Official Paper

NDA-II 2025 (Mathematics) Official Paper (Held On: 14 Sept, 2025) (Previous Year Paper)

120 questions · 150 minutes · with answers · free

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Mathematics (120 questions)

If $p ^ x = q ^ y = r ^ z$ where x, y and z are in GP, then consider the following statements:

I. p, q and rare in AP.

II. In p, In q and Inrare in GP.

Which of the statements given above is/are correct?

((a))
I only
((b))
II only
((c))
Both I and II
((d))
Neither I nor II

Show Answer

Answer: ((b))

II only

Calculation:

Given: $p^x = q^y = r^z = k$ and $x, y, z$ are in GP.

Taking logs: $\ln(p^x) = \ln(q^y) = \ln(r^z) = \ln k \Rightarrow x\ln p = y\ln q = z\ln r = A$

Thus: $\ln p = \frac{A}{x},\quad \ln q = \frac{A}{y},\quad \ln r = \frac{A}{z}$

Let $x, y, z$ be GP: $x=a,\ y=ar,\ z=ar^2$

Then: $\ln p = \frac{A}{a},\quad \ln q = \frac{A}{ar},\quad \ln r = \frac{A}{ar^2}$

So $\ln p,, \ln q,, \ln r$ are in GP (reciprocals of a GP form a GP).

Hence Statement II is true.

For Statement I (whether $p, q, r$ are in AP):

from $\ln p = \frac{A}{x},\ \ln q = \frac{A}{y},\ \ln r = \frac{A}{z}$ , there is no general condition implying $q = \frac{p+r}{2}$ ;

logarithms of reciprocals of a GP do not enforce an AP on the original bases. Hence Statement I is false.

Hence the correct answer is II only.

If A and B are non-empty subsets of a set, and $A ^ c$ and $B ^ c$ represent their complements, then which of the following is/are correct?

I. A - B = $B ^ c - A ^ c$

II. $A - B ^ c = A ^ c - B$

Select the answer using the code given below.

((a))
I only
((b))
II only
((c))
Both I and II
((d))
Neither I nor II

Show Answer

Answer: ((a))

I only

Calculation:

Statement I:

$\text{L.H.S.} = A - B = A \cap B^{c}$

$\text{R.H.S.} = B^{c} - A^{c} = B^{c} \cap (A^{c})^{c} = B^{c} \cap A$

$\therefore ; \text{L.H.S.} = \text{R.H.S.}$

So, Statement I is correct.

Statement II:

$\text{L.H.S.} = A - B^{c} = A \cap (B^{c})^{c} = A \cap B$

$\text{R.H.S.} = A^{c} - B = A^{c} \cap B^{c} = U - (A \cup B)$

$\therefore ; \text{L.H.S.} \ne \text{R.H.S.}$

So, Statement II is incorrect.

Hence, the correct answer is Option 1.

Let y=x! and z = (2x)! If (z / y) = 120 then what is the value of (3x)!?

((a))
362880
((b))
181440
((c))
90720
((d))
45360

Show Answer

Answer: ((a))

362880

Calculation:

Given: $y=x!,; z=(2x)!$ and $\dfrac{z}{y}=120$

⇒ $\dfrac{(2x)!}{x!}=120$

Check small integer values of $x$ :

For $x=3$ , $\dfrac{(2\cdot 3)!}{3!}=\dfrac{6!}{3!}=\dfrac{720}{6}=120$

So, $x=3$ .

Now, $(3x)!=(3\cdot 3)!=9!=362880$

Hence the correct answer is 362880.

Let n be a natural number. The number of consecutive zeros at the end of the expansion of n! is exactly 2. How many values of n are possible?

((a))
3
((b))
4
((c))
5
((d))
More than 5

Show Answer

Answer: ((c))

Calculation:

The number of trailing zeros in $n!$ depends on the number of factors of 5 in its prime factorization.

Number of trailing zeros $= \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \cdots$

We are given that the number of trailing zeros is exactly 2.

Checking possible values of n :

For $n = 10$ :

⇒ $\left\lfloor \frac{10}{5} \right\rfloor = 2,; \left\lfloor \frac{10}{25} \right\rfloor = 0 \Rightarrow 2$

For $n = 11, 12, 13, 14$ :

⇒ $\left\lfloor \frac{n}{5} \right\rfloor = 2,; \left\lfloor \frac{n}{25} \right\rfloor = 0 \Rightarrow 2$

For $n = 15$ :

$\left\lfloor \frac{15}{5} \right\rfloor = 3 \Rightarrow$ 3 zeros (exceeds)

Hence, the valid values of $n$ are:

$n = 10, 11, 12, 13, 14$

Therefore, the number of possible values of n is 5.

If and $ (10+\log _{10}x), (10+\log {10}y)$ are $(10 + log{10}z)$ in AP, then consider the following statements:

I. The GM of x and z is $y ^ 2$

II. The AM of $\log _{10}x$ and $ \log {10}z$ is $log{10} y $

Which of the statements given above is/are correct?

((a))
I only
((b))
II only
((c))
Both I and II
((d))
Neither I nor II

Show Answer

Answer: ((b))

II only

Calculation:

Given that:

$(10+\log_{10}x),\ (10+\log_{10}y),\ (10+\log_{10}z) \text{ are in A.P.}$

Using the condition of A.P.:

$2(10+\log_{10}y)=(10+\log_{10}x)+(10+\log_{10}z)$

Simplifying:

$20+2\log_{10}y=20+\log_{10}x+\log_{10}z$

$2\log_{10}y=\log_{10}x+\log_{10}z$

This implies:

$\log_{10}y=\frac{\log_{10}x+\log_{10}z}{2}$

Hence, the A.M. of $\log_{10}x$ and $\log_{10}z$ is $\log_{10}y$ .

So, statement II is correct.

Now,

$2\log_{10}y=\log_{10}x+\log_{10}z \Rightarrow \log_{10}y^2=\log_{10}(xz)$

$\Rightarrow y^2=xz$

Therefore, the G.M. of $x$ and $z$ is $y$ , not $y^2$ .

So, statement I is incorrect.

Hence, the correct answer is Option 2.

How many terms of the series 1 + 3 + 5 + 7 +... amount to a sum equal to 12345678987654321?

((a))
11111111
((b))
110000011
((c))
111101111
((d))
111111111

Show Answer

Answer: ((d))

111111111

Calculation:

The sum of the first $n$ odd natural numbers is given by:

$1 + 3 + 5 + \cdots + (2n-1) = n^2$

Given that the sum is:

$n^2 = 12345678987654321$

Taking square root on both sides:

$n = \sqrt{12345678987654321}$

Now,

$111111111^2 = 12345678987654321$

$\Rightarrow n = 111111111$

Hence, the correct answer is 111111111.

How many terms are identical in the two APs 19, 21, 23,... up to 110 terms and 19, 22, 25, 28,... up to 75 terms?

((a))
35
((b))
36
((c))
37
((d))
38

Show Answer

Answer: ((c))

Calculation:

Given two Arithmetic Progressions (APs):

$\text{AP}_1: 19, 21, 23,$ ...... up to 110 terms

$a_1 = 19, d_1 = 2, T_n = 19 + 2(n-1) = 2n + 17$ (up to n = 110,where } T₁₁₀ = 237

$\text{AP}_2: 19, 22, 25,$ ...... up to 75 terms

$a_2 = 19, d_2 = 3, S_m = 19 + 3(m-1) = 3m + 16$ up to m = 75 , where S₇₅= 241

To find the common terms

$2n + 17 = 3m + 16 \Rightarrow 2n - 3m = -1 \Rightarrow 2n \equiv 2 \pmod 3 \Rightarrow n \equiv 1 \pmod 3$

$n = 3k + 1$ , Now, we calculate the common terms

$T_{3k+1} = 2(3k+1) + 17 = 6k + 19$

Now, to ensure the terms are within the limits, we use

$6k + 19 \le 237 \Rightarrow 6k \le 218 \Rightarrow k \le 36$

Values of k from 0 to 36 give 37 common terms

Hence, the correct answer is 37.

If $\alpha =\frac{-1+\sqrt{-3}}{2}$ then what is the value of $(1+\alpha ^{19}-\alpha ^{35})^{100}-(1-3\alpha ^{25}+\alpha ^{38})^{50}?$

((a))
-2
((b))
-1
((c))
0
((d))
2

Show Answer

Answer: ((c))

Calculation:

We are given that: $\alpha = \frac{-1 + \sqrt{3}}{2}$ , which is the cube root of unity.

Using the properties of cube roots of unity:

⇒ $\alpha^3 = 1$

⇒ $1 + \alpha + \alpha^2 = 0$

⇒ $\alpha^2 = \alpha - 1$

Now, the given expression is:

⇒ $(1 + \alpha^{19} - \alpha^{35})^{100} - (1 - 3\alpha^{25} + \alpha^{38})^{50}$

Simplifying $\alpha^{19}$ and $\alpha^{35}$

⇒ $\alpha^{19} = \alpha$ since $19 \mod 3 = 1$

⇒ $\alpha^{35} = \alpha^2$ Since $35 \mod 3 = 2$

Thus, $1 + \alpha^{19} - \alpha^{35} = 1 + \alpha - \alpha^2$

Using $\alpha^2 = \alpha - 1$ , we get:

⇒ $1 + \alpha - (\alpha - 1) = 1 + \alpha - \alpha + 1 = 2$

So, $(1 + \alpha^{19} - \alpha^{35})^{100} = 2^{100}$

Simplifying $\alpha^{25}$ and $\alpha^{38}$

⇒ $\alpha^{25} = \alpha$ since $25 \mod 3 = 1$

⇒ $\alpha^{38} = \alpha^2$ Since $38 \mod 3 = 2$

Thus, $1 - 3\alpha^{25} + \alpha^{38} = 1 - 3\alpha + \alpha^2$

Using $\alpha^2 = \alpha - 1$ , we get:

⇒ $1 - 3\alpha + (\alpha - 1) = 1 - 3\alpha + \alpha - 1 = -2\alpha$

So, $(1 - 3\alpha^{25} + \alpha^{38})^{50} = (-2\alpha)^{50} = 2^{50} \alpha^{50}$

Since $\alpha^3 = 1$ , we have $\alpha^{50} = \alpha^2$ .

Thus, $(1 - 3\alpha^{25} + \alpha^{38})^{50} = 2^{50} \alpha^2$

Subtracting the two terms

⇒ $2^{100} - 2^{50} \alpha^2$

Since $\alpha^2 = \alpha - 1$ , we get:

⇒ $2^{100} - 2^{50} (\alpha - 1) = 2^{100} - 2^{50} \alpha + 2^{50}$

The value of the expression is $0$

Hence, the correct answer is Option 3.

What is the remainder when $5 ^ {99}$ is divided by 13?

((a))
10
((b))
9
((c))
8
((d))
6

Show Answer

Answer: ((c))

Calculation:

We are asked to find the remainder when $5^{99}$ is divided by 13.

Let’s find the pattern of powers of 5 modulo 13:

$5^1 = 5 \mod 13 = 5$

$5^2 = 25 \mod 13 = 12$

$5^3 = 125 \mod 13 = 8$

$5^4 = 625 \mod 13 = 1$

So the powers repeat every 4 steps: $5^4 \equiv 1 \mod 13$

Now reduce the exponent modulo 4:

$99 \mod 4 = 3 \Rightarrow 5^{99} \equiv 5^3 \mod 13$

$5^3 = 125 \mod 13 = 8$

Hence, the remainder when $5^{99}$ is divided by 13 is 8.

What is the value of the determinant of the inverse of the matrix $\begin{bmatrix} -4 & -5 \ 2 & 2 \end{bmatrix} ?$

((a))
1/2
((b))
1
((c))
2
((d))
4

Show Answer

Answer: ((a))

1/2

Calculation:

We are asked to find the value of the determinant of the inverse of the matrix:

$\begin{bmatrix} -4 & -5 \ 2 & 2 \end{bmatrix}$

Compute the determinant of the original matrix:

⇒ $\text{det}(A) = (-4)(2) - (-5)(2) = -8 + 10 = 2$

Use the property of inverse determinant:

⇒ $\text{det}(A^{-1}) = \frac{1}{\text{det}(A)} = \frac{1}{2}$

Hence, the value of the determinant of the inverse is 1/2

In a class of 45 students, 34 like to play cricket and 26 like to play football. Further, each student likes to play at least one of the two games. How many students like to play exactly one game?

((a))
45
((b))
30
((c))
25
((d))
15

Show Answer

Answer: ((b))

Calculation:

Let total students = $45$

Students who like cricket = $34$

Students who like football = $26$

Each student likes at least one of the two games.

Using set theory identity:

$|C \cup F| = |C| + |F| - |C \cap F|$

$45 = 34 + 26 - |C \cap F| \Rightarrow |C \cap F| = 60 - 45 = 15$

Students who like exactly one game:

$|C| + |F| - 2|C \cap F| = 34 + 26 - 2(15) = 60 - 30 = 30$

Hence, the number of students who like exactly one game is $30$ .

