NDA-II 2024 (Maths) Official Paper (Held On: 01 Sept, 2024) (Previous Year Paper)

Concept:

Matrix Multiplication:

• Matrix multiplication is defined only when the number of columns in the first matrix is equal to the number of rows in the second matrix.
• For two matrices $A_{m \times n}$ and $B_{n \times p}$, their product $AB$ will result in a matrix $C_{m \times p}$.
• In general, the product of $X$ and $Y$ is defined if and only if the number of columns in $X$ equals the number of rows in $Y$.

Calculation:

We have the following matrix operations:

$A_{m \times n} B_{n \times p} = (AB)_{m \times p}$

Now, consider the matrix multiplications:

$[Z_{3 \times 2} . Y_{2\times3}].X_{3 \times 3}] = [ZYX]_{3 \times 3}$

$Y_{2 \times 3} [X_{3 \times 3} Z_{3 \times 2}] = [YXZ]_{2 \times 2}$

Finally, we observe that:

$X_{3 \times 3} [Y_{2 \times 3} Z_{3 \times 2}] = X_{3 \times 3} [YZ]_{2 \times 2}$

Conclusion:

$\text{No. of columns in X} \neq \text{No. of rows in (YZ)}$

Hence, X(YZ) is not defined.

∴ the Correct answer is Option 4

Concept:

Infinite and Finite Sets:

• A finite set contains a limited number of elements that can be counted to an end.
• An infinite set contains endlessly many elements; you cannot finish counting them.
• Between any two different real numbers there are infinitely many irrational numbers.
• A set described without a lower bound (for integers) typically includes infinitely many negative members.

Calculation:

Given statements:

I. All irrational numbers between √12 and √15

II. All odd integers less than 1000

Step wise check:

⇒ For I: The interval between √12 and √15 is a normal real interval.

⇒ For I: Any real interval contains infinitely many irrational numbers.

⇒ So I is true.

⇒ For II: "Less than 1000" gives no lower limit.

⇒ For II: This includes ..., -5, -3, -1, 1, 3, ..., 999.

⇒ For II: We can continue to more negative odd integers without end.

⇒ So II is not finite (it is infinite).

∴ Only statement I is correct.

Concept:

Counting 4-digit Numbers with All Odd Digits:

• Each digit in a 4-digit number can be one of the odd digits: 1, 3, 5, 7, or 9.
• There are 5 choices for each digit since there are 5 odd digits available.
• For a 4-digit number, we need to select a digit for each of the 4 positions independently.
• The total number of 4-digit numbers with all odd digits is the product of choices for each digit.

Calculation:

Given,

Choices for each digit = 5 (as the digits are 1, 3, 5, 7, 9)

Number of digits, n = 4

The total number of 4-digit numbers with all odd digits is,

$5^4$ =  5 × 5 ×5 × 5 = 625

∴ 625 4-digit numbers have all digits as odd.5×5×5×5 5×5×5×\there

∴ The total number of 4-digit numbers with all odd digits is 625.

Explanation:

(1+ω -ω²)¹⁰⁰+ (1-ω +ω2)100

= (– ω² – ω² )¹⁰⁰ + (–ω – ω)¹⁰⁰  {∵ 1 + ω  + ω² = 0}

= 2¹⁰⁰(ω²⁰⁰ + ω¹⁰⁰)

= 2¹⁰⁰(ω² + ω)     (∵ ω³ = 1)

= –2¹⁰⁰ 

∴ Option (d) is correct

Explanation:

Given AB = O

Let  A = $\left[\begin{array}{lll} a & b \ c & d \end{array}\right]$

B = $\left[\begin{array}{lll} p & q \ r & s \end{array}\right]$

Now AB = $\left[\begin{array}{lll} a & b \ c & d \end{array}\right] \left[\begin{array}{lll} p & q \ r & s \end{array}\right] = \left[\begin{array}{lll} 0 & 0 \ 0 & 0 \end{array}\right]$

$\left[\begin{array}{lll} ap+br & aq+bs \ cp+dr & cq+ds \end{array}\right]= \left[\begin{array}{lll} 0 & 0 \ 0 & 0 \end{array}\right]$

Comparing both sides, we get

⇒ ap + br = 0...(1)

⇒ aq + bs = 0...(2)

⇒ cp + dr = 0...(3)

⇒ cq + ds = 0...(4)

From Eqs. (i) and (iii), we get:

⇒ (ad – bc)p = 0 and (ad – bc)r = 0 

From Eqs. (ii) and (iv), we get:

⇒ (ad – bc) q = 0 and (ad – bc)s = 0 

Now, if A is non-singular

ad – bc ≠ 0

So p = q = r = s = 0.

