 = 164 + \left( \frac{5}{24} \right) \times 2 = 164 + 0.42 = 164.42

∴ The median height is approximately 164.41 cm.

Hence, the correct answer is Option 3.

103

The height which occurs most frequently in the class is

  1. ((a))

    163.5 cm 

  2. ((b))

    163.9 cm 

  3. ((c))

    164.5 cm 

  4. ((d))

    164.9 cm 

Show Answer
Answer: ((d))

164.9 cm 

Calculation:

Given the frequency distribution of height of students:

Height (in cm)Number of Students
160–16212
162–16415
164–16624
166–16813

The class interval with the highest frequency is 164–166, with 24 students.

The formula for calculating the mode for grouped data is:

Mode=L+(f1−f02f1−f0−f2)×h \text{Mode} = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h

Where:

L: Lower boundary of the modal class = 164

f1: Frequency of the modal class = 24

f0: Frequency of the class before the modal class = 15

f2: Frequency of the class after the modal class = 13

h: Class width = 2

Substituting the values into the formula:

Mode=164+(24−152(24)−15−13)×2 \text{Mode} = 164 + \left( \frac{24 - 15}{2(24) - 15 - 13} \right) \times 2

=164+(948−28)×2 = 164 + \left( \frac{9}{48 - 28} \right) \times 2

=164+(920)×2 = 164 + \left( \frac{9}{20} \right) \times 2

=164+0.9 = 164 + 0.9

=164.9 = 164.9

∴ The mode is approximately 164.9 cm.

Hence, the correct answer is Option 4.

104

The most appropriate graphical representation of the given frequency distribution is

  1. ((a))

    bar chart

  2. ((b))

    percentage bar chart

  3. ((c))

    histogram

  4. ((d))

    pie chart

Show Answer
Answer: ((c))

histogram

Solution:

The appropriate graphical representation for the given frequency distribution is a Histogram. The bar chart above shows the frequency of students' heights within the specified ranges. The highest frequency is for the height range of 164-166 cm, which has 24 students.

Consider the following for the two (02) items that follow:

The sum and the sum of squares of the observations corresponding to length X (in cm) and weight Y (in gm) of 50 tropical tubers are given as \(\Sigma X = 200$, \Sigma Y = 250, \Sigma X^2 = 900 and \Sigma Y^2 = 1400\)

105

Which one of the following is correct?

  1. ((a))

    Variance (X) > Variance (Y)

  2. ((b))

    Variance (X) < Variance (Y)

  3. ((c))

    Variance (X) = Variance (Y)

  4. ((d))

    Cannot be determined from the given data

Show Answer
Answer: ((b))

Variance (X) < Variance (Y)

Calculation:

Given,

The sum and sum of squares of the observations corresponding to length X in cm and weight Y in gm of 50 tropical tubers are given as:

ΣX\Sigma X = 200, ΣY\Sigma Y = 250, ΣX2\Sigma X^2= 900, ΣY2\Sigma Y^2 = 1400 

The formula for variance is:

Variance=ΣX2N−(ΣXN)2 \text{Variance} = \frac{\Sigma X^2}{N} - \left( \frac{\Sigma X}{N} \right)^2

Where N = 50 is the number of observations.

Variance of X:

Variance of X=ΣX2N−(ΣXN)2 \text{Variance of } X = \frac{\Sigma X^2}{N} - \left( \frac{\Sigma X}{N} \right)^2

Substituting the given values:

Variance of X=90050−(20050)2 \text{Variance of } X = \frac{900}{50} - \left( \frac{200}{50} \right)^2

Variance of X=18−16=2 \text{Variance of } X = 18 - 16 = 2

Variance of Y:

Variance of Y=ΣY2N−(ΣYN)2 \text{Variance of } Y = \frac{\Sigma Y^2}{N} - \left( \frac{\Sigma Y}{N} \right)^2

Substituting the given values:

Variance of Y=140050−(25050)2 \text{Variance of } Y = \frac{1400}{50} - \left( \frac{250}{50} \right)^2

Variance of Y=28−25=3 \text{Variance of } Y = 28 - 25 = 3

Conclusion:

Variance of X<Variance of Y \text{Variance of } X < \text{Variance of } Y

Hence, the correct answer is Option 2.

106

Which one of the following statements is correct?

  1. ((a))

    Coefficient of variation of X is strictly more than coefficient of variation of Y.