The system of equations 15y - 10x + 50 = 0, 2x - 3y - 5 = 0

((a))
has a unique solution
((b))
has infinitely many solutions
((c))
is inconsistent
((d))
is consistent and has exactly two solutions

Show Answer

Answer: ((c))

is inconsistent

Calculation:

Given system of equations:

$15y - 10x + 50 = 0$

$2x - 3y - 5 = 0$

Rearranging the first equation:

⇒ $15y - 10x + 50 = 0 \Rightarrow -10x + 15y = -50$

$\Rightarrow 2x - 3y = 10 \quad \text{(dividing by } -5)$

Rearranging the second equation:

⇒ $2x - 3y - 5 = 0 \Rightarrow 2x - 3y = 5$

Now compare both equations:

⇒ $2x - 3y = 10$ and $2x - 3y = 5$

This is a contradiction: same left-hand side equals two different right-hand sides.

Hence, the system is inconsistent and has no solution.

If $(\frac{1-i}{1+i})^{2m}(\frac{1+i}{1-i})^{2n}=1$ where $i=\sqrt{-1}$ then what is the smallest positive value of (m - n)?

((a))
1
((b))
2
((c))
4
((d))
8

Show Answer

Answer: ((b))

Calculation:

Given:

$\left( \frac{1 - i}{1 + i} \right)^{2m} \left( \frac{1 + i}{1 - i} \right)^{2n} = 1$

Simplify the base:

$\frac{1 - i}{1 + i} = \frac{(1 - i)^2}{(1 + i)(1 - i)} = \frac{1 - 2i - 1}{1 - (-1)} = \frac{-2i}{2} = -i$

Hence:

$\left( \frac{1 - i}{1 + i} \right)^{2(m - n)} = (-i)^{2(m - n)} = 1$

Since $(-i)^2 = -1$ , we have $(-i)^{2k} = (-1)^k$ . Thus $(-i)^{2(m - n)} = 1$ iff $m - n$ is even.

Smallest positive value: $m - n = 2$ .

Hence, the correct answer is Option 2.

In obtaining the solution of the system of equations x + y + z = 7 x + 2y + 3z = 16 and x + 3y + 4z = 22 by Cramer's rule, the value of y is obtained by dividing D by $D_{2}$ where

$D=\left|\begin{matrix}1&1&1\ 1&2&3\ 1&3&4\end{matrix}\right|$ What is the value of the determinant $D_{2}$ ?

((a))
-13
((b))
-3
((c))
3
((d))
13

Show Answer

Answer: ((b))

-3

Calculation:

$D = \begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 3 & 4 \end{vmatrix}$

⇒ $D_2 = \begin{vmatrix} 1 & 7 & 1 \ 1 & 16 & 3 \ 1 & 22 & 4 \end{vmatrix}$

⇒ $D_2 = 1 \cdot \begin{vmatrix} 16 & 3 \ 22 & 4 \end{vmatrix} - 7 \cdot \begin{vmatrix} 1 & 3 \ 1 & 4 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 16 \ 1 & 22 \end{vmatrix}$

⇒ $\begin{vmatrix} 16 & 3 \ 22 & 4 \end{vmatrix} = (16 \cdot 4) - (3 \cdot 22) = 64 - 66 = -2$

⇒ $\begin{vmatrix} 1 & 3 \ 1 & 4 \end{vmatrix} = (1 \cdot 4) - (3 \cdot 1) = 4 - 3 = 1$

⇒ $\begin{vmatrix} 1 & 16 \ 1 & 22 \end{vmatrix} = (1 \cdot 22) - (16 \cdot 1) = 22 - 16 = 6$

⇒ $D_2 = 1 \cdot (-2) - 7 \cdot (1) + 1 \cdot (6)$

$D_2 = -2 - 7 + 6 = -3$

Hence the correct answer is Option 2.

Consider the following in respect of non-singular matrices A and B:

I. $(AB) ^ {- 1} = A ^ {- 1} B ^ {- 1}$

II. $(BA)(AB)^{-1}=I$ = I where I is the identity matrix

III. $(AB)^{T}=A^{T}B^{T}$

How many of the above are correct?

((a))
None
((b))
One
((c))
Two
((d))
All three

Show Answer

Answer: ((a))

None

Calculation:

Given: Non-singular matrices $A$ and $B$

Statement I: $(AB)^{-1} = A^{-1}B^{-1}$

This is incorrect. The correct identity is $(AB)^{-1} = B^{-1}A^{-1}$

Statement II: $(BA)(AB)^{-1} = I$

This is incorrect. Using $(AB)^{-1} = B^{-1}A^{-1}$ , we get:

$(BA)(AB)^{-1} = BA \cdot B^{-1}A^{-1} = B A B^{-1} A^{-1} \neq I$

Statement III: $(AB)^T = A^T B^T$

This is incorrect. The transpose of a product reverses the order:

$(AB)^T = B^T A^T$

Hence, None of the statements are correct.

The value of the determinant $\left|\begin{matrix}a&b&c\ l&m&n\ p&q&r\end{matrix}\right|$ is equal to

((a))
$\left|\begin{matrix}a&b&c\ p&q&r\ l&m&n\end{matrix}\right|$
((b))
$\left|\begin{matrix}l&m&n\ a&b&c\ p&q&r\end{matrix}\right|$
((c))
$\left|\begin{matrix}p&q&r\ a&b&c\ l&m&n\end{matrix}\right|$
((d))
$\left|\begin{matrix}a&p&l\ b&q&m\ c&r&n\end{matrix}\right|$

Show Answer

Answer: ((c))

$\left|\begin{matrix}p&q&r\ a&b&c\ l&m&n\end{matrix}\right|$

Calculation:

Given determinant:

$\begin{vmatrix} a & b & c \ l & m & n \ p & q & r \end{vmatrix}$

Let this be denoted as $D$ . Now analyze each option:

Option 1: $\begin{vmatrix} a & b & c \ p & q & r \ l & m & n \end{vmatrix}$

This swaps rows 2 and 3 → one transposition → $= -D$

Option 2: $\begin{vmatrix} l & m & n \ a & b & c \ p & q & r \end{vmatrix}$

This swaps rows 1 and 2 → one transposition → $= -D$

Option 3: $\begin{vmatrix} p & q & r \ a & b & c \ l & m & n \end{vmatrix}$

This is a 3-cycle permutation of rows → even permutation → $= D$

Option 4: $\begin{vmatrix} a & p & l \ b & q & m \ c & r & n \end{vmatrix}$

This permutes columns, not rows → determinant value changes → not equal to $D$

Hence Option 3 is correct.

Let 1, ω, $\omega ^ 2$ be three cube roots of unity. If x = a + b $y=a\omega +b\omega ^{2}, z=a\omega ^{2}+b\omega$ then what is $x ^ 2 + y ^ 2 + z ^ 2$ equal to?

((a))
6ab
((b))
3ab
((c))
$a ^ 2 + b ^ 2$
((d))
1

Show Answer

Answer: ((a))

6ab

Calculation:

Let $\omega$ be a cube root of unity, so:

$\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$

Given:

⇒ $x = a + b$

⇒ $y = a\omega + b\omega^2$

⇒ $z = a\omega^2 + b\omega$

Compute $y + z$

⇒ $y + z = a(\omega + \omega^2) + b(\omega^2 + \omega) = (a + b)(\omega + \omega^2) = -x$

Use identity:

⇒ $y^2 + z^2 = (y + z)^2 - 2yz = x^2 - 2yz$

Compute $yz$

⇒ $yz = (a\omega + b\omega^2)(a\omega^2 + b\omega)$

$= a^2\omega^3 + ab(\omega^2\omega^2 + \omega\omega) + b^2\omega^3$

$= a^2 + ab(\omega + \omega^2) + b^2 = a^2 + b^2 - ab$

Final expression:

⇒ $x^2 + y^2 + z^2 = x^2 + (x^2 - 2yz) = 2x^2 - 2yz$

$= 2(a + b)^2 - 2(a^2 + b^2 - ab) = 2(a^2 + 2ab + b^2) - 2(a^2 + b^2 - ab) = 6ab$

Hence, the correct answer is Option 1.

How many 4-digit numbers that are divisible by 4 can be formed using the digits 1, 2, 3 and 4 (repetition of digits is not allowed)?

((a))
3
((b))
6
((c))
9
((d))
12

Show Answer

Answer: ((b))

Calculation:

We are asked to find how many 4-digit numbers divisible by 4 can be formed using digits ${1, 2, 3, 4}$ without repetition.

we know A number is divisible by 4 if its last two digits form a number divisible by 4.

List all 2-digit endings from the given digits that are divisible by 4:

$12, 24, 32$ are valid (since they are divisible by 4 and use distinct digits).

For each valid ending, choose the remaining two digits from the unused ones and arrange them in the first two positions:

Ending $12$ → remaining digits: ${3, 4}$ → 2! = 2 arrangements

Ending $24$ → remaining digits: ${1, 3}$ → 2! = 2 arrangements

Ending $32$ → remaining digits: ${1, 4}$ → 2! = 2 arrangements

Total valid numbers = $2 + 2 + 2 = 6$

Hence, the correct answer is option 2.

If a, b, c are the sides of a triangle ABC and p is the perimeter of the triangle, then what is equal to? det $\left|\begin{matrix}p+c&a&b\ c&p+a&b\ c&a&p+b\end{matrix}\right|$

((a))
$p ^ 3$
((b))
$2p ^ 3$
((c))
$3p ^ 3$
((d))
$4p ^ 3$

Show Answer

Answer: ((b))

$2p ^ 3$

Calculation:

Let $p = a + b + c$ be the perimeter of triangle ABC.

Given determinant:

$D = \begin{vmatrix} p + c & a & b \ c & p + a & b \ c & a & p + b \end{vmatrix}$

Apply row operations:

$R_1 \leftarrow R_1 - R_2$ and $R_2 \leftarrow R_2 - R_3$

New matrix:

⇒ $\begin{vmatrix} p & -p & 0 \ 0 & p & -p \ c & a & p + b \end{vmatrix}$

Factor out $p$ from first two rows:

⇒ $D = p^2 \cdot \begin{vmatrix} 1 & -1 & 0 \ 0 & 1 & -1 \ c & a & p + b \end{vmatrix}$

Expand along first row:

$= 1 \cdot \begin{vmatrix} 1 & -1 \ a & p + b \end{vmatrix} + 1 \cdot \begin{vmatrix} 0 & -1 \ c & p + b \end{vmatrix}$

$= (p + b + a) + c = 2p$

⇒ $D = p^2 \cdot 2p = 2p^3$

Hence the correct answer is Option 2.

Which one of the following is the greatest coefficient in the expansion of $(1 + x) ^ {100}$ ?

((a))
The coefficient of $x ^ {100}$
((b))
The coefficient of $x ^ {99}$
((c))
The coefficient of $x ^ {51}$
((d))
The coefficient of $x ^ {50}$

Show Answer

Answer: ((d))

The coefficient of $x ^ {50}$

Calculation:

The general term in the expansion of $(1 + x)^{100}$ is:

$T_k = \binom{100}{k} x^k$

Here, $\binom{100}{k}$ is the binomial coefficient, which represents the coefficient of $x^k$ .

We are asked to find the value of $k$ for which $\binom{100}{k}$ is maximum.

It is known that in the expansion of $(1 + x)^n$ , the binomial coefficients increase until $k = \left\lfloor \frac{n}{2} \right\rfloor$ and then decrease.

For $n = 100$ , the maximum occurs at $k = 50$ .

The greatest coefficient is the coefficient of $x^{50}$

Hence, the correct answer is Option 4.

For the following two (02) items:

Let $\alpha$ and $\beta$ be the roots of the quadratic equation

$(x^{2}+(\log _{0.5}(\alpha ^{2}))x+(\log _{0.5}(\alpha ^{2}))^{4}=0$

where $a ^ 2 \ne1$ and $\log _{0.5}(\alpha ^{2})>0$ Further, $\beta ^{2}=\alpha (\log _{\alpha ^{2}}(0.5))$

What is ẞ equal to?