Then,  B is the null matrix.

∴ Option (c) is correct.

7Concept:

Binomial Expansion:

• The binomial theorem is used to expand expressions of the form $(1 + x)^n$.
• The general term in the expansion of $(1 + x)^n$ is given by $T_k = C(n, k) \cdot x^k$, where ⁿC_k is the binomial coefficient.
• To find the coefficient of x^3, we identify the corresponding terms from the expansion and set the coefficient equal to 35.

Calculation:

Given the expansion of $(1 + x)^p \cdot (1 + x)^q$, we have:

The coefficient of x³ in  the expansion is 35.

We use the binomial expansion formula for the term x³

⇒ (p+ q)_c3 = 35 = ⁷C₃

⇒ p+q =7

∴ the correct answer is Option C

Explanation:

Let 'a' and 'd' be the first term and common difference of AP

(p + q)^th Term = T_p+q 

= a + (p + q - 1)......(1)

Given

p. T_p = q.T_p

⇒ p.[ a + (p -1) d] = q. [ a + (q - 1) d]

⇒ (p - q) [ a + (p + q -1) d] =0

⇒ [ a + (p + q -1) d] =0           (∵ p ≠ q)

 ⇒Tp+q = 0 ......(From i)

∴ Option (a) is correct

Concept:

Arithmetic Progression (AP):

• The terms are in AP if the difference between consecutive terms is constant.
• The common difference is given by: $d = q - p = r - q$

Geometric Progression (GP):

• The terms are in GP if the ratio between consecutive terms is constant.
• The common ratio is given by: $r = q \times p$

Calculation:

Given

p = ln(x),

q = ln(x3)  = 3 lnx

and r = ln(x5) = 5 lnx

Clearly, p - q  = r - q = $2lnx$

⇒ p, q, r are in AP

Also $\frac{q}{p} ≠ \frac{r}{p} \because \frac{q}{p} = 3 , \frac{r}{p} = \frac{5}{3}$

∴ p, q, r can never be in GP

∴ Option (c) is correct.

Explanation:

Given

 $\rm z=\frac{1}{3}\begin{vmatrix}i&2i&1\\ 2i&3i&2\\ 3&1&3\end{vmatrix}$ 

$Z = \frac{1}{3}[[ (i9 – 2) – 2i (6i – 6) 1(2i – 9i)]$

$Z = \frac{1}{3}[3+3i] =1 + i$

|Z| = $√{1^2+1^2} =$ √2 

∴ Option (b) is correct.

Explanation:

Let I =  $\rm \Sigma_{n=1}^{20}(i^{n-1}+i^n+i^{n+1})$

⇒ I = $\rm \Sigma_{n=1}^{20}(i^n (\frac{1}{i} +1+ i)$

⇒ I = $\rm \Sigma_{n=1}^{20}(i^n (1)$

⇒ I = $i^1+ i^2+i^3+i^4i^5+........+ i^{20}$

⇒ I = $5.(i^1+ i^2+i^3+i^4) = 0$ ( {∵  i + i² + i³ + i⁴ = 0} 

∴ Option (b) is correct.

Explanation:

Given,

x, y, z are in GP

Thus, 

⇒ y² =xz

⇒ $2lny = lnx + lnz$

lnx, ln y and lnz are in AP

⇒ (1 + lnx), (1 + lny) and (1 + lnz) are in AP

⇒ $\frac{1}{lnx}, \frac{1}{lny}, \frac{1}{lnz}$ are in HP

Concept:

Determinant and Column Operations:

• The determinant of a matrix is a scalar value representing area, volume, or invertibility.
• Column operations like $C_1 \rightarrow C_1 + C_2 + C_3$ can simplify a determinant.
• If one column (or row) becomes a linear combination of others or zero, the determinant becomes 0.
• $\omega$ is a cube root of unity with identity: $1 + \omega + \omega^2 = 0$.