  2. ((b))

    Coefficient of variation of X is strictly less than coefficient of variation of Y

  3. ((c))

    Coefficient of variation of X is same as coefficient of variation of Y.

  4. ((d))

    Coefficient of variation cannot be determined from the given data.

Show Answer
Answer: ((b))

Coefficient of variation of X is strictly less than coefficient of variation of Y

Calculation:

Explanation:

The Coefficient of Variation (C.V.) formula is given as:

C.V.(X)=σXMX×100 C.V.(X) = \frac{\sigma_X}{M_X} \times 100

Substituting values:

C.V.(X)=250×100=22 C.V.(X) = \frac{\sqrt{2}}{50} \times 100 = 2\sqrt{2}

The Coefficient of Variation for Y is:

C.V.(Y)=350×100=23 C.V.(Y) = \frac{\sqrt{3}}{50} \times 100 = 2\sqrt{3}

∴ C.V.(X) < C.V.(Y)

Hence, the correct answer is Option 2.

Consider the following for the two (02) items that follow:

Let X be a random variable following binomial distribution with parameters n = 6 and p = k Further, 9P(X = 4) = P(X = 2) .

107

What is the value of k?

  1. ((a))

    1/2

  2. ((b))

    1/3

  3. ((c))

    1/4

  4. ((d))

    1/5

Show Answer
Answer: ((c))

1/4

Calculation:

Given,

Let X be a random variable following a binomial distribution with parameters n = 6 and p = k.

Further, it is given that:

9P(X=4)=P(X=2) 9P(X = 4) = P(X = 2)

The probability mass function for a binomial distribution is:

P(X=x)=(nx)px(1−p)n−x P(X = x) = \binom{n}{x} p^x (1 - p)^{n - x}

For P(X = 4) , we have:

P(X=4)=(64)k4(1−k)2=15k4(1−k)2 P(X = 4) = \binom{6}{4} k^4 (1 - k)^2 = 15 k^4 (1 - k)^2

For P(X = 2) , we have:

P(X=2)=(62)k2(1−k)4=15k2(1−k)4 P(X = 2) = \binom{6}{2} k^2 (1 - k)^4 = 15 k^2 (1 - k)^4

We are given that:

9P(X=4)=P(X=2) 9P(X = 4) = P(X = 2)

Substitute the expressions for P(X = 4)  and P(X = 2) :

9×15k4(1−k)2=15k2(1−k)4 9 \times 15 k^4 (1 - k)^2 = 15 k^2 (1 - k)^4

Cancel the common factor of 15:

9k4(1−k)2=k2(1−k)4 9 k^4 (1 - k)^2 = k^2 (1 - k)^4

9k2=(1−k)2 9 k^2 = (1 - k)^2

9k2=1−2k+k2 9 k^2 = 1 - 2k + k^2

 

8k2+2k−1=0 8 k^2 + 2k - 1 = 0

Use the quadratic formula to solve for k :

k=−2±22−4×8×(−1)2×8 k = \frac{-2 \pm \sqrt{2^2 - 4 \times 8 \times (-1)}}{2 \times 8}

k=−2±4+3216 k = \frac{-2 \pm \sqrt{4 + 32}}{16}

k=−2±3616 k = \frac{-2 \pm \sqrt{36}}{16}

k=−2±616 k = \frac{-2 \pm 6}{16}

Thus, the two possible values for k are:

k=416=14 k = \frac{4}{16} = \frac{1}{4} or k=−816=−12 k = \frac{-8}{16} = -\frac{1}{2}

Since k  represents a probability, it must be between 0 and 1. Therefore, the valid solution is:

k=14 k = \frac{1}{4}

Hence, the correct answer is Option 3.

108

What is the value of P (X = 3) 

  1. ((a))

    135/1024

  2. ((b))

    5/128

  3. ((c))

    45/1024

  4. ((d))

    70/1024

Show Answer
Answer: ((a))

135/1024

Calculation:

Given,

Let X  be a random variable following a binomial distribution with parameters n = 6 and p = 14\frac{1}{4}

The probability mass function for a binomial distribution is:

P(X=x)=(nx)px(1−p)n−x P(X = x) = \binom{n}{x} p^x (1 - p)^{n - x}

For P(X = 3) , we have:

P(X=3)=(63)(14)3(34)3 P(X = 3) = \binom{6}{3} \left( \frac{1}{4} \right)^3 \left( \frac{3}{4} \right)^3

Substitute the values:

P(X=3)=20×164×2764 P(X = 3) = 20 \times \frac{1}{64} \times \frac{27}{64}

Now simplify the equation:

P(X=3)=20×274096=5404096 P(X = 3) = 20 \times \frac{27}{4096} = \frac{540}{4096}

Simplifying the fraction:

P(X=3)=1351024 P(X = 3) = \frac{135}{1024}

Hence, the correct answer is Option 1

Consider the following for the two (02) items that follow:

A committee of 6 members is formed from a group of 7 gentlemen and 4 ladies.