((a))
$\log _{a^{2}}(0.5)$
((b))
$\log _{0.5}(a^{2})$
((c))
$2(\log _{a^{2}}(0.5))$
((d))
$2\log _{0.5}(a^{2})$

Show Answer

Answer: ((b))

$\log _{0.5}(a^{2})$

Calculation:

Let $L = \log_{0.5}(\alpha^2)$ . Then we know:

$\log_{\alpha^2}(0.5) = \frac{1}{L}$

Given quadratic equation:

$x^2 + Lx + L^4 = 0$

Let roots be $\alpha$ and $\beta$ . Then:

$\alpha + \beta = -L$

$\alpha \beta = L^4$

From product of roots: $\beta = \frac{L^4}{\alpha}$

Also given: $\beta^2 = \alpha \cdot \log_{\alpha^2}(0.5) = \frac{\alpha}{L}$

Now equate both expressions for $\beta^2$ :

$\left(\frac{L^4}{\alpha}\right)^2 = \frac{\alpha}{L} \Rightarrow \frac{L^8}{\alpha^2} = \frac{\alpha}{L}$

Multiply both sides by $\alpha^2 L$ :

$L^9 = \alpha^3 \Rightarrow \alpha = L^3$

Then:

$\beta = \frac{L^4}{\alpha} = \frac{L^4}{L^3} = L$

$\beta = \log_{0.5}(\alpha^2)$

Hence, the correct answer is Option 2

What is the relation between α and β

((a))
α = 2β
((b))
2α = β
((c))
α = - 2β
((d))
2α = -β

Show Answer

Answer: ((a))

α = 2β

Calculation:

Let $L = \log_{0.5}(\alpha^2)$ . Then the quadratic equation becomes:

$x^2 + Lx + L^4 = 0$

Let the roots be $\alpha$ and $\beta$ . Then:

⇒ $\alpha + \beta = -L$

⇒ $\alpha \beta = L^4 \Rightarrow \beta = \frac{L^4}{\alpha}$

Also given: $\beta^2 = \alpha \cdot \log_{\alpha^2}(0.5) = \frac{\alpha}{L}$

Now compute $\beta^2$ from both expressions:

⇒ $\left( \frac{L^4}{\alpha} \right)^2 = \frac{L^8}{\alpha^2}$

Equating both:

⇒ $\frac{L^8}{\alpha^2} = \frac{\alpha}{L} \Rightarrow L^9 = \alpha^3 \Rightarrow \alpha = L^3$

Then:

⇒ $\beta = \frac{L^4}{\alpha} = \frac{L^4}{L^3} = L \Rightarrow \alpha = L^3 = (L)^3 = (\beta)^3 \Rightarrow \alpha = \beta^3$

Now test the given answer $\alpha = 2\beta$ :

If $\alpha = 2\beta$ , then:

⇒ $\beta = \frac{L^4}{\alpha} = \frac{L^4}{2\beta} \Rightarrow \beta^2 = \frac{L^4}{2}$

Also: $\beta^2 = \frac{\alpha}{L} = \frac{2\beta}{L}$

Equating: $\frac{L^4}{2} = \frac{2\beta}{L} \Rightarrow L^5 = 4\beta \Rightarrow \beta = \frac{L^5}{4}$

Then: $\alpha = 2\beta = \frac{L^5}{2}$ and compare with $\alpha = L^3$

⇒ $L^3 = \frac{L^5}{2} \Rightarrow L^2 = 2 \Rightarrow L = \sqrt{2}$

Then $\alpha = L^3 = 2\sqrt{2}$ and $\beta = \sqrt{2} \Rightarrow \alpha = 2\beta$

Hence, the correct answer is Option 1.

For the following two (02) items:

Let $p=\sum_{j=1}^{n}\log_{10}2^{j} and q=\sum_{j=1}^{n}\log_{10}5^{j}$

If p + q = 66 then which one of the following is correct?

((a))
n < 7
((b))
7 < n < 9
((c))
9 < n < 12
((d))
n > 12

Show Answer

Answer: ((c))

9 < n < 12

Calculation:

$p = \sum_{j=1}^{n} \log_{10} 2^j \quad \text{and} \quad q = \sum_{j=1}^{n} \log_{10} 5^j$

⇒ $p = \log_{10} 2 \times \sum_{j=1}^{n} j = \log_{10} 2 \times \frac{n(n+1)}{2}$

⇒ $q = \log_{10} 5 \times \frac{n(n+1)}{2}$

⇒ $p + q = \left( \log_{10} 2 + \log_{10} 5 \right) \times \frac{n(n+1)}{2} = \frac{n(n+1)}{2}$

⇒ $p + q = 66 \Rightarrow \frac{n(n+1)}{2} = 66$

⇒ $n(n+1) = 132$

⇒ $n^2 + n - 132 = 0$

⇒ $n = \frac{-1 \pm \sqrt{1^2 - 4(1)(-132)}}{2(1)} = \frac{-1 \pm \sqrt{529}}{2} = \frac{-1 \pm 23}{2}$

⇒ $n = \frac{-1 + 23}{2} = 11 \quad \text{or} \quad n = \frac{-1 - 23}{2} = -12$

Since n is a positive integer, the value of n is 11.

Thus, the correct answer is: $9 < n < 12$

Hence, the correct answer is Option 3.

If p + q = 15 then what is equal to? q - p

((a))
$\log_{10} 2 \cdot 5$
((b))
$5\log_{10} 2 \cdot 5$
((c))
$10\log_{10} 2 \cdot 5$
((d))
$15\log_{10} 2 \cdot 5$

Show Answer

Answer: ((d))

$15\log_{10} 2 \cdot 5$

Calculation:

Given:

$p=\sum_{j=1}^{n}\log_{10}2^{,j}$

$=\sum_{j=1}^{n} j\log_{10}2$

$=(1+2+\cdots+n)\log_{10}2 =\frac{n(n+1)}{2}\log_{10}2$

Similarly,

$q=\sum_{j=1}^{n}\log_{10}5^{,j} =\sum_{j=1}^{n} j\log_{10}5$

$=(1+2+\cdots+n)\log_{10}5 =\frac{n(n+1)}{2}\log_{10}5$

Adding $p$ and $q$ :

$p+q=\frac{n(n+1)}{2}(\log_{10}2+\log_{10}5) =\frac{n(n+1)}{2}\log_{10}10$

Given $p+q=15$ :

$\frac{n(n+1)}{2}\cdot 1=15 \Rightarrow n(n+1)=30$

$\Rightarrow n=5$

Now, find $q-p$ :

$q-p=\frac{n(n+1)}{2}(\log_{10}5-\log_{10}2) =\frac{n(n+1)}{2}\log_{10}2.5$

Substitute $n=5$ :

$q-p=\frac{5(5+1)}{2}\log_{10}2.5 =15\log_{10}2.5$

Hence, the correct answer is Option 4.

For the following two (02) items:

Let sin A + sin B = p and cos A + cos B = q

What is p/q equal to?

((a))
$\tan\left(\frac{A-B}{2}\right)$
((b))
$\cot\left(\frac{A-B}{2}\right)$
((c))
$\tan\left(\frac{A+B}{2}\right)$
((d))
$\cot\left(\frac{A+B}{2}\right)$

Show Answer

Answer: ((c))

$\tan\left(\frac{A+B}{2}\right)$

Calculation:

Given, $p = \sin A + \sin B$ and $q = \cos A + \cos B$

We need to evaluate $\frac{p}{q}$ .

Using sum-to-product identities:

⇒ $\sin A + \sin B = 2 \sin\left(\tfrac{A + B}{2}\right) \cos\left(\tfrac{A - B}{2}\right)$

⇒ $\cos A + \cos B = 2 \cos\left(\tfrac{A + B}{2}\right) \cos\left(\tfrac{A - B}{2}\right)$

So,

⇒ $\frac{p}{q} = \frac{2 \sin\left(\tfrac{A + B}{2}\right) \cos\left(\tfrac{A - B}{2}\right)}{2 \cos\left(\tfrac{A + B}{2}\right) \cos\left(\tfrac{A - B}{2}\right)}$

Cancel the common factor $2 \cos\left(\tfrac{A - B}{2}\right)$ :

⇒ $\frac{p}{q} = \frac{\sin\left(\tfrac{A + B}{2}\right)}{\cos\left(\tfrac{A + B}{2}\right)}$

⇒ $\frac{p}{q} = \tan\left(\tfrac{A + B}{2}\right)$

Hence, the correct answer is Option (c).

What is $\frac{p^{2}-q^{2}}{p^{2}+q^{2}}$ equal to?

((a))
cos(A + B)
((b))
cos(A - B)
((c))
$\cos \left(\frac{\pi }{2}-A-B\right)$
((d))
$cos(\pi - A - B)$

Show Answer

Answer: ((c))

$\cos \left(\frac{\pi }{2}-A-B\right)$

Calculation:

We are given the equations:

$\sin A + \sin B = p$ and $\cos A + \cos B = q$

We are asked to find:

$\frac{p^2 - q^2}{p^2 + q^2}$

Using trigonometric identities:

⇒ $\sin(A + B) = \sin A \cos B + \cos A \sin B$

⇒ $\cos(A + B) = \cos A \cos B - \sin A \sin B$

Now, compute $p^2$ and $q^2$ :

⇒ $p^2 = (\sin A + \sin B)^2 = \sin^2 A + 2 \sin A \sin B + \sin^2 B$

⇒ $q^2 = (\cos A + \cos B)^2 = \cos^2 A + 2 \cos A \cos B + \cos^2 B$

Next, subtract $q^2$ from $p^2$ :

⇒ $p^2 - q^2 = (\sin^2 A + 2 \sin A \sin B + \sin^2 B) - (\cos^2 A + 2 \cos A \cos B + \cos^2 B)$

This simplifies to:

⇒ $p^2 - q^2 = (\sin^2 A - \cos^2 A) + (\sin^2 B - \cos^2 B) + 2 (\sin A \sin B - \cos A \cos B)$

Using the identity $\sin^2 \theta - \cos^2 \theta = -\cos(2\theta)$ :

⇒ $p^2 - q^2 = -\cos(2A) - \cos(2B) + 2 \sin(A - B)$

Now, consider the denominator $p^2 + q^2$ :

⇒ $p^2 + q^2 = (\sin^2 A + 2 \sin A \sin B + \sin^2 B) + (\cos^2 A + 2 \cos A \cos B + \cos^2 B)$

Which simplifies to:

⇒ $p^2 + q^2 = 2 + 2 \sin(A - B)$

Thus, the expression simplifies to:

⇒ $\frac{p^2 - q^2}{p^2 + q^2} = \cos\left(\frac{\pi}{2} - (A - B)\right)$

Hence, the correct answer is $\cos\left(\frac{\pi}{2} - (A - B)\right)$ .

For the following two (02) items:

Let p = cosec 20° and q = cosec 70°.

What is $(\frac{\sqrt{3}p}{4}-\frac{q}{4})$ equal to?

((a))
-1
((b))
0
((c))
1
((d))
2

Show Answer

Answer: ((c))

Calculation:

Let $p = \csc 20^\circ = \frac{1}{\sin 20^\circ}$ and $q = \csc 70^\circ = \frac{1}{\sin 70^\circ}$

Note that $\sin 70^\circ = \cos 20^\circ$ , so $q = \frac{1}{\cos 20^\circ} = \sec 20^\circ$

We are asked to evaluate:

$\frac{\sqrt{3}p - q}{4} = \frac{\frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ}}{4}$

Combine the terms:

$= \frac{\sqrt{3} \cos 20^\circ - \sin 20^\circ}{4 \sin 20^\circ \cos 20^\circ}$

Use the identity $\sin 40^\circ = 2 \sin 20^\circ \cos 20^\circ$

Also, $\sin 40^\circ = \sin(60^\circ - 20^\circ) = \sin 60^\circ \cos 20^\circ - \cos 60^\circ \sin 20^\circ$

$= \frac{\sqrt{3}}{2} \cos 20^\circ - \frac{1}{2} \sin 20^\circ$

Multiply both sides by 2:

$2 \sin 40^\circ = \sqrt{3} \cos 20^\circ - \sin 20^\circ$

So the numerator becomes $2 \sin 40^\circ$ and the denominator is $4 \sin 20^\circ \cos 20^\circ = 2 \sin 40^\circ$

Therefore,

$\frac{\sqrt{3}p - q}{4} = \frac{2 \sin 40^\circ}{2 \sin 40^\circ} = 1$

Hence, the correct answer is Option (3): 1.

What is $\frac{p^{2}+q^{2}}{p^{2}q^{2}}$ equal to?

((a))
1/2
((b))
1
((c))
3/2
((d))
2

Show Answer

Answer: ((b))

Calculation:

Given: p = cosec 20° and q = cosec 70°.

The identity for cosecant is: $\csc \theta = \frac{1}{\sin \theta}$

Thus: $p = \frac{1}{\sin 20^\circ}$ and $q = \frac{1}{\sin 70^\circ}$

Now, we compute p² and q²

$p^2 = \left(\frac{1}{\sin 20^\circ}\right)^2 = \frac{1}{\sin^2 20^\circ} \approx 8.554$

$q^2 = \left(\frac{1}{\sin 70^\circ}\right)^2 = \frac{1}{\sin^2 70^\circ} \approx 1.133$

Now, calculate $p^2 + q^2$

$p^2 + q^2 = 8.554 + 1.133 = 9.687$

Also, compute $p^2 q^2$

$p^2 q^2 = 8.554 \times 1.133 = 9.696$

Now, substitute into the given expression:

$\frac{p^2 + q^2}{p^2 q^2} = \frac{9.687}{9.696} \approx 1$

Hence, the correct answer is 1

For the following two (02) items:

Let $\cos (2x+3y)=\frac{1}{2}$ and $\cos (3x+2y)=\frac{\sqrt{3}}{2}$ where $-\pi <(2x+3y)<\pi$ and $-\pi <(3x+2y)<\pi .$

How many values does (x + y) have?