Calculation:

Let $I = \begin{bmatrix} 1+\omega & 1+\omega^2 & \omega + \omega^2 \ 1 & \omega & \omega^2 \ \frac{1}{\omega} & \frac{1}{\omega^2} & 1 \end{bmatrix}$

⇒ Apply column operation: $C_1 \rightarrow C_1 + C_2 + C_3$

⇒ First column becomes:

$C_1 = \begin{bmatrix} (1+\omega) + (1+\omega^2) + (\omega + \omega^2) \ 1 + \omega + \omega^2 \ \frac{1}{\omega} + \frac{1}{\omega^2} + 1 \end{bmatrix}$

⇒ Using identity: $1 + \omega + \omega^2 = 0$

⇒ First column becomes: $\begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}$

⇒ New matrix:

$I = \begin{bmatrix} 0 & 1+\omega^2 & \omega + \omega^2 \ 0 & \omega & \omega^2 \ 0 & \frac{1}{\omega^2} & 1 \end{bmatrix}$

⇒ Since first column is all zero,

∴ Determinant I = 0

Explanation:

Given

S_n = n(2n + 1), S_n–1 = (n – 1)(2n – 1) 

n^th term = S_n – S_{n – 1}

= n(2n + 1) – (n – 1)(2n–1)

=  4n – 1

∴ Option (a) is correct.

Explanation:

Word 'INDIA' contains three vowels (I, I and A) and two consonants (N and D).

Three vowels (I, I and A) can be arranged at odd position in $\frac{3!}{2!} = 3$ ways

2 consonants(N and D) can be arranged in 2! ways = 2 ways. 

Thus, required numbers of ways = (3) (2) = 6.

∴ Option (b) is correct.

Concept:

Permutations and Arrangements:

• A permutation is an arrangement of objects in a specific order. The number of ways to arrange 'n' objects is given by the factorial of 'n' denoted by n!.
• For a set of distinct vowels and consonants, the total number of ways to arrange them can be calculated by multiplying the factorials of the number of vowels and consonants separately.

Calculation:

Given the word 'EQUATION', we have:

• 5 vowels: A, E, I, O, U
• 3 consonants: Q, T, N

⇒ Number of ways = (3!) . (5!)

⇒ Numbers of ways = (3!) . (5!)

⇒ Total required number of ways = (3!) . (5!) + (3!) . (5!)

= 1440

∴ Option (c) is correct.

Calculation:

n is a root of the equation x2 + px + m = 0

So, n² + pn + n = 0

 ..... (1)

Also, m is the root of the equation of x² + px + n = 0

So, m² + pm + n = 0 ..... (2)

Now Eq. (i) – Eq. (ii), we get

⇒ $(n^2 – m^2) + p(n – m) – (n – m) = 0$

⇒ $(n – m) [(n + m) + p – 1] = 0$

⇒ $n + m + p – 1 = 0$ ..... ( n ≠ m)

⇒ n + m + p = 1

∴ Option (3) is correct.

Explanation:

Given,

Number of courses, n > 4

The student needs to choose (n - 2) courses out of n courses, with 2 courses being compulsory

∴Total courses is n – 2 and total remaining courses that can be chosen from (n – 2) courses is ( n – 4).

∴ Required number of ways ^(n-2) C _(n-4).

⇒ $\frac{(n-2)(n-3)}{2}$

∴ Option d is correct.

Concept:

Determinant of a 3×3 Matrix:

• The determinant of a 3×3 matrix can be found by the cofactor expansion method.
• It is expressed as: $\text{det}(A) = a(ei − fh) − b(di − fg) + c(dh − eg)$
• If the matrix has a common factor across rows or columns, it can be factored out to simplify calculations.
• $\sum_{n=1}^{k} f(n)$ is a summation notation indicating summing function values from n = 1 to n = k.
• Important identities used:
• $\sum_{n=1}^{k} n = \frac{k(k+1)}{2}$
• $\sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6}$
• $\sum_{n=1}^{k} n^3 = \left( \frac{k(k+1)}{2} \right)^2$

Calculation:

Given,

$D_n = \begin{bmatrix} n & 20 & 30 \ n^2 & 40 & 50 \ n^3 & 60 & 70 \ \end{bmatrix}$

⇒ Extract common factors 20 and 10:

$D_n = (20)(10) \begin{bmatrix} n & 1 & 3 \ n^2 & 2 & 5 \ n^3 & 3 & 7 \ \end{bmatrix}$

⇒ Expand determinant using first column:

$D_n = 200(-n + 2n^2 - n^3)$

Now,

$\sum_{n=1}^{4} D_n = 200 \sum_{n=1}^{4} (-n + 2n^2 - n^3)$

⇒ $= 200 \left[ -\frac{4(5)}{2} + 2 \cdot \frac{4(5)(9)}{6} - \left( \frac{4(5)}{2} \right)^2 \right]$

⇒ $= 200 (-10 + 60 - 100)$

∴ The value is -10000.