109

What is the probability that the committee includes exactly 3 gentlemen?

  1. ((a))

    10/33

  2. ((b))

    30/77

  3. ((c))

    100/231

  4. ((d))

    5/11

Show Answer
Answer: ((a))

10/33

Calculation:

Given,

A committee of 6 members is to be formed from a group of 7 gentlemen and 4 ladies.

The total number of ways to form the committee of 6 members is:

(116)=11×10×9×8×7×66×5×4×3×2×1=462 \binom{11}{6} = \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 462

The number of ways to form a committee with exactly 3 gentlemen and 3 ladies is:

(73)×(43)=35×4=140 \binom{7}{3} \times \binom{4}{3} = 35 \times 4 = 140

The probability of forming a committee with exactly 3 gentlemen is:

Probability=140462=1033 \text{Probability} = \frac{140}{462} = \frac{10}{33}

Hence, the correct answer is Option 1.

110

What is the probability that the committee includes at least 2 ladies?

  1. ((a))

    41/66

  2. ((b))

    47/66

  3. ((c))

    49/66

  4. ((d))

    53/66

Show Answer
Answer: ((d))

53/66

Calculation:

Given,

A committee of 6 members is formed from a group of 7 gentlemen and 4 ladies.

The total number of ways to form the committee is:

(116)=462 \binom{11}{6} = 462

The favorable outcomes (at least 2 ladies) are calculated by the sum of the cases:

(42)×(74)+(43)×(73)+(44)×(72) \binom{4}{2} \times \binom{7}{4} + \binom{4}{3} \times \binom{7}{3} + \binom{4}{4} \times \binom{7}{2}

This simplifies to:

(42)×(74)=6×35=210 \binom{4}{2} \times \binom{7}{4} = 6 \times 35 = 210

(43)×(73)=4×35=140 \binom{4}{3} \times \binom{7}{3} = 4 \times 35 = 140

(44)×(72)=1×21=21 \binom{4}{4} \times \binom{7}{2} = 1 \times 21 = 21

The total favorable outcomes are:

210+140+21=371 210 + 140 + 21 = 371

The probability is:

P(E)=371462=5366 P(E) = \frac{371}{462} = \frac{53}{66}

Hence, the correct answer is Option 4.

Consider the following for the two (02) items that follow:

The probabilities that A, B and C become managers are 3/10 1/2 and 4/5 respectively. The probabilities that bonus scheme will be introduced if A, B and C become managers are 4/9, 2/9 and 1/3 respectively.

111

What is the probability that the bonus scheme will be introduced?

  1. ((a))

    17/45

  2. ((b))

    19/45

  3. ((c))

    23/45

  4. ((d))

    26/45

Show Answer
Answer: ((c))

23/45

Calculation:

Given,

The probability that A , B , and C  become managers are:

P(A)=310,,P(B)=12,,P(C)=45 P(A) = \frac{3}{10}, , P(B) = \frac{1}{2}, , P(C) = \frac{4}{5}

The conditional probabilities that the bonus scheme is introduced are:

P(D∣A)=49,,P(D∣B)=29,,P(D∣C)=13 P(D|A) = \frac{4}{9}, , P(D|B) = \frac{2}{9}, , P(D|C) = \frac{1}{3}

The total probability that the bonus scheme will be introduced is given by:

P(D)=P(D∣A)P(A)+P(D∣B)P(B)+P(D∣C)P(C) P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C)

Substituting the values:

P(D)=(49×310)+(29×12)+(13×45) P(D) = \left(\frac{4}{9} \times \frac{3}{10}\right) + \left(\frac{2}{9} \times \frac{1}{2}\right) + \left(\frac{1}{3} \times \frac{4}{5}\right)

Now, simplifying the terms:

P(D)=1290+218+415 P(D) = \frac{12}{90} + \frac{2}{18} + \frac{4}{15}

The least common denominator (LCD) is 90. So:

P(D)=1290+1090+2490 P(D) = \frac{12}{90} + \frac{10}{90} + \frac{24}{90}

Adding the fractions gives:

P(D)=4690=2345 P(D) = \frac{46}{90} = \frac{23}{45}

Hence, the correct answer is Option 3.