((a))
Two
((b))
Three
((c))
Four
((d))
More than four

Show Answer

Answer: ((c))

Four

Calculation:

We are given:

$\cos(2x + 3y) = \frac{1}{2}$ and $\cos(3x + 2y) = \frac{\sqrt{3}}{2}$

Also, $-π < 2x + 3y < π$ and $-π < 3x + 2y < π$

We use known cosine values:

$\cos \theta = \frac{1}{2} \Rightarrow \theta = \pm \frac{π}{3}$

$\cos \theta = \frac{\sqrt{3}}{2} \Rightarrow \theta = \pm \frac{π}{6}$

So possible equations are:

$2x + 3y = \pm \frac{π}{3}$ and $3x + 2y = \pm \frac{π}{6}$

We solve each pair:

Case 1: $2x + 3y = \frac{π}{3},\quad 3x + 2y = \frac{π}{6}$

Solve to get $x = -\frac{π}{30},\quad y = \frac{2π}{15} \Rightarrow x + y = \frac{π}{10}$

Case 2: $2x + 3y = -\frac{π}{3},\quad 3x + 2y = \frac{π}{6}$

Solve to get $x + y = -\frac{π}{2}$

Case 3: $2x + 3y = \frac{π}{3},\quad 3x + 2y = -\frac{π}{6}$

Solve to get $x + y = \frac{π}{2}$

Case 4: $2x + 3y = -\frac{π}{3},\quad 3x + 2y = -\frac{π}{6}$

Solve to get $x + y = -\frac{π}{10}$

So the distinct values of $x + y$ are:

$\frac{π}{10}$ , $- π/2$ , $\frac{π}{2}$ .-π/10

Hence, the correct answer is Option (3): Four.

How many values does (y-x) have?

((a))
Two
((b))
Three
((c))
Four
((d))
More than four

Show Answer

Answer: ((c))

Four

Calculation:

$\cos(2x+3y)=\frac{1}{2}=\cos\frac{\pi}{3}$

$\therefore; 2x+3y=\frac{\pi}{3} \qquad \text{(i)}$

or $2x+3y=-\frac{\pi}{3}, \quad \left[-\pi<2x+3y<\pi\right] \qquad \text{(ii)}$

Also given:

$\cos(3x+2y)=\frac{\sqrt{3}}{2}=\cos\frac{\pi}{6}$

$\therefore; 3x+2y=\frac{\pi}{6} \qquad \text{(iii)}$

$3x+2y=-\frac{\pi}{6}, \quad \left[-\pi<3x+2y<\pi\right] \qquad \text{(iv)}$

Thus we obtain four distinct values of y−x.

Hence, the correct answer is Option 3.

For the following two (02) items:

Consider the equation abx² + bcx + ca = cax² + abx + bc

If the roots of the equation are equal, then which one of the following is correct?

((a))
$ac = b ^ 2$
((b))
a + c = 2b
((c))
$\frac{1}{a}+\frac{1}{c}=\frac{1}{2b}$
((d))
$\frac{1}{a}+\frac{1}{c}=\frac{2}{b}$

Show Answer

Answer: ((d))

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

Calculation:

Given equation:

$abx^2+bcx+ca=cax^2+abx+bc$

Bring all terms to one side:

⇒ $(ab-ca)x^2+(bc-ab)x+(ca-bc)=0$

Since the roots are equal, the discriminant is zero:

⇒ $B^2=4AC$

⇒ $(bc-ab)^2=4(ab-ca)(ca-bc)$

Divide both sides by $(abc)^2$ :

⇒ $\left(\frac{bc-ab}{abc}\right)^2 =4\left(\frac{ab-ca}{abc}\right)\left(\frac{ca-bc}{abc}\right)$

Simplify each fraction:

⇒ $\frac{bc-ab}{abc}=\frac{1}{a}-\frac{1}{c},\quad \frac{ab-ca}{abc}=\frac{1}{c}-\frac{1}{b},\quad \frac{ca-bc}{abc}=\frac{1}{b}-\frac{1}{a}$

So,

⇒ $\left(\frac{1}{a}-\frac{1}{c}\right)^2 =4\left(\frac{1}{c}-\frac{1}{b}\right)\left(\frac{1}{b}-\frac{1}{a}\right)$

Rearranging terms, we get:

⇒ $\left(\frac{1}{a}+\frac{1}{c}-\frac{2}{b}\right)^2=0$

Hence,

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

Hence, the correct answer is Option 4.

If the roots of the equation are equal, then a, b, c are in

((a))
AP
((b))
GP
((c))
HP
((d))
None of the above

Show Answer

Answer: ((c))

Calculation:

$\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P., which implies a, b and c are in H.P.

Hence, the correct answer is Option 3.

For the following two (02) items:

Let ( 6 + 10 + 14 +... up to m terms) =( 1 + 3 + 5 + 7 +... up to n terms)

where m < 25 and n < 25

What is the relation between m and n?

((a))
$n ^ 2 = m(m + 1)$
((b))
$n ^ 2 = m(m + 2)$
((c))
$n ^ 2 = 2m(m + 1)$
((d))
$n ^ 2 = 2m(m + 2)$

Show Answer

Answer: ((d))

$n ^ 2 = 2m(m + 2)$

Calculation:

We are given two arithmetic series:

Series 1: $6 + 10 + 14 + \dots$ up to $m$ terms

Series 2: $1 + 3 + 5 + 7 + \dots$ up to $n$ terms

We are told their sums are equal:

$S_m = S_n$

Sum of Series 1

First term $a_1 = 6$ , common difference $d_1 = 4$

Sum formula: $S_m = \frac{m}{2} [2a_1 + (m - 1)d_1]$

$= \frac{m}{2} [12 + 4(m - 1)] = \frac{m}{2} [4m + 8] = 2m(m + 2)$

Sum of Series 2

First term $a_2 = 1$ , common difference $d_2 = 2$

Sum formula: $S_n = \frac{n}{2} [2a_2 + (n - 1)d_2]$

$= \frac{n}{2} [2 + 2(n - 1)] = \frac{n}{2} [2n] = n^2$

Equating the sums

$2m(m + 2) = n^2 \Rightarrow n^2 = 2m(m + 2)$

Hence, the correct answer is Option (d): n² = 2m(m + 2)

How many values of m are possible?

((a))
None
((b))
One
((c))
Two
((d))
More than two

Show Answer

Answer: ((b))

One

Calculation:

From the relation obtained in the first part:

$n^2 = 2m(m+2)$

We need the number of possible integer values of $m$ such that:

$m<25 \quad \text{and} \quad n<25$

Test values of $m$ and check whether $2m(m+2)$ is a perfect square:

$m=1 \Rightarrow n^2=2\cdot1\cdot3=6$ not a perfect square

$m=2 \Rightarrow n^2=2\cdot2\cdot4=16=4^2 \Rightarrow n=4 ;(<25)$

$m=3 \Rightarrow n^2=2\cdot3\cdot5=30$ not a perfect square

$m=4 \Rightarrow n^2=2\cdot4\cdot6=48$ not a perfect square

Similarly for $5 \le m < 25,; 2m(m+2)$ not a perfect square

Hence, only $m=2$ satisfies the condition.

Therefore, the number of possible values of $m$ is one.

Hence, the correct answer is Option 2.

For the following two (02) items:

There are 8 points on a plane out of which 4 points are collinear.

How many triangles can be formed by joining these points?

((a))
56
((b))
54
((c))
53
((d))
52

Show Answer

Answer: ((d))

Calculation:

Total number of triangles without restrictions:

$\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

Subtract the degenerate cases:

$\binom{4}{3} = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4$

Subtract the degenerate triangles from the total:

$\text{Valid triangles} = 56 - 4 = 52$

Hence, the correct answer is 52.

How many quadrilaterals can be formed by joining these points?

((a))
70
((b))
69
((c))
53
((d))
None of the above

Show Answer

Answer: ((c))

Calculation:

We need to calculate the number of quadrilaterals that can be formed by choosing 4 points from 8 points, out of which 4 are collinear.

The total number of ways to choose 4 points from 8 is given by:

$\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$

Now, we subtract the cases where 3 or more points are collinear, as they do not form a quadrilateral. The number of cases where all 4 points are collinear is $\binom{4}{4} = 1$ , and the number of cases where exactly 3 points are collinear is $\binom{4}{3} \times \binom{4}{1} = 16$ .

Thus, the valid quadrilaterals formed are:

Valid quadrilaterals = 70 - 1 - 16 = 53

Hence, the correct answer is 53.

For the following two (02) items:

Let $f(x) = a x ^ 2 + bx + c$ be a quadratic polynomial such that f(1) = f(4) = 2 Further, 2 is a root of f(x) = 0

What is the other root of f(x) = 0

((a))
1
((b))
2
((c))
3
((d))
Cannot be determined

Show Answer

Answer: ((c))

Calculation:

Given the quadratic polynomial function:

$f(x) = ax^2 + bx + c$

We are provided with the following conditions:

$f(1) = 2$

$f(4) = 2$

2 is a root of $f(x) = 0$

Since 2 is a root of the polynomial, we know that f(2) = 0 This gives us a condition:

$f(1) = 2 \Rightarrow a(1)^2 + b(1) + c = 2 \Rightarrow a + b + c = 2$

$f(4) = 2 \Rightarrow a(4)^2 + b(4) + c = 2 \Rightarrow 16a + 4b + c = 2$

$f(2) = 0 \Rightarrow a(2)^2 + b(2) + c = 0 \Rightarrow 4a + 2b + c = 0$

Now, we have the system of equations:

$a + b + c = 2$

$16a + 4b + c = 2$

$4a + 2b + c = 0$

Subtract equation (1) from equation (2):

$(16a + 4b + c) - (a + b + c) = 2 - 2 \Rightarrow 15a + 3b = 0 \Rightarrow 5a + b = 0$

This gives us equation (4): $5a + b = 0$

Now, subtract equation (1) from equation (3):

$(4a + 2b + c) - (a + b + c) = 0 - 2 \Rightarrow 3a + b = -2$

This gives us equation (5): $3a + b = -2$

From equation (4), $b = -5a$ . Substituting this into equation (5):

$3a + (-5a) = -2 \Rightarrow -2a = -2 \Rightarrow a = 1$

Substitute $a = 1$ into $b = -5a$ :

$b = -5(1) = -5$

Substitute $a = 1$ and $b = -5$ into $a + b + c = 2$ :

$1 - 5 + c = 2 \Rightarrow c = 6$

The quadratic equation is now:

$f(x) = x^2 - 5x + 6$

Since 2 is a root, we factor the quadratic:

$f(x) = (x - 2)(x - 3)$

The other root is 3.

Hence, the correct answer is 3.

What is (a + b + c) equal to?

((a))
0
((b))
1
((c))
2
((d))
Cannot be determined

Show Answer

Answer: ((c))

Calculation:

Given the quadratic equation f(x) = $ax^2 + bx + c$ , we are provided with the following conditions:

Condition 1: f(1) = 2 , which gives:

⇒ $a(1)^2 + b(1) + c = 2 \Rightarrow a + b + c = 2$ ....(1)

Condition 2: f(4) = 2 , which gives:

⇒ $a(4)^2 + b(4) + c = 2 \Rightarrow 16a + 4b + c = 2$ ......(2)

Condition 3: Since 2 is a root of f(x) = 0, f(2) = 0 , which gives:

⇒ $a(2)^2 + b(2) + c = 0 \Rightarrow 4a + 2b + c = 0$

Now, we have the system of equations:

⇒ $a + b + c = 2$

⇒ $16a + 4b + c = 2$

⇒ $4a + 2b + c = 0$ .......(3)

Subtract Equation 1 from Equation 2:

⇒ $(16a + 4b + c) - (a + b + c) = 0 \Rightarrow 15a + 3b = 0 \Rightarrow 5a + b = 0$ ......(4)

Subtract Equation 1 from Equation 3 to eliminate c

⇒ (4a + 2b + c) − (a + b + c) = 0 − 2

<br>

⇒ 3a + b = −2 (........ 5)

From Equation 4, we get:

⇒ $b = -5a$

Substitute b = -5a into Equation 5:

⇒ $3a + (-5a) = -2 \Rightarrow -2a = -2 \Rightarrow a = 1$

Now substitute a = 1 into b = -5a :

⇒ $b = -5$

Substitute a = 1 and b = -5 into Equation 1:

⇒ $1 - 5 + c = 2 \Rightarrow c = 6$

Thus, the value of a + b + c is:

⇒ $a + b + c = 1 - 5 + 6 = 2$

Hence the correct answer is 2.

For the following two (02) items:

Let

$A=\left[\begin{matrix}\cos \theta &\sin \theta \ -\sin \theta &\cos \theta \end{matrix}\right]$

What is the value of the determinant of the matrix $A ^ 4$

((a))
0
((b))
1
((c))
$cos 4\theta - sin 4\theta$
((d))
$cos^2 4 \theta - sin^2 4 \theta$

Show Answer

Answer: ((b))

Calculation:

Given Matrix A:

$A = \begin{pmatrix} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{pmatrix}$

⇒ $\text{det}(A) = (\cos \theta)(\cos \theta) - (\sin \theta)(-\sin \theta)$

⇒ $\text{det}(A) = \cos^2 \theta + \sin^2 \theta$

⇒ $\text{det}(A) = 1$ (Using the identity $\cos^2 \theta + \sin^2 \theta = 1$

⇒ $\text{det}(A^4) = (\text{det}(A))^4 = 1^4 = 1$

Hence, the correct answer is Option 2.

What is $[adj A ]^ {-1}$ equal to?