Concept:

Matrix Multiplication and Properties:

• Matrix multiplication involves the dot product of rows and columns.
• A null matrix is a matrix in which all elements are zero.
• An identity matrix is a square matrix with 1's on the diagonal and 0's elsewhere.
• For matrices P and Q, PQ = QP does not generally hold unless P and Q commute.

Matrix Definitions:

• Null Matrix: A matrix where all elements are zero.
• Identity Matrix: A square matrix with 1's on the main diagonal and 0's elsewhere.

Calculation:

$\rm P=\begin{bmatrix}0&c&-b\\ -c&0&a\\ b&-a&0\end{bmatrix}\ and\ \rm Q=\begin{bmatrix}a^2&ab&ac\\ ab&b^2&bc\\ ac&bc&c^2\end{bmatrix}$

⇒ PQ = $=\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{bmatrix}\$

⇒QP = $=\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{bmatrix}\$

Then PQ = QP

∴ Option (c) is correct

Explanation:

The determinant of any skew-symmetric matrix with odd order is always zero

∴ Option (b) is correct.

Explanation:

Given:

4 sin-1 x + cos-1 x = π 

⇒ $4( \frac{\pi }{2} - Cos^{-1}x) + (\frac{\pi }{2} - sin^{-1}x) = \pi$

⇒ $2\pi - 4cos^{-1}x + \frac{\pi }{2} - sin^{-1}x = \pi$

⇒ 4cos^-1 x + sin^-1x = $\frac{3\pi}{2}$

∴ Option (c) is correct

Explanation:

Let I = cot² (sec^-1 2) + tan²(cosec^-1 3)

⇒ I = cot² (60°) + tan ² (cot^-1 $\sqrt 9 -1$

⇒ I = $\frac{1}{3} + tan^2( tan^{-1}\frac{1}{\sqrt8})$

⇒ I = $\frac{1}{3} + \frac{1}{8} = \frac{11}{24}$

∴ Option (b) is correct

Explanation:

Given,

$\rm \frac{a}{\cos A}=\frac{b}{\cos B}=\frac{c}{\cos C}$

⇒ $\frac{a}{(\frac{b^2 + c^2 -a^2}{2bc})} = \frac{b}{(\frac{a^2 + c^2 -b^2}{2ac})} =\frac{c}{(\frac{a^2 + b^2 -c^2}{2ab})}$

⇒b²+ c ² – a ² = c ² + a ² – b ² = a ² + b ² – c ²

⇒ a ² = b ² = c ²  

⇒ a = b = c

Hence, ABC is an equilateral triangle

Side of the equilateral triangle, a = 6 cm

The area of the equilateral triangle is,

$\text{Area} = \frac{\sqrt{3}}{4} a^2$

⇒ $\text{Area} = \frac{\sqrt{3}}{4} \times 6^2$

⇒ $\text{Area} = \frac{\sqrt{3}}{4} \times 36$

⇒ $\text{Area} = 9\sqrt{3} , \text{cm}^2$

∴ The area of the triangle is 9 × √3 square cm.

Explanation:

Given:

tanα and tanβ are roots of 7x² – 6x + 1 = 0

Hence, 

⇒ tanα + tanβ = 6/7...(1)

⇒ tanα.tanβ = 1/7...(2)

⇒ tan(α+β) = $\frac{tanα + tanβ}{1- tanα . tanβ} = \frac{\frac{6}{7}}{1- \frac{1}{7}} = 1$

⇒ α +β = 45°

⇒ 2α +2β = 90°

It means the third angle of a triangle is 90°.

Hence, the triangle is right-angled.

Since

⇒tanα - tanβ = $\sqrt{(tanα + tanβ)^2 - 4tanα .tanβ} = \frac{2\sqrt2}{7}$

Also

⇒ tan(α -β) = $\frac{tanα - tanβ}{1+ tanα .tanβ}$ = $\frac{1}{2\sqrt2}$

⇒ tan2(α -β) = $\frac{2tan(α -β)}{1- tan^2(α-β)}$

 = $\frac{4\sqrt7}{7}$

Thus 2α≠2β 

∴ Option (c) is correct.