112

If the bonus scheme has been introduced, then what is the probability that the manager appointed was B?

  1. ((a))

    5/23

  2. ((b))

    6/23

  3. ((c))

     7/23

  4. ((d))

    8/23

Show Answer
Answer: ((a))

5/23

Calculation:

Given,

The probability that A , B , and C  become managers are:

P(A)=310,,P(B)=12,,P(C)=45 P(A) = \frac{3}{10}, , P(B) = \frac{1}{2}, , P(C) = \frac{4}{5}

The conditional probabilities that the bonus scheme is introduced are:

P(D∣A)=49,,P(D∣B)=29,,P(D∣C)=13 P(D|A) = \frac{4}{9}, , P(D|B) = \frac{2}{9}, , P(D|C) = \frac{1}{3}

We need to find the probability that B  is the manager given that the bonus scheme has been introduced using Bayes' Theorem:

P(B∣D)=P(D∣B)P(B)P(D) P(B|D) = \frac{P(D|B)P(B)}{P(D)}

First, calculate the total probability P(D)  that the bonus scheme is introduced:

P(D)=P(D∣A)P(A)+P(D∣B)P(B)+P(D∣C)P(C) P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C)

Substitute the values:

P(D)=(49×310)+(29×12)+(13×45) P(D) = \left( \frac{4}{9} \times \frac{3}{10} \right) + \left( \frac{2}{9} \times \frac{1}{2} \right) + \left( \frac{1}{3} \times \frac{4}{5} \right)

Simplifying:

P(D)=1290+218+415 P(D) = \frac{12}{90} + \frac{2}{18} + \frac{4}{15}

Find a common denominator (LCD = 90):

P(D)=1290+1090+2490=4690=2345 P(D) = \frac{12}{90} + \frac{10}{90} + \frac{24}{90} = \frac{46}{90} = \frac{23}{45}

Now, use Bayes' Theorem to find P(B|D) :

P(B∣D)=P(D∣B)P(B)P(D)=(29×12)2345 P(B|D) = \frac{P(D|B)P(B)}{P(D)} = \frac{\left( \frac{2}{9} \times \frac{1}{2} \right)}{\frac{23}{45}}

Simplifying:

P(B∣D)=2182345=192345=523 P(B|D) = \frac{\frac{2}{18}}{\frac{23}{45}} = \frac{\frac{1}{9}}{\frac{23}{45}} = \frac{5}{23}

The probability that the manager appointed was B , given that the bonus scheme has been introduced, is 5/23.

Hence, the correct answer is Option 1.

113

The arithmetic mean of 100 observations is 50. If 5 is subtracted from each observation and then divided by 20, then what is the new arithmetic mean?

  1. ((a))

    2.25

  2. ((b))

    3.5

  3. ((c))

    4.25

  4. ((d))

    5.5

Show Answer
Answer: ((a))

2.25

Calculation:

Given,

Number of observations, n = 100

Original arithmetic mean, xˉ=50 \bar{x} = 50

Transformation applied to each observation:

New value, x′=x−520 x' = \frac{x - 5}{20}

The new arithmetic mean is,

xˉ′=xˉ−520 \bar{x}' = \frac{\bar{x} - 5}{20}

xˉ′=50−520=4520=2.25 \bar{x}' = \frac{50 - 5}{20} = \frac{45}{20} = 2.25

∴ The new arithmetic mean is 2.25.

Hence, the correct answer is Option 1.

114

The standard deviation of 100 observations is 10. If 5 is added to each observation and then divided by 20, then what will be the new standard deviation?

  1. ((a))

    0.25

  2. ((b))

    0.5

  3. ((c))

    0.75

  4. ((d))

    1.00

Show Answer
Answer: ((b))

0.5

Calculation:

Given,

Number of observations, n = 100

Original standard deviation, σ=10 \sigma = 10

Transformation applied to each observation:

New value, x′=x+520 x' = \frac{x + 5}{20}

The additive constant (5) does not affect the standard deviation, and scaling by 20 divides the standard deviation by 20, so the new standard deviation is,

σ′=σ20 \sigma' = \frac{\sigma}{20}

σ′=1020=0.5 \sigma' = \frac{10}{20} = 0.5

∴ The new standard deviation is 0.5.