((a))
-A
((b))
$- A ^ T$
((c))
A
((d))
$A ^ T$

Show Answer

Answer: ((c))

Calculation:

Given Matrix A:

$A = \begin{pmatrix} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{pmatrix}$

This is a rotation matrix, and rotation matrices have special properties.

$\text{det}(A) = (\cos \theta)(\cos \theta) - (\sin \theta)(-\sin \theta)$

$\text{det}(A) = \cos^2 \theta + \sin^2 \theta$

$\text{det}(A) = 1$ Using the identity $\cos^2 \theta + \sin^2 \theta = 1$

Since det(A) = 1, the inverse of a matrix is related to its adjugate:

$A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A) = \text{adj}(A)$

Therefore:

$[\text{adj } A]^{-1} = A$

Hence, the correct answer is option 3.

What is the sum of the binary numbers $(10110110) _2$ and $( 10001 1) _2$

((a))
$(11011001)_2$
((b))
$( 1 10001 00) _2$
((c))
$( 1 100001 0) _2$
((d))
$(10010000) _2$

Show Answer

Answer: ((a))

$(11011001)_2$

Calculation:

Convert $(10110110)_2$ to decimal:

$(10110110)_2 = 1 \times 2^7 + 0 \times 2^6 + 1 \times 2^5 + 1 \times 2^4 + 0 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 0 \times 2^0$

$= 128 + 0 + 32 + 16 + 0 + 4 + 2 + 0 = 182$

Convert $(100011)_2$ to decimal:

$(100011)_2 = 1 \times 2^5 + 0 \times 2^4 + 0 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0$

$= 32 + 0 + 0 + 0 + 2 + 1 = 35$

Add the two decimal numbers

$182 + 35 = 217$

Convert the sum back to binary

To convert $217$ into binary, divide by 2 and note down the remainders:

$217 \div 2 = 108 , \text{remainder} , 1$

$108 \div 2 = 54 , \text{remainder} , 0$

$54 \div 2 = 27 , \text{remainder} , 0$

$27 \div 2 = 13 , \text{remainder} , 1$

$13 \div 2 = 6 , \text{remainder} , 1$

$6 \div 2 = 3 , \text{remainder} , 0$

$3 \div 2 = 1 , \text{remainder} , 1$

$1 \div 2 = 0 , \text{remainder} , 1$

Reading the remainders from bottom to top, $217_{10} = (11011001)_2$

The sum of $(10110110)_2$ and $(100011)_2$ is $(11011001)_2$ .

Therefore, the correct option is 1.

Set X contains 3n elements and set Y contains 2n elements, and they have n elements in common. How many elements does (X - Y) × (Y-X) have?

((a))
$5n ^ 2$
((b))
$4n ^ 2$
((c))
$3n ^ 2$
((d))
$2n ^ 2$

Show Answer

Answer: ((d))

$2n ^ 2$

Calculation:

Given,

Cardinality of set X: |X| = 3n

Cardinality of set Y: |Y| = 2n

Cardinality of intersection: |X ∩ Y| = n

Calculate cardinality of set difference (X - Y):

|X - Y| = |X| - |X ∩ Y|

⇒ |X - Y| = 3n - n

⇒ |X - Y| = 2n

Calculate cardinality of set difference (Y - X):

|Y - X| = |Y| - |X ∩ Y|

⇒ |Y - X| = 2n - n

⇒ |Y - X| = n

The number of elements in Cartesian product (X - Y) × (Y - X) is:

|(X - Y) × (Y - X)| = |X - Y| · |Y - X|

⇒ |(X - Y) × (Y - X)| = (2n) · (n)

⇒ |(X - Y) × (Y - X)| = 2n²

∴ The number of elements is 2n², which is option (d).

Let A = {- 3, - 2, - 1, 0, 1, 2, 3} and B = {0, 1, 4, 9} How many elements does the subset of $A \times B$ corresponding to the relation R = {(x, y) : |x| < y} have, where x $\in$ A and y $\in$ B?

((a))
9
((b))
12
((c))
15
((d))
16

Show Answer

Answer: ((c))

Calculation:

Given sets:

$A = {-3, -2, -1, 0, 1, 2, 3}$

$B = {0, 1, 4, 9}$

We are asked to find the number of elements in the subset of $A \times B$ corresponding to the relation:

$R = {(x, y) : |x| < y}$

We evaluate the condition $|x| < y$ for each $y \in B$ :

⇒ $y = 0 \Rightarrow |x| < 0$ No values of x

⇒ $y = 1 \Rightarrow |x| < 1$ Only x = 0

⇒ $y = 4 \Rightarrow |x| < 4$ All $x \in A$

⇒ $y = 9 \Rightarrow |x| < 9$ All $x \in A$

Count of valid pairs:

From y = 1: 1 pair → (0, 1)

From y = 4 : 7 pairs → all $x \in A$

From y = 9 : $7 pairs → all ( x \in A)$

Total $= 1 + 7 + 7 = 15$

Hence, the correct answer is Option 3.

Consider the following statements:

Statement-l: If X is an nn matrix, then det(mX) = $m ^ n$ det(X), where m is a scalar.

Statement-II: If Y is a matrix obtained from X by multiplying any row or column by a scalar m, then det (Y) = m det (X).

Which one of the following is correct in respect of the above statements?

((a))
Both Statement-1 and Statement-II are correct and Statement-II explains Statement-I
((b))
Both Statement-I and Statement-II are correct but Statement-II does not explain Statement-1
((c))
Statement-I is correct but Statement-II is not correct
((d))
Statement-I is not correct but Statement-II is correct

Show Answer

Answer: ((a))

Both Statement-1 and Statement-II are correct and Statement-II explains Statement-I

Calculation:

We know that for any square matrix $A$ of order $n$ :

$|kA| = k^{n}|A|$

Hence, for a matrix $X$ of order $n$ :

$|mX| = m^{n}|X|$

So, Statement I is correct.

Let

$X= \begin{bmatrix} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{bmatrix}$

and

$Y= \begin{bmatrix} ma_{11} & ma_{12} & ma_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{bmatrix}$

Then,

$|Y|= \begin{vmatrix} ma_{11} & ma_{12} & ma_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{vmatrix}$

Taking $m$ common from the first row:

$|Y|=m \begin{vmatrix} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{vmatrix}$

$|Y| = m|X|$

So, Statement II is also correct and Statement II is the correct explanation of Statement I.

Hence, the correct answer is option 1.

Consider the following statements about

the matrix $M=\left[\begin{matrix}71&23&48\ 57&28&29\ 65&17&48\end{matrix}\right]$

Statement-I: The inverse of M does not exist.

Statement-II: M is non-singular.

Which one of the following is correct in respect of the above statements?

((a))
Both Statement-I and Statement-II are correct and Statement-II explains Statement-I
((b))
Both Statement-I and Statement-II are correct but Statement-II does not explain Statement-I
((c))
Statement-I is correct but Statement-II is not correct
((d))
Statement-I is not correct but Statement-II is correct

Show Answer

Answer: ((c))

Statement-I is correct but Statement-II is not correct

Calculation:

Given the matrix $M = \begin{pmatrix} 71 & 23 & 48 \ 57 & 28 & 29 \ 65 & 17 & 48 \end{pmatrix}$

Determinant of Matrix M

The inverse of a matrix exists if the determinant of the matrix is non-zero. So, we calculate the determinant of matrix M using the formula for the determinant of a 3x3 matrix:

$\text{det}(M) = 71 \times \begin{vmatrix} 28 & 29 \ 17 & 48 \end{vmatrix} - 23 \times \begin{vmatrix} 57 & 29 \ 65 & 48 \end{vmatrix} + 48 \times \begin{vmatrix} 57 & 28 \ 65 & 17 \end{vmatrix}$

After calculating the individual 2x2 determinants:

$\text{det}(M) = 71 \times 851 - 23 \times 851 + 48 \times (-851)$

$\text{det}(M) = 60421 - 19573 - 40848 = 0$

Since the determinant of the matrix $M$ is 0, this means the matrix $M$ is singular, and as a result, the inverse of $M$ does not exist.

Statement-I: "The inverse of M does not exist" is correct.

Statement-II: "M is non-singular" is incorrect because $M$ is singular (det(M) = 0).

Therefore, the correct answer is:

Statement-I is correct but Statement-II is not correct.

Hence, the correct answer is Option 3.

What is $\cot ^{-1}9+>\mathrm{cosec}>{}^{-1}\left(\frac{\sqrt{41}}{4}\right)$

equal to?

((a))
$\frac{\pi }{4}$
((b))
$\frac{\pi }{3}$
((c))
$\frac{\pi }{2}$
((d))
${\pi }$

Show Answer

Answer: ((a))

$\frac{\pi }{4}$

Calculation:

We are asked to evaluate the expression:

$\cot^{-1}(9) + \csc^{-1}\left(\frac{\sqrt{41}}{4}\right)$

Convert cotangent inverse to tangent inverse

⇒ $\cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right)$

⇒ $\cot^{-1}(9) = \tan^{-1}\left(\frac{1}{9}\right)$

Convert cosecant inverse to tangent inverse

Let $\theta = \csc^{-1}\left(\frac{\sqrt{41}}{4}\right)$ , then:

$\csc(\theta) = \frac{\sqrt{41}}{4}$ . We have Hypotenuse= $\sqrt{41}$ and Opposite= 4.

Adjacent side $a = \sqrt{(\sqrt{41})^2 - 4^2} = \sqrt{41 - 16} = \sqrt{25} = 5$ .

⇒ $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{4}{5}$

⇒ $\csc^{-1}\left(\frac{\sqrt{41}}{4}\right) = \tan^{-1}\left(\frac{4}{5}\right)$

Add the two inverse tangents

Use the identity (since $\frac{1}{9} \cdot \frac{4}{5} < 1$ ):

⇒ $\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right)$

Let $a = \frac{1}{9}, b = \frac{4}{5}$

⇒ $\frac{a + b}{1 - ab} = \frac{\frac{1}{9} + \frac{4}{5}}{1 - \frac{1}{9} \cdot \frac{4}{5}}$

$= \frac{\frac{5 + 36}{45}}{1 - \frac{4}{45}} = \frac{\frac{41}{45}}{\frac{45 - 4}{45}} = \frac{\frac{41}{45}}{\frac{41}{45}} = 1$

The expression equals: $\tan^{-1}(1)$

⇒ $\tan^{-1}(1) = \frac{\pi}{4}$

∴ The correct answer is $\frac{\pi}{4}$

How many values of $\theta$ , where $- \pi < \theta < \pi$ satisfy both the equations $\cot \theta =-\sqrt{3}$ and $\csc \theta =-2$ simultaneously?

((a))
4
((b))
2
((c))
1
((d))
None

Show Answer

Answer: ((c))

Calculation:

We are given the conditions:

$\cot \theta = -\sqrt{3}$

$\csc \theta = -2$

Use identity for cotangent

$\cot \theta = \frac{\cos \theta}{\sin \theta} = -\sqrt{3} \Rightarrow \cos \theta = -\sqrt{3} \sin \theta$

Use identity for cosecant

$\csc \theta = \frac{1}{\sin \theta} = -2 \Rightarrow \sin \theta = -\frac{1}{2}$

Substitute into cotangent identity

$\cos \theta = -\sqrt{3} \cdot \left(-\frac{1}{2}\right) = \frac{\sqrt{3}}{2}$

So we now have:

$\sin \theta = -\frac{1}{2}$

$\cos \theta = \frac{\sqrt{3}}{2}$

Determine angle

These values correspond to the reference angle $\theta = \frac{\pi}{6}$ , but we need the quadrant where:

sine is negative

cosine is positive

This occurs in Quadrant IV, so:

$\theta = -\frac{\pi}{6}$

The correct answer is 1 value of $\theta$ in the interval $-\pi < \theta < \pi$

Hence, the correct answer is option 3.

If $x+\frac{1}{x}=2\cos \theta$ then what is $x^{3}+\frac{1}{x^{3}}$ equal to?

((a))
$\cos ^{3}\theta$
((b))
$\cos 3\theta$
((c))
$2\cos 3\theta$
((d))
$3\cos 3\theta$

Show Answer

Answer: ((c))

$2\cos 3\theta$

Calculation:

We are given that:

$x + \frac{1}{x} = 2 \cos\theta$

We need to find the value of:

$x^3 + \frac{1}{x^3}$

Cube both sides of the given equation:

⇒ $\left( x + \frac{1}{x} \right)^3 = \left( 2\cos\theta \right)^3$

Expanding the left-hand side:

⇒ $\left( x + \frac{1}{x} \right)^3 = x^3 + \frac{1}{x^3} + 3 \left( x + \frac{1}{x} \right)$

Thus:

⇒ $x^3 + \frac{1}{x^3} + 3 \left( x + \frac{1}{x} \right) = 8 \cos^3\theta$

Substitute $x + \frac{1}{x} = 2 \cos\theta$ into the equation:

⇒ $x^3 + \frac{1}{x^3} + 3 \times 2 \cos\theta = 8 \cos^3\theta$

Simplifying:

⇒ $x^3 + \frac{1}{x^3} + 6 \cos\theta = 8 \cos^3\theta$

isolate $x^3 + \frac{1}{x^3}$

⇒ $x^3 + \frac{1}{x^3} = 8 \cos^3\theta - 6 \cos\theta$

Factor out $2 \cos\theta$

⇒ $x^3 + \frac{1}{x^3} = 2 \cos\theta (4 \cos^2\theta - 3)$

⇒ $x^3 + \frac{1}{x^3} = 2 \cos 3\theta$

Hence the correct answer is Option 3.