Concept:

Triangle Angle Sum Property:

• In a triangle, the sum of all interior angles is always equal to 180º.
• Given angles can be used to find the unknown angle using the formula: $\angle A + \angle B + \angle C = 180^\circ$
• Once all angles are known, trigonometric ratios can be used to relate the sides of the triangle.
• Law of Sines:
• The Law of Sines states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
• Here, a, b, and c are the sides opposite to angles A, B, and C respectively.

Calculation:

Given,

Angle A = 75º

Angle B = 45º

Using the triangle angle sum property, find angle C:

⇒ ∠C = 180º - (∠A + ∠B)

⇒ ∠C = 180º - (75º + 45º)

⇒ ∠C = 180º - 120º

⇒ ∠C = 60º

Now, using the Law of Sines:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

⇒ $a = \frac{sin 75°}{sin 60°}.c = \frac{\sqrt3 +1}{\sqrt6}.c$

⇒ $b = \frac{sin 45°}{sin 60°}.c = \frac{2c}{\sqrt6}$

Now

⇒  2a - b = $\frac{2\sqrt3 c}{\sqrt6} = \sqrt2 c$

∴ Option (b) is correct.

Explanation:

Given

cot2x. cot3x = 1

⇒ cos2x. cos3x – sin2x.sin3x= 0

⇒ cos(2x+3x) = cos5x = 0    0 < x < π 

Hence 5x = $\frac{\pi }{2}, \frac{3\pi }{2}, \frac{5\pi }{2}, \frac{7\pi }{2}, \frac{9\pi }{2}$

⇒ x = $\frac{\pi }{10}, \frac{3\pi }{10}, \frac{5\pi }{10}, \frac{7\pi }{10}, \frac{9\pi }{10}$

Thus, only five solutions are possible

∴ the correct answer is option C

Explanation:

cos¹⁰⁰x – sin¹⁰⁰x = 1 is only possible when cos² x = 1 or sin² x = 0

because cos² x ∈ [0, 1]

⇒ cos¹⁰⁰x ∈  [0, 1]

Also  sin² x ∈ [0, 1] 

⇒  sin¹⁰⁰x ∈ [0, 1] 

⇒ Cos²x = 1

⇒ x = nπ ± 0°

x = nπ 

∴ The correct answer is option 1.

Explanation:

In triangle ABC, A + B + C = π 

⇒ tan(A + B + C) = tanπ  = 0

⇒ $\frac{tan A+tan B+ tan C - tan A tan Btan C}{1- tan A.tan B- tan B.tan C - tan C.tan A} = 0$

⇒  tanA. tanB. tanC = tanA + tanB + tanC

⇒ cotA. cotB. cotC = $\frac{1}{tan A+tan B+tan C }$

= 1/k

∴ Option (b) is correct

Explanation:

⇒ $\frac{cos17°- sin17°}{cos17°+ sin17°} = \frac{cos17°(1-\frac{sin 17°}{Cos17°})}{cos17°(1+\frac{sin 17°}{Cos17°})}$ 

= $\frac{1-\tan17°}{1+\tan17°}$ 

= $\frac{tan 45° - tan17°}{1 + tan 45°. tan17°}$

= tan 28° = Cot 62°

∴ the Correct answer is Option 4.

Explanation:

(I) Let tan 22.5° = tan 45/2 = t

⇒ tan45° = $tan( 2 \times\frac{45}{2})$

= $\frac{2tan\frac{45}{2}}{1 -tan^2\frac{45}{2}}$

⇒ 1 = $\frac{2t}{1-t^2}$ ⇒  t² + 2t –1 = 0 

⇒ t = $\frac{-2 ± √4+4}{2} = -1±√2$

⇒ tan $\frac{45}{2} = -1 ± √2 = -1 +√2$   (In first quadrant tan is positive}

This is an irrational number. 

(II) cot22.5° = $\frac{1}{tan 22.5} = \frac{1}{-1 +√2} = 1+√2$

This is also an irrational number

(III) tan22.5° – cot22.5° = -1+ √2 -(1 +√ 2)

= -2

This is not an irrational number

∴ Option (c) is correct.