Hence, the correct answer is Option 2.

115

If P(A)=1/3, P(B)=1/2 and P(A∩B)=1/4, then what is the value of P(B|AC)?

  1. ((a))

    1/8

  2. ((b))

    3/8

  3. ((c))

    5/8

  4. ((d))

    7/8

Show Answer
Answer: ((b))

3/8

Calculation:

Given,

P(A) = 1/3

P(B) = 1/2

P(A ∩ B) = 1/4

We need to calculate the probability P(B | AC), the probability that B occurs given that A does not occur.

Using the formula for conditional probability:

P(B | AC) = P(B ∩ AC) / P(AC)

Now,

P(AC) = 1 - P(A) = 1 - 1/3 = 2/3

P(B ∩ AC) = P(B) - P(A ∩ B) = 1/2 - 1/4 = 1/4

Substitute the values:

P(B | AC) = (1/4) / (2/3) = (1/4) × (3/2) = 3/8

Hence, the correct answer is Option 2.

116

If P(A)=1/3, P(B)=1/2 and P(A∩B)=1/4, then what is the value of P(AC∩BC)?

  1. ((a))

    1/4

  2. ((b))

     5/12

  3. ((c))

    7/12

  4. ((d))

    11/12

Show Answer
Answer: ((b))

 5/12

Calculation:

Given,

P(A)=13 P(A) = \frac{1}{3}

P(B)=12 P(B) = \frac{1}{2}

P(A∩B)=14 P(A \cap B) = \frac{1}{4}

We need to calculateP(AC∩BC) P(A^C \cap B^C) , the probability that neither A  nor B  occurs.

Using the complement rule:

P(AC∩BC)=1−P(A∪B) P(A^C \cap B^C) = 1 - P(A \cup B)

Now, apply the inclusion-exclusion formula to find (P(A∪B)( P(A \cup B) :

P(A∪B)=P(A)+P(B)−P(A∩B) P(A \cup B) = P(A) + P(B) - P(A \cap B)

Substitute the given values:

P(A∪B)=13+12−14 P(A \cup B) = \frac{1}{3} + \frac{1}{2} - \frac{1}{4}

To simplify:

P(A∪B)=412+612−312=712 P(A \cup B) = \frac{4}{12} + \frac{6}{12} - \frac{3}{12} = \frac{7}{12}

Now, calculate P(AC∩BC)P(A^C \cap B^C) :

P(AC∩BC)=1−712=512 P(A^C \cap B^C) = 1 - \frac{7}{12} = \frac{5}{12}

Hence, the correct answer is Option 2.

117

If two fair dice are tossed, then what is the probability that the sum of the numbers on the faces of the dice is strictly greater than 7?

  1. ((a))

    1/3

  2. ((b))

    5/12

  3. ((c))

    7/12

  4. ((d))

    3/4

Show Answer
Answer: ((b))

5/12

Calculation:

Given,

Two fair six-sided dice are rolled.

Each die has faces numbered 1 through 6.

Total number of outcomes:

Since each die has 6 faces, the sample space size is

N=6×6=36N = 6 \times 6 = 36.

Favourable outcomes (sum > 7):

We list all ordered pairs ((i, j)) with (i + j > 7):

Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes

Sum = 9: (3,6), (4,5), (5,4), (6,3) → 4 outcomes

Sum = 10: (4,6), (5,5), (6,4) → 3 outcomes

Sum = 11: (5,6), (6,5) → 2 outcomes

Sum = 12: (6,6) → 1 outcome

Total favourable outcomes = (5 + 4 + 3 + 2 + 1 = 15).

Probability calculation:

The probability that the sum is strictly greater than 7 is

P(sum>7)=favourable outcomestotal outcomes=1536=512P(\text{sum} > 7) = \frac{\text{favourable outcomes}}{\text{total outcomes}} = \frac{15}{36} = \frac{5}{12}.

Hence, the correct answer is Option 2.

118

The probability of a man hitting a target is 1/5. If the man fires 7 times, then what is the probability that he hits the target at least twice?