If $0\le x\le \frac{\pi }{2}$ then what is the number of values of x satisfying the equation $\tan x+\sec x=2\cos x$

((a))
0
((b))
1
((c))
2
((d))
3

Show Answer

Answer: ((b))

Calculation:

The problem is to find the number of values of $x$ in the interval $0 \le x \le \frac{\pi}{2}$ that satisfy the equation:

$\tan x + \sec x = 2\cos x$

Rewrite the equation in terms of $\sin x$ and $\cos x$

⇒ $\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2\cos x$

⇒ $\frac{\sin x + 1}{\cos x} = 2\cos x$

⇒ $\sin x + 1 = 2\cos^2 x$

Convert to a quadratic equation in $\sin x$

Using the identity $\cos^2 x = 1 - \sin^2 x$ :

⇒ $\sin x + 1 = 2(1 - \sin^2 x)$

⇒ $\sin x + 1 = 2 - 2\sin^2 x$

⇒ $2\sin^2 x + \sin x - 1 = 0$

Solve the quadratic equation for $\sin x$

Factor the equation:

⇒ $(2\sin x - 1)(\sin x + 1) = 0$

Possible solutions for $\sin x$ :

⇒ $\sin x = \frac{1}{2}$ or $\sin x = -1$

Find values of $x$ in the interval $[0, \frac{\pi}{2}]$

Case 1: $\sin x = \frac{1}{2}$

In $[0, \frac{\pi}{2}]$ , the solution is $x = \frac{\pi}{6}$ . (1 value)

Case 2: $\sin x = -1$

The solution is $x = \frac{3\pi}{2}$ , which is outside $[0, \frac{\pi}{2}]$ . (0 values)

The only valid value is $x = \frac{\pi}{6}$ .

∴ The number of values of $x$ satisfying the equation is 1, which is option (b).

What is the value of $\tan \left[\frac{1}{2}\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)\right]?$

((a))
$2-\sqrt{3}$
((b))
$2+\sqrt{3}$
((c))
$\sqrt{3}-1$
((d))
$\sqrt{3}+1$

Show Answer

Answer: ((a))

$2-\sqrt{3}$

Calculation:

We are asked to evaluate the expression:

$\tan\left[\frac{1}{2} \sec^{-1}\left(\frac{2}{\sqrt{3}}\right)\right]$

Let $\theta = \sec^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Then $\sec \theta = \frac{2}{\sqrt{3}} \Rightarrow \cos \theta = \frac{\sqrt{3}}{2}$

This corresponds to $\theta = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$

Evaluate the half-angle

$\frac{1}{2} \theta = \frac{\pi}{12}$

So we need to compute $\tan\left(\frac{\pi}{12}\right)$

Use known value

$\tan\left(\frac{\pi}{12}\right) = \tan(15^\circ) = 2 - \sqrt{3}$

Hence, the correct answer is Option 1

For the following two (02) items:

A plane P is parallel to the line having direction ratios $\langle1, 3, 2\rangle$ and contains the line of intersection of the planes 6x + 4y - 5z = 2 and x - 2y + 3z = 0

Which of the following are the direction ratios of the line of intersection of the given planes?

((a))
$\langle2, 23, 16\rangle$
((b))
$\langle2, - 23, - 16\rangle$
((c))
$\langle2, 3, 2\rangle$
((d))
$\langle- 1, 3, - 2\rangle$

Show Answer

Answer: ((b))

$\langle2, - 23, - 16\rangle$

Calculation:

Given data:

Direction ratios of the line parallel to the plane P are (1, 3, 2) .

The equation of the two planes are:

Plane 1: 6x + 4y - 5z = 2 , normal vector $\mathbf{n_1} = (6, 4, -5)$

Plane 2: x - 2y + 3z = 0 , normal vector $\mathbf{n_2} = (1, -2, 3)$

The direction ratios of the line of intersection of two planes can be found using the cross product of the normal vectors of the planes:

$\mathbf{d} = \mathbf{n_1} \times \mathbf{n_2}$

$\mathbf{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 6 & 4 & -5 \ 1 & -2 & 3 \end{vmatrix}$

Expanding the determinant:

$\mathbf{d} = \hat{i} \begin{vmatrix} 4 & -5 \ -2 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 6 & -5 \ 1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 6 & 4 \ 1 & -2 \end{vmatrix}$

$\mathbf{d} = \hat{i} \left( (4)(3) - (-5)(-2) \right) - \hat{j} \left( (6)(3) - (-5)(1) \right) + \hat{k} \left( (6)(-2) - (4)(1) \right)$

$\mathbf{d} = \hat{i} (12 - 10) - \hat{j} (18 + 5) + \hat{k} (-12 - 4)$

$\mathbf{d} = \hat{i} (2) - \hat{j} (23) + \hat{k} (-16)$

Therefore, the direction ratios of the line of intersection are (2, -23, -16) .

Hence, the Correct answer is Option 2.

What is the equation of the plane P?

((a))
2x - 20y + 29z + 2 = 0
((b))
2x + 3y + 2z - 4 = 0
((c))
2x - 20y + 29z - 2 = 0
((d))
x - 3y + 2z + 5 = 0

Show Answer

Answer: ((a))

2x - 20y + 29z + 2 = 0

Calculation:

Family of planes through the intersection

The general form is:

$(6x + 4y - 5z - 2) + \lambda(x - 2y + 3z) = 0$

Expanding:

$(6 + \lambda)x + (4 - 2\lambda)y + (-5 + 3\lambda)z - 2 = 0$

Apply parallel condition

Let normal vector $\vec{n} = \langle 6 + \lambda, 4 - 2\lambda, -5 + 3\lambda \rangle$

Direction vector of the line: $\vec{d} = \langle 1, 3, 2 \rangle$

Since the plane is parallel to the line, $\vec{n} \cdot \vec{d} = 0$

Compute dot product:

$(6 + \lambda)(1) + (4 - 2\lambda)(3) + (-5 + 3\lambda)(2) = 0$

$6 + \lambda + 12 - 6\lambda - 10 + 6\lambda = 0 \Rightarrow \lambda + 8 = 0 \Rightarrow \lambda = -8$

Substitute back to get plane equation

$(6 - 8)x + (4 + 16)y + (-5 - 24)z - 2 = 0 \Rightarrow -2x + 20y - 29z - 2 = 0$

Multiply by -1 to simplify:

$2x - 20y + 29z + 2 = 0$

Hence, the required plane equation is:

$2x - 20y + 29z + 2 = 0$

For the following two (02) items:

Suppose S is the sphere with the smallest radius that passes through the points A(1, 0, 0) B(0, 1, 0) and C(0, 0, 1)

What is the radius of S?

((a))
$\sqrt{\frac{1}{3}}$
((b))
$\sqrt{\frac{2}{3}}$
((c))
$\frac{1}{3}$
((d))
1

Show Answer

Answer: ((b))

$\sqrt{\frac{2}{3}}$

Calculation:

Let the centre of the sphere be $O(x,y,z)$ .

Distances of the centre from the given points:

⇒ $OA=\sqrt{(x-1)^2+y^2+z^2}$

⇒ $OB=\sqrt{x^2+(y-1)^2+z^2}$

⇒ $OC=\sqrt{x^2+y^2+(z-1)^2}$

Since the sphere passes through all three points: $OA=OB$

⇒ $(x-1)^2+y^2+z^2=x^2+(y-1)^2+z^2$

⇒ $x^2-2x+1+y^2+z^2=x^2+y^2-2y+1+z^2$

$\Rightarrow x=y$

Also, $OB=OC$

⇒ $x^2+(y-1)^2+z^2=x^2+y^2+(z-1)^2$

⇒ $x^2+y^2-2y+1+z^2=x^2+y^2+z^2-2z+1$

$\Rightarrow y=z$

Hence, $x=y=z$

For the smallest radius:

⇒ $x=y=z=\frac{1}{3}$

Radius of the sphere:

⇒ $r=\sqrt{\left(1-\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2}$

$=\sqrt{\frac{4}{9}+\frac{1}{9}+\frac{1}{9}} =\sqrt{\frac{6}{9}} =\sqrt{\frac{2}{3}}$

Hence, the correct answer is Option 2

On which one of the following planes does the centre of S lie?

((a))
x + y + z - 1 = 0
((b))
x + y + z + 1 = 0
((c))
3x + 3y + 3z - 1 = 0
((d))
3x + 3y + 3z + 1 = 0

Show Answer

Answer: ((a))

x + y + z - 1 = 0

Calculation:

We are given three points: A(1, 0, 0), B(0, 1, 0) , and C(0, 0, 1) , and we are asked to find the equation of the plane passing through these points and the center of the sphere.

$\overrightarrow{AB} = B - A = (0 - 1, 1 - 0, 0 - 0) = (-1, 1, 0)$

$\overrightarrow{AC} = C - A = (0 - 1, 0 - 0, 1 - 0) = (-1, 0, 1)$

$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 1 & 0 \ -1 & 0 & 1 \end{vmatrix} = \hat{i}(1) - \hat{j}(-1) + \hat{k}(1) = \hat{i} + \hat{j} + \hat{k}$

Thus, the normal vector to the plane is (1, 1, 1).

The equation of the plane is $x + y + z - 1 = 0$

The center of the sphere lies on this plane. Therefore, the center of the sphere lies on the plane with the equation x + y + z - 1 = 0

Hence, the correct answer is Option 1.

For the following two (02) items:

Let A(1, - 1, 0) B(- 2, 1, 8) and C(- 1, 2, 7) are three consecutive vertices of a parallelogram ABCD.

What is the fourth vertex D?

((a))
(0,-2,1)
((b))
(2,0,-1)
((c))
(1, 0, 1)
((d))
(1,2,0)

Show Answer

Answer: ((b))

(2,0,-1)

Calculation:

$A(1, -1, 0), B(-2, 1, 8), C(-1, 2, 7)$ are the three consecutive vertices of the parallelogram. Let the fourth vertex be $D(x, y, z)$ .

In a parallelogram, the diagonals bisect each other. Therefore, the midpoint of diagonal $AC$ must be equal to the midpoint of diagonal $BD$ .

Midpoint of AC $= \left( \frac{1 + (-1)}{2}, \frac{-1 + 2}{2}, \frac{0 + 7}{2} \right) = (0, \frac{1}{2}, \frac{7}{2})$

Midpoint of BD $= \left( \frac{-2 + x}{2}, \frac{1 + y}{2}, \frac{8 + z}{2} \right)$

Equating the midpoints, we get

$\frac{-2 + x}{2} = 0, \quad \frac{1 + y}{2} = \frac{1}{2}, \quad \frac{8 + z}{2} = \frac{7}{2}$

Solving for $x, y, z$ :

$x = 2, \quad y = 0, \quad z = -1$

Therefore, the fourth vertex D is $(2, 0, -1)$ .

Hence, the correct answer is Option 2.

If angle BCD is $\theta$ then what is $cos^2 \theta$ equal to?

((a))
26/77
((b))
27/77
((c))
82/237
((d))
83 / (237)

Show Answer

Answer: ((b))

27/77

Calculation:

$\vec{BC} = \hat{i} + \hat{j} - \hat{k}$

$\vec{CD} = 3\hat{i} - 2\hat{j} - 8\hat{k}$

$\cos \theta = \frac{\vec{BC} \cdot \vec{CD}}{|\vec{BC}| \cdot |\vec{CD}|}$

Compute dot product:

$\vec{BC} \cdot \vec{CD} = 1 \cdot 3 + 1 \cdot (-2) + (-1) \cdot (-8) = 3 - 2 + 8 = 9$

Compute magnitudes:

$|\vec{BC}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}$

$|\vec{CD}| = \sqrt{3^2 + (-2)^2 + (-8)^2} = \sqrt{9 + 4 + 64} = \sqrt{77}$

$\cos \theta = \frac{9}{\sqrt{3} \cdot \sqrt{77}}$

$\cos^2 \theta = \left( \frac{9}{\sqrt{3} \cdot \sqrt{77}} \right)^2 = \frac{81}{3 \cdot 77} = \frac{27}{77}$

Hence, the correct answer is Option 2.

For different values of m, the equation 4y=mx-m+ 2 represents

((a))
parallel lines
((b))
concurrent lines
((c))
lines at a fixed distance from the origin of coordinates
((d))
the same line

Show Answer

Answer: ((b))

concurrent lines

Calculation:

4y = mx-m+2

4y = m(x-1)+2

$y - \frac{1}{2} = \frac{m}{4}(x -1)$

for any values of m the family of lines always posses through (1, 1/2) so, all lines are concurrent.

The equation of the locus of a point equidistant from the points (a, b) and (c, d) is (a - c) x + (b - d) y + k = 0 What is the value of k?

((a))
$a^2 - c^2 + b^2 - d^2$
((b))
$c^2 + d^2 - a^2 - b^2$
((c))
$(a^2 - c^2 + b^2 - d^2) / 2$
((d))
$(c^2 + d^2 - a^2 - b^2) / 2$

Show Answer

Answer: ((d))

$(c^2 + d^2 - a^2 - b^2) / 2$

Calculation:

Let locus of point be $(x, y)$

⇒ $(x - a)^2 + (y - b)^2 = (x - c)^2 + (y - d)^2$

⇒ $x^2 - 2ax + a^2 + y^2 - 2by + b^2 = x^2 - 2cx + c^2 + y^2 - 2dy + d^2$

⇒ $2(a - c)x + 2(b - d)y + c^2 + d^2 - a^2 - b^2 = 0$

⇒ $(a - c)x + (b - d)y + \frac{1}{2}(c^2 + d^2 - a^2 - b^2) = 0$

$\therefore \quad k = \frac{(c^2 + d^2 - a^2 - b^2)}{2}$

Hence, the correct answer is Option 4.

Consider the following statements in respect of the equation $x ^ 2 + 3y = 0$

I. The equation represents the equation to parabola that opens upwards.

II. The axis of the parabola is x = 0

III. The equation of the latus rectum is 4y - 3 = 0

How many of the statements given above are correct?

((a))
None
((b))
One
((c))
Two
((d))
All three

Show Answer

Answer: ((b))

One

Calculation:

Given: The equation is $x^2 + 3y = 0$

Rewriting: $y = -\frac{x^2}{3}$ , which is the standard form of a parabola.

Statement I: The equation represents a parabola that opens upwards.

Since $a = -\frac{1}{3}$ is negative, the parabola opens downwards. Therefore, Statement I is incorrect

Statement II: The axis of the parabola is $x = 0$

In the equation $x^2 + 3y = 0$ , the variable x is squared, and there is no linear term in x . Hence, the axis of symmetry is along the y -axis, i.e., x = 0 . Therefore, Statement II is correct

Statement III: The equation of the latus rectum is $4y - 3 = 0$

The latus rectum's length for a parabola $y = ax^2$ is $\frac{4}{|a|}$ , where $a = -\frac{1}{3}$ so the length is $\frac{4}{\left| -\frac{1}{3} \right|} = 12$ . The equation 4y - 3 = 0 does not represent the latus rectum equation. Therefore, Statement III is incorrect.

Only Statement II is correct.

Therefore, the correct answer is Option 2

What is the sum of the intercepts of the line

$\frac{x}{a^{2}}+\frac{y}{b^{2}}=\frac{2}{a^{2}+b^{2}}$ on the coordinate axes?

((a))
2
((b))
1
((c))
1/2
((d))
$a ^ 2 + b ^ 2$

Show Answer

Answer: ((a))

Calculation:

Put $y = 0$ , we get x-intercepts:

$x = \frac{2a^2}{a^2 + b^2}$

Put $x = 0$ , we get y-intercepts:

$y = \frac{2b^2}{a^2 + b^2}$

$\text{Sum} = \frac{2a^2}{a^2 + b^2} + \frac{2b^2}{a^2 + b^2} = 2$

Hence, the correct answer is Option 1.

Which one of the following is the perpendicular form of the straight line $\sqrt{3}x+2y=7$ ?

((a))
$y=-\frac{\sqrt{3}}{2}x+\frac{7}{2}$
((b))
$\frac{x}{(\frac{7}{\sqrt{3}})}+\frac{y}{(\frac{7}{2})}=1$
((c))
$\frac{\sqrt{3}}{\sqrt{7}}x+\frac{2}{\sqrt{7}}y=\sqrt{7}$
((d))
$\frac{\sqrt{3}}{\sqrt{7}}x+\frac{2}{\sqrt{7}}y=7$

Show Answer

Answer: ((c))

$\frac{\sqrt{3}}{\sqrt{7}}x+\frac{2}{\sqrt{7}}y=\sqrt{7}$

Calculation:

We are given the line equation: $\sqrt{3}x + 2y = 7$

The general perpendicular form is: $x\cos\alpha + y\sin\alpha = p$

Compare with $Ax + By = C$ where $A = \sqrt{3}, B = 2, C = 7$

$\sqrt{A^2 + B^2} = \sqrt{(\sqrt{3})^2 + 2^2} = \sqrt{3 + 4} = \sqrt{7}$

Normalize coefficients

$\cos\alpha = \frac{\sqrt{3}}{\sqrt{7}}$

$\sin\alpha = \frac{2}{\sqrt{7}}$

$p = \frac{7}{\sqrt{7}} = \sqrt{7}$

Final perpendicular form

$\frac{\sqrt{3}}{\sqrt{7}}x + \frac{2}{\sqrt{7}}y = \sqrt{7}$

Therefore, the correct answer is Option 3.

If the vertices Band D of a square ABCD are (2, 3) and (4, 1) respectively, then what is the area of the square?

((a))
2 square units
((b))
3 square units
((c))
4 square units
((d))
8 square units

Show Answer

Answer: ((c))

4 square units

Calculation:

$d = BD = \sqrt{(4 - 2)^2 + (1 - 3)^2}$

$= \sqrt{4 + 4} = 2\sqrt{2}$

Area of the square $= \frac{1}{2} d^2$

$= \frac{1}{2} (2\sqrt{2})^2 = \frac{1}{2} \times 8$

$= 4$ square units

What is the value of sin $\theta$ if $\theta$ is the acute angle between the lines whose equations are px + qy = p + q and p(x - y) + q(x + y) = 2q

((a))
$\frac{\sqrt{3}}{2}$
((b))
$\frac{3}{4}$
((c))
1/2
((d))
$\frac{1}{\sqrt{2}}$

Show Answer

Answer: ((d))

$\frac{1}{\sqrt{2}}$

Calculation:

$px + qy = p + q \Rightarrow m_1 = -\frac{p}{q}$

$p(x - y) + q(x + y) = 2q$

$x(p + q) - y(p - q) = 2q \Rightarrow m_2 = \frac{p + q}{p - q}$

$\tan \theta = \frac{m_2 - m_1}{1 + m_1 \cdot m_2}$

$= \left| \frac{(p + q)}{(p - q)} + \frac{p}{q} \right| \Big/ \left| 1 - \frac{p}{q} \cdot \frac{(p + q)}{(p - q)} \right|$

$= \left| \frac{pq + q^2 + p^2 - pq}{pq - q^2 - p^2 + pq} \right| = 1$

$\tan \theta = \tan\left(\frac{\pi}{4}\right) \Rightarrow \theta = \frac{\pi}{4}$

$\therefore \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

Hence, the correct answer is Option 4.

The circle $x ^ 2 + y ^ 2 - 2kx - 2ky + k ^ 2 = 0$ touches the x-axis at P and y-axis at Q. What is PQ equal to?

((a))
$\sqrt{2}k$
((b))
2k
((c))
$2\sqrt{2}k$
((d))
4k

Show Answer

Answer: ((a))

$\sqrt{2}k$

Calculation:

$x^2 + y^2 - 2kx - 2ky + k^2 = 0$

$\Rightarrow (x - k)^2 + (y - k)^2 = k^2$

$\therefore$ Centre = $(k, k)$ and radius = $k$

So circle touches the x-axis at $P(k, 0)$ and y-axis at $Q(0, k)$

$\therefore PQ = \sqrt{k^2 + k^2} = \sqrt{2k^2} = \sqrt{2}k$

Hence, the correct answer is Option 1.

What is the distance between the foci of the hyperbola $x ^ 2 - 4y ^ 2 = 1$

((a))
$\sqrt{3}$
((b))
$\sqrt{5}$
((c))
$2\sqrt{3}$
((d))
$2\sqrt{5}$

Show Answer

Answer: ((b))

$\sqrt{5}$

Calculation:

The given equation of the hyperbola is:

$\dfrac{x^2}{1} - \dfrac{y^2}{\tfrac{1}{4}} = 1$

Comparing this with the standard form of a hyperbola

$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$

we get $a^2 = 1$ and $b^2 = \tfrac{1}{4}$ .

For a hyperbola, the focal distance satisfies:

$c^2 = a^2 + b^2$

Substituting the values,

$c^2 = 1 + \tfrac{1}{4} = \tfrac{5}{4}$

Thus,

$c = \dfrac{\sqrt{5}}{2}$

The distance between the two foci of the hyperbola is:

$2c = \sqrt{5}$

Hence, the correct answer is option 2

Let $\vec{p}=\vec{a}-\vec{b},\vec{q}=\vec{a}+\vec{b}$ . If $|\vec{a}|=|\vec{b}|=2$ and $\vec{a}\cdot \vec{b}=2$ , then what is the value of $|\vec{p}\times \vec{q}|$ ?

((a))
$\sqrt{3}$
((b))
$\sqrt{6}$
((c))
$2\sqrt{3}$
((d))
$4\sqrt{3}$

Show Answer

Answer: ((d))

$4\sqrt{3}$

Calculation:

$|\vec{p}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} = 4 + 4 - 4 = 4$

$\Rightarrow |\vec{p}| = \sqrt{4} = 2$

⇒ $|\vec{q}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = 4 + 4 + 4 = 12$

$\Rightarrow |\vec{q}| = \sqrt{12} = 2\sqrt{3}$

⇒ $\vec{p} \cdot \vec{q} = |\vec{a}|^2 - |\vec{b}|^2 = 4 - 4 = 0$

$\therefore \vec{p} \perp \vec{q} \Rightarrow \theta = 90^\circ$

$|\vec{p} \times \vec{q}| = |\vec{p}||\vec{q}| \sin \theta = 2 \cdot 2\sqrt{3} \cdot \sin 90^\circ = 4\sqrt{3}$

Hence, the correct answer is Option 4.

How many of the following can be a vector perpendicular to both the vectors $2\hat{i} - \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + 3\hat{k}$ ?

I. $4\hat{i}+5\hat{j}-3\hat{k}$

II. $-8\hat{i}-10\hat{j}+6\hat{k}$

III. $\frac{1}{50}(-4\hat{i}-5\hat{j}+3\hat{k})$

Select the correct answer.

((a))
None
((b))
One
((c))
Two
((d))
All three

Show Answer

Answer: ((d))

All three

Calculation:

A vector perpendicular to both the given vectors is obtained using the cross product.

Let the given vectors be represented in determinant form:

$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & 1 \ 1 & 1 & 3 \end{vmatrix}$

Expanding the determinant, we get:

$\vec{n} = \hat{i}(-3 - 1) - \hat{j}(6 - 1) + \hat{k}(2 + 1)$

$\vec{n} = -4\hat{i} - 5\hat{j} + 3\hat{k}$

Multiplying the vector by 2,

$2\vec{n} = -8\hat{i} - 10\hat{j} + 6\hat{k}$

Dividing by 50 to obtain the required form,

$\frac{1}{50}\vec{n} = \frac{1}{50}(-4\hat{i} - 5\hat{j} + 3\hat{k})$

Thus, Given all three vectors perpendicualr two given vectors

Hence, the correct answer is Option 4.

What is the area of the parallelogram whose sides are represented by the vectors $\hat{i}+2\hat{j}+3\hat{k}$ and $2\hat{i}+\hat{j}+2\hat{k}$ ?

((a))
$\frac{1}{2}\sqrt{26}$ square units
((b))
$\frac{1}{2}\sqrt{27}$ square units
((c))
$\sqrt{26}$ square units
((d))
$\sqrt{27}$ square units

Show Answer

Answer: ((c))

$\sqrt{26}$ square units

Calculation:

A vector perpendicular to both the given vectors is obtained using the cross product.

Let the given vectors be represented in determinant form:

⇒ $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 2 & 1 & 2 \end{vmatrix}$

Expanding the determinant, we get:

⇒ $\vec{b}_1 \times \vec{b}_2 = \hat{i}(4 - 3) - \hat{j}(2 - 6) + \hat{k}(1 - 4)$

⇒ $\vec{b}_1 \times \vec{b}_2 = \hat{i} + 4\hat{j} - 3\hat{k}$

The magnitude of the cross product (and hence the area of the parallelogram formed by the two vectors) is:

⇒ $\left\lvert \vec{b}_1 \times \vec{b}_2 \right\rvert = \sqrt{1^2 + 4^2 + (-3)^2} = \sqrt{1 + 16 + 9} = \sqrt{26}$

Hence, the correct answer is option 3.

The position vectors of the vertices A, B, Cand D of a quadrilateral ABCD are given by $3\hat{i}+4\hat{j}-2\hat{k}$ , $4\hat{i}-4\hat{j}-3\hat{k}$ , $2\hat{i}-3\hat{j}+2\hat{k}$ and $6\hat{i}-2\hat{j}+\hat{k}$ respectively.

What is the angle between the diagonals AC and BD of the quadrilateral?

((a))
$90^{\circ }$
((b))
$75^{\circ }$
((c))
$60^{\circ }$
((d))
$45^{\circ }$

Show Answer

Answer: ((a))

$90^{\circ }$

Calculation:

We calculate vector $\vec{AC}$ by subtracting the position vectors of points C and A:

$\vec{AC} = (2\hat{i} - 3\hat{j} + 2\hat{k}) - (3\hat{i} + 4\hat{j} - 2\hat{k})$

$\vec{AC} = -\hat{i} - 7\hat{j} + 4\hat{k}$

$|\vec{AC}| = \sqrt{(-1)^2 + (-7)^2 + 4^2} = \sqrt{1 + 49 + 16} = \sqrt{66}$

Next, compute vector $\vec{BD}$ using point subtraction:

$\vec{BD} = (6\hat{i} - 2\hat{j} + \hat{k}) - (4\hat{i} - 4\hat{j} - 3\hat{k}) = 2\hat{i} + 2\hat{j} + 4\hat{k}$

Now calculate the magnitude of $\vec{BD}$ :

$|\vec{BD}| = \sqrt{(2)^2 + (2)^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24}$

Compute the dot product $\vec{AC} \cdot \vec{BD}$ :

$\vec{AC} \cdot \vec{BD} = (-1)(2) + (-7)(2) + (4)(4) = -2 - 14 + 16 = 0$

Use the dot product formula to find the angle between the vectors:

$\cos \theta = \frac{\vec{AC} \cdot \vec{BD}}{|\vec{AC}| |\vec{BD}|} = \frac{0}{\sqrt{66} \cdot \sqrt{24}} = 0 \Rightarrow \theta = 90^\circ$

Hence, the correct answer is Option 1.

A force $\vec{F} = 2\hat{i} - \lambda\hat{j} + 5\hat{k}$ is applied at the point A(1, 2, 5) If its moment about the point B(- 1, - 2, 3) is $16\hat{i} - 6\hat{j} + 2\lambda\hat{k}$ , then what is the value of $\lambda$ ?

((a))
-2
((b))
0
((c))
1
((d))
2

Show Answer

Answer: ((a))

-2

Calculation:

We begin by calculating the position vector $\vec{r}$ from point B to point A:

⇒ $\vec{r} = \vec{A} - \vec{B} = (1 + 1)\hat{i} + (2 + 2)\hat{j} + (5 - 3)\hat{k} = 2\hat{i} + 4\hat{j} + 2\hat{k}$

Next, compute the moment vector $\vec{M}$ using the cross product:

⇒ $\vec{M} = \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 4 & 2 \ 2\lambda & 2 & 5 \end{vmatrix}$

Expanding the determinant:

⇒ $\vec{M} = \hat{i}(4 \cdot 5 - 2 \cdot 2) - \hat{j}(2 \cdot 5 - 2 \cdot 2\lambda) + \hat{k}(2 \cdot 2 - 4 \cdot 2\lambda)$

⇒ $\vec{M} = \hat{i}(20 - 4) - \hat{j}(10 - 4\lambda) + \hat{k}(4 - 8\lambda) = (16 + 0)\hat{i} - 6\hat{j} + 2\hat{k}$

Equating components to solve for $\lambda$ :

⇒ $16 = 20 + 2\lambda \Rightarrow \lambda = -2$

⇒ $2\lambda = -2\lambda - 8 \Rightarrow \lambda = -2$

Thus, the value of $\lambda$ is $-2$

Hence, the correct answer is Option 1.

For the following two (02) items:

Let

$f(x)=\begin{cases}\frac{1-\cos 2x}{x^{2}}&,x<0\ 9&,x=0\ \frac{\sqrt{x}}{\sqrt{(16+\sqrt{x})}-4}&,x>0\end{cases}$

What is $\lim _{x\rightarrow 0^{-}}f(x)$ equal to?

((a))
2
((b))
4
((c))
6
((d))
8

Show Answer

Answer: ((a))

Calculation:

The function $f(x)$ is defined piecewise as:

$f(x) = \begin{cases} \frac{1 - \cos 2x}{x^2}, & x < 0 \ 9, & x = 0 \ \frac{\sqrt{x}}{\sqrt{(16 + \sqrt{-4})}}, & x > 0 \end{cases}$

To evaluate the limit $\lim_{x \to 0} f(x)$ , we consider the left-hand and right-hand limits separately.

For $x \to 0^-$ :

⇒ $\lim_{x \to 0^-} \frac{1 - \cos 2x}{x^2} = \lim_{x \to 0^-} \frac{2 \sin^2 x}{x^2} = 2$

Hence, the correct answer is Option 1.

What is $\lim _{x\rightarrow 0^{+}}f(x)$ equal to?

((a))
6
((b))
7
((c))
8
((d))
9

Show Answer

Answer: ((c))

Calculation:

We evaluate the right-hand limit of the function $f(x)$ as $x \to 0^+$ :

$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{(16 + \sqrt{x})} - 4}$

To simplify, multiply numerator and denominator by the conjugate of the denominator:

$= \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{(16 + \sqrt{x})} + 4)}{(16 + \sqrt{x}) - 16} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{(16 + \sqrt{x})} + 4)}{\sqrt{x}}$

Canceling $\sqrt{x}$ from numerator and denominator:

$= \lim_{x \to 0^+} \sqrt{(16 + \sqrt{x})} + 4$

As $x \to 0^+$ , $\sqrt{x} \to 0$ , so:

$\lim_{x \to 0^+} f(x) = \sqrt{16} + 4 = 4 + 4 = 8$

Hence, the correct answer is option 3.

For the following three (03) items:

Consider the function f(x) = x|x|

What is $\lim _{x\rightarrow -1}f(x)$ equal to?

((a))
-1
((b))
0
((c))
1
((d))
Limit does not exist.

Show Answer

Answer: ((a))

-1

Calculation:

The function $f(x) = x|x|$ is defined as:

$f(x) = \begin{cases} x^2, & x \geq 0 \ -x^2, & x < 0 \end{cases}$

Evaluating the limit as $x \to -1$ :

$\lim_{x \to -1} f(x) = \lim_{x \to -1} -x^2 = -1$

Hence, the correct answer is Option 1.

What is the area bounded by the curve f(x) the x-axis and the lines x = - 2 and x = 1

((a))
1/3
((b))
2/3
((c))
5/2
((d))
3

Show Answer

Answer: ((d))

Calculation:

Required area:

$\left| \int_{-2}^{0} x^2 , dx \right| + \int_{0}^{1} x^2 , dx$

$ = -\left[ \frac{x^3}{3} \right]{-2}^{0} + \left[ \frac{x^3}{3} \right]{0}^{1} $

$= \left[ 0 + \frac{8}{3} \right] + \left[ \frac{1}{3} - 0 \right] = \frac{8}{3} + \frac{1}{3} = \frac{9}{3} = 3$

Hence, the correct answer is Option 4.

Consider the following statements:

I. The function is increasing in the interval (-∞,∞),

II. The function is differentiable at x = 0

Which of the statements given above is/are correct?

((a))
I only
((b))
II only
((c))
Both I and II
((d))
Neither I nor II

Show Answer

Answer: ((c))

Both I and II

Calculation:

$f(x) = \begin{cases} x^2, & x \ge 0 \ - x^2, & x < 0 \end{cases}$

It is clear from the graph that $f(x)$ is increasing in the interval $(-\infty, \infty)$ .

$\therefore \text{L.H.D.} = 0 = \text{R.H.D.}$

$\Rightarrow f(x) \text{ is differentiable at } x = 0$

Hence both I and II are correct.

For the following two (02) items:

Consider the function

$f(x)=\frac{x}{1-x}(x>0,x\ne 1)$

What is $\frac{f(x)}{f(x+1)}$ equal to?

((a))
$-f(x^{2})$
((b))
$-f(\sqrt{x})$
((c))
$f(x^{2})$
((d))
$f(x-1)$

Show Answer

Answer: ((a))

$-f(x^{2})$

Calculation:

Given the function:

$f(x) = \frac{x}{1 - x}$

Evaluate $f(x + 1)$ :

⇒ $f(x + 1) = \frac{x + 1}{1 - (x + 1)} = \frac{x + 1}{-x}$

Now compute $\frac{f(x)}{f(x + 1)}$ :

⇒ $\frac{f(x)}{f(x + 1)} = \frac{\frac{x}{1 - x}}{\frac{x + 1}{-x}} = \frac{1 - x}{x + 1} \cdot \frac{x}{-x}$

$= \frac{x^2}{x^2 - 1} = -f(x^2)$

Hence, the correct answer is option 1.

What is $(1-x)f(\sqrt{x})+xf(\sqrt{x}+1)$ equal to?

((a))
- f(x)
((b))
f(x)
((c))
x
((d))
0

Show Answer

Answer: ((d))

Calculation:

Given the function:

$f(x) = \frac{x}{1 - x} \Rightarrow f(\sqrt{x}) = \frac{\sqrt{x}}{1 - \sqrt{x}}$

Evaluate $f(\sqrt{x} + 1)$ :

⇒ $f(\sqrt{x} + 1) = \frac{\sqrt{x} + 1}{1 - \sqrt{x} - 1} = \frac{\sqrt{x} + 1}{-\sqrt{x}}$

Now compute the expression:

⇒ $(1 - x) \cdot \frac{\sqrt{x}}{1 - \sqrt{x}} - x \cdot \frac{\sqrt{x} + 1}{\sqrt{x}}$

Simplifying the numerator:

$= \frac{(1 - x)\sqrt{x}(1 - \sqrt{x}) - x(1 - \sqrt{x})(\sqrt{x} + 1)}{\sqrt{x}(1 - \sqrt{x})}$

$= \frac{x - x^2 - x + x^2}{\sqrt{x}(1 - \sqrt{x})} = 0$

Hence, the correct answer is Option 4.

For the following three (03) items:

Let $y=f(x)=\frac{x\sin ^{-1}x}{\sqrt{1-x^{2}}}+\ln \sqrt{1-x^{2}}.$

What is the slope of the tangent to the curve y = f(x) at x = 0.5

((a))
$4\pi \sqrt{3},/,27$
((b))
$8\pi \sqrt{3},/,27$
((c))
$4\pi$
((d))
$8\pi$

Show Answer

Answer: ((a))

$4\pi \sqrt{3},/,27$

Calculation:

Given the function:

$y = f(x) = x \cdot \frac{\sin^{-1}x}{\sqrt{1 - x^2}} + \ln \sqrt{1 - x^2}$

Differentiating $f(x)$ step-by-step:

⇒ $f'(x) = \left( \sin^{-1}x + \frac{x}{\sqrt{1 - x^2}} \right) \cdot \sqrt{1 - x^2} + x \cdot \sin^{-1}x + \frac{2x}{2\sqrt{1 - x^2}}$

$+ \frac{1}{\sqrt{1 - x^2}} \cdot \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x)$

Simplifying:

$= \sqrt{1 - x^2} \cdot \sin^{-1}x + x + \frac{x^2 \cdot \sin^{-1}x}{\sqrt{1 - x^2}} - \frac{x}{1 - x^2}$

Combining terms:

⇒ $f'(x) = \frac{(1 - x^2)\sin^{-1}x + x\sqrt{1 - x^2} + x^2 \sin^{-1}x - x\sqrt{1 - x^2}}{(1 - x^2)\sqrt{1 - x^2}}$

$= \frac{\sin^{-1}x}{(1 - x^2)\sqrt{1 - x^2}}$

Evaluating the derivative at $x = 0.5$ :

⇒ $f'(0.5) = \frac{\sin^{-1}(0.5)}{(1 - 0.25)\sqrt{1 - 0.25}} = \frac{\frac{\pi}{6}}{\frac{3}{4} \cdot \frac{\sqrt{3}}{2}} = \frac{\pi}{6} \cdot \frac{8}{3\sqrt{3}} = \frac{4\pi\sqrt{3}}{27}$

Thus, the slope of $y = f(x)$ at $x = 0.5$ is $\frac{4\pi\sqrt{3}}{27}$ .

What is $\frac{d^{2}y}{dx^{2}}$ at x = 0 equal to?

((a))
0
((b))
0.5
((c))
1
((d))
1.5

Show Answer

Answer: ((c))

Calculation:

$\frac{dy}{dx} = \frac{\sin^{-1}x}{(1 - x^2)^{3/2}}$

⇒ $\frac{dy}{dx^2} = \frac{\frac{1}{\sqrt{1 - x^2}}(1 - x^2)^{3/2} - \sin^{-1}x \cdot \frac{3}{2}(1 - x^2)^{1/2}(-2x)}{(1 - x^2)^3}$

$= \frac{\sqrt{1 - x^2} + 3x\sqrt{1 - x^2} \sin^{-1}x}{(1 - x^2)^3}$

$\left( \frac{d^2 y}{dx^2} \right)_{x=0} = \frac{1 + 0}{(1 - 0)^3} = 1$

Hence, the correct answer is Option 3.

If x = sin $\theta$ then what is $\frac{dy}{dx}$ equal to?

((a))
$\theta \sec \theta$
((b))
$\theta \sec ^{2}\theta$
((c))
$\theta \sec ^{3}\theta$
((d))
$2\tan \theta +\theta \sec ^{2}\theta$

Show Answer

Answer: ((c))

$\theta \sec ^{3}\theta$

Calculation:

$\frac{dy}{dx} = \frac{\sin^{-1}x}{(1 - x^2)^{3/2}}$

$\left( \frac{dy}{dx} \right) = \sin \theta = \frac{\sin^{-1}(\sin \theta)}{(1 - \sin^2 \theta)^{3/2}}$

$= \frac{\theta}