Explanation:

Given:

⇒ $\rm \frac{x}{\cos θ}=\frac{y}{\cos \left(\frac{2\pi}{3}-θ\right)}=\frac{z}{\cos\left(\frac{2\pi}{3}+θ\right)}$ = k (say)

⇒ x = kcosθ

⇒ y = k $cos\frac{2\pi }{3} -θ$

⇒ z = k $cos\frac{2\pi }{3} +θ$

Now, x + y + z = k [cosθ + cos $(\frac{2\pi }{3} -θ) + cos(\frac{2\pi }{3} +θ)$]

= k [cosθ + 2cos$\frac{2\pi }{3}cosθ$]

= k[cosθ + 2 ($-\frac{1}{2}) cosθ$]

= k[cosθ- cosθ ] =0

⇒ x + y + z =0

∴ Option (b) is correct.

Explanation:

Given:

p tan (θ - 30°) = q tan (θ + 120°)

⇒ p\q = $\frac{tan(\theta + 120°)}{tan(\theta -30°)}$

= $\frac{sin(\theta+120)cos(\theta -30)}{cos(\theta+120)sin(\theta-30)}$

⇒ $\frac{p+q}{p-q} =$(\frac{sin(\theta+120)cos(\theta -30)+cos(\theta+120)sin(\theta-30)}{sin(\theta+120)cos(\theta -30)-cos(\theta+120)sin(\theta-30)}\)

= $\frac{sin[(\theta+120)+(\theta-30)]}{sin[(\theta+120)-(\theta-30)]}$

= $\frac{sin(90+2\theta)}{sin 150}$

⇒ $\frac{p+q}{p-q} = \frac{sin(90+2\theta)}{\frac{1}{2}}$

= 2 cos2θ

∴ Option (d) is correct

Explanation:

(I) Given that P and Q are reflexive relations of A.

⇒ For any a∈A, (a, a) ∈ P and (a, a) ∈ Q 

⇒ (a, a )∈ P∩Q

⇒ P∩Q is reflexive relation.

So statement (I)is true

(II) Given that P and Q are symmetric relation

Let (a, b) ∈ (P∪Q)

⇒ (a, b) ∈ P or (a, b) ∈ Q.

⇒ (b, a) ∈ P or (b, a) ∈ Q

⇒  (b, a) ∈ (P∪Q)

⇒P∪Q is symmetric relation.

So statement (II) is true.

(III) Given that P and Q are transitive

Let (a, b)∈ (P∩Q) and (b, c)∈ (P∩Q)

⇒ {(a, b)∈ P, (b, c) ∈ P} and {(a, b)∈Q, (b, c) ∈Q}.....     {∵ P and Q are transitive}

⇒ (a, c) ∈ P∩Q

Thus P∩Q is a transitive relation.

So statement (III) is also true.

∴ Option (d) is correct.

Explanation:

Given:

⇒A∩B =10

Now, (A × B) ∩  (B × A) = 10² =100

Thus, there are 100 elements Common in A x B and B x A 

∴ Option (d) is correct

Explanation:

Let p(n) = 7ⁿ – 6n

= (6+1)ⁿ – 6n

= 1+6n +ⁿC₂6² + ⁿC₃6³+.......... 6ⁿ -6n)

= 1+ 6²(nC2+ nC3 × 6 +....... 6^n-2)

= 1+ 36(¹⁰⁰C₂ + 100C3 × 6 +........6⁹⁸).....(n =100)

= 1 + 36 k

where k = (100C2 + 100C3 × 6 +........698)

Thus When 7ⁿ – 6n is divided by 36, the remainder will be 1

∴ Option (b) is correct

Explanation:

Since there are 4 lines, the maximum number of intersection points are ⁴C₂ = 6.

The maximum number of intersections of the circle with 4 lines is 4 × 2 = 8.

Required intersection points = 6 + 8 = 14

∴ the correct answer is Option 3 ie 14

Explanation:

Given:

$\frac{S_p}{S_q} =\frac{p^2}{q^2}$

⇒ $\frac{\frac{p}{2}[2a +(p-1)d]}{\frac{q}{2} [2a+(q-1)d]}$ = p²/q²

⇒ $\frac{2a+(p-1)d}{2a+(q-1)d} =\frac{p}{q}$

⇒ 2aq +q(p – 1)d = 2ap + p(q – 1)d

⇒ 2a(q – p) = d(q – p) 

⇒ 2a = d or d = 2a

Thus, The common difference is equal to twice of the first term

∴ The Correct answer is Option C

Explanation:

Given:

(x - 1)2 + (x - 3)2 + (x - 5)2 = 0

⇒ x² – 2x + 1 + x² – 6x + 9 + x² – 10x + 25 = 0

⇒ 3x² – 18x + 35 = 0

Now, discriminant D =  (–18)² –4 × 3 × 35 

= – 96 < 0

Thus, The given equation does not have any real roots. 