  1. ((a))

    1−(35)(45)61-(\frac{3}{5})(\frac{4}{5})^6

  2. ((b))

    1−(35)(45)71-(\frac{3}{5})(\frac{4}{5})^7

  3. ((c))

    1−(115)(45)61-(\frac{11}{5})(\frac{4}{5})^6

  4. ((d))

    1−(115)(45)71-(\frac{11}{5})(\frac{4}{5})^7

Show Answer
Answer: ((c))

1−(115)(45)61-(\frac{11}{5})(\frac{4}{5})^6

Calculation:

Given,

Probability of hitting the target per shot, p=15p = \frac{1}{5}

Number of shots fired, n=7n = 7

The number of hits follows a binomial distribution:

X∼Binomial(n=7,;p=15)X \sim \mathrm{Binomial}(n=7,;p=\tfrac15)

Probability of zero hits:

P(X=0)=(45)7P(X=0)=\left(\frac45\right)^{7}

Probability of exactly one hit:

P(X=1)=(71)!(15)!(45)6=75(45)6P(X=1)=\binom71!\left(\frac15\right)!\left(\frac45\right)^{6} =\frac{7}{5}\left(\frac45\right)^{6}

Probability of at least two hits:

P(X≥2)=1−[P(X=0)+P(X=1)]=1−115(45)6P(X\ge2)=1-\bigl[P(X=0)+P(X=1)\bigr] =1-\frac{11}{5}\left(\frac45\right)^{6}

∴ The probability of hitting the target at least twice is  1−115(45)61-\dfrac{11}{5}\left(\dfrac45\right)^{6} .

Hence, the correct answer is Option 3.

119

Let X be a random variable following binomial distribution whose mean and variance are 200 and 160 respectively. What is the value of the number of trials (n)?

  1. ((a))

    500

  2. ((b))

    1000

  3. ((c))

    1500

  4. ((d))

    2000

Show Answer
Answer: ((b))

1000

Calculation:

Given,

Mean of the binomial distribution, μ=200 \mu = 200

Variance of the binomial distribution, σ2=160 \sigma^{2} = 160

For a binomial random variable,

μ=npandσ2=np(1−p). \mu = np \quad\text{and}\quad \sigma^{2} = np(1-p).

From the mean,

np=200;⟹;p=200n. np = 200 ;\Longrightarrow; p = \dfrac{200}{n}.

Using the variance,

np(1−p)=160;⟹;n(200n)!(1−200n)=160;⟹;200(1−200n)=160. np(1-p) = 160 ;\Longrightarrow; n\bigl(\tfrac{200}{n}\bigr)!\Bigl(1-\tfrac{200}{n}\Bigr)=160 ;\Longrightarrow; 200\Bigl(1-\tfrac{200}{n}\Bigr)=160.

Simplifying,

1−200n=0.8;⟹;200n=0.2;⟹;n=2000.2=1000. 1-\tfrac{200}{n} = 0.8 ;\Longrightarrow; \tfrac{200}{n} = 0.2 ;\Longrightarrow; n = \dfrac{200}{0.2} = 1000.

∴ The number of trials is n=1000 n = 1000 .

120

What is the arithmetic mean of 82,92,102,...,152?

  1. ((a))

    133.5

  2. ((b))

    135.5

  3. ((c))

    137.5

  4. ((d))

    139.5

Show Answer
Answer: ((c))

137.5

Calculation:

Given,

The sequence is 82,,92,,102,,…,,1528^{2},,9^{2},,10^{2},,\dots,,15^{2}

Number of terms, n=15−8+1=8n = 15 - 8 + 1 = 8

The sum of the first mm squares is  ∑k=1mk2=m(m+1)(2m+1)6 \displaystyle \sum_{k=1}^{m} k^{2} = \frac{m(m+1)(2m+1)}{6}

Sum up to 15215^{2}:

S15=15×16×316=1240S_{15} = \frac{15 \times 16 \times 31}{6} = 1240

Sum up to 727^{2}:

S7=7×8×156=140S_{7} = \frac{7 \times 8 \times 15}{6} = 140

Sum of the required terms:

S=S15−S7=1240−140=1100S = S_{15} - S_{7} = 1240 - 140 = 1100

The arithmetic mean is,

Mean=Sn=11008=137.5 \displaystyle \text{Mean} = \frac{S}{n} = \frac{1100}{8} = 137.5

∴ The arithmetic mean is 137.5137.5.

Hence, the correct answer is option 3.

Attempt this paper under real exam conditions

Timed interface, section switching, instant scoring, and question-by-question analytics — free.

Start Timed Attempt