∴ Option (a) is correct

Explanation:

⇒ Total student = 240

⇒10 students failed in every subject Hence, remaining total = 230

According to question

⇒ a + b + c + d + e + f + g = 230 ..... (1)

⇒ b + c + d = 110.... (2)

⇒ e+ f + g = 60 ..... (3)

Eq. (i) – Eqs. [(ii) + (iii)], we get

⇒ a = 230 –(110 + 60) = 60  

∴ Option (a) is correct.

Explanation:

$\rm Z_1^2+Z_2^2+Z_1Z_2=0$

⇒ Z ₁  = ω and Z ₂  = ω ²

Now

⇒ $|\frac{Z_1}{Z_2}| =| \frac{\omega }{(\omega)^2}| =|\frac{1}{\omega|}= 1$

Option (a) is correct.

Explanation:

$\rm \frac{1}{2}+Re\left(\frac{Z_1}{Z_2}\right) = \frac{1}{2}+ Re (\frac{\omega }{\omega^2})$

= $\frac{1}{2}Re(\omega)^2$

= $\frac{1}{2}+ Re [ -\frac{1}{2} - \frac{\sqrt3}{2}i$

= $\frac{1}{2} + (-\frac{1}{2}) = 0$

∴ Option (b) is correct.

Explanation:

Let the required terms of AP are a – 2d, a – d, a, a + d, a + 2d

⇒ (a – 2d) (a – d) a (a + d) (a + 2d) = 2,29,635

⇒ a(a ² – d² ) (a ² – 4d² ) = 2,29,635.....(i)

Now  a – 2d, a – d and a + 2d are in GP

⇒ (a – d)² = a² – 4d²

⇒ 5d² – 2ad = 0 

⇒ d(5d – 2a) = 0

⇒ 5d = 2a......(ii)

Solving equation (i) and (ii), we get

⇒ d = 6

∴ Option (d) is correct.

Explanation:

Putting the value of d in equation (ii), we get

⇒ 5 × 6 = 2a 

⇒ a = 15

Thus, The terms are 3, 9, 15, 21 and 27

So required sum = 3 + 9 + 15 + 21 + 27 = 75 

∴ Option (c) is correct

Explanation:

Given:

(8 + 3√7) 20  = U + V...(i)

(8 - 3√7) 20  = W...(ii)

Here, 0 < W < 1 

Now, adding Eqs. (i) and (ii), we get

U + V +W = (8 + 3√7) 20 + (8 - 3√7) 20

= 2[²⁰C₀8²⁰+ ²⁰C₂8¹⁸. (3√7 )²+........+(3√7)²⁰

⇒ It is an even number.

Also, 0 < V < 1, 0 < W < 1 and U is an integer

Thus,  V + W is an integer

V + W = 1

∴ Option (d) is correct.

Explanation:

Given:

(8 + 3√7) 20  = U + V...(i)

(8 - 3√7) 20  = W...(ii)

⇒ (U + V)W =] (8 + 3√7) 20][(8 - 3√7) 20]

= (64 – 63)²⁰ = 1²⁰ = 1

∴ Option (b) is correct.

Explanation:

Given:

 ⇒ a² (b² – c ² ) x² + b ² (c ² – a ² ) x + c ² (a ² – b ² ) = 0...(i)

Now  Equation has equal roots:

Then D =0

⇒ [b² (c ² – a² )]² – 4a ² c ² (b ² – c ² ) (a ² – b ² ) = 0

⇒  b⁴ (c ² – a ² )² – 4a ² c ² (b ² – c ² ) (a ² – b ² ) = 0

Solving above equation, we get 

⇒  b⁴ (c² + a² )² = 4a ⁴ c ⁴ 

⇒ b ² (a ² + c ² ) = 2a ² c ²

⇒ $\frac{2}{b^2} = \frac{1}{a^2}+ \frac{1}{c^2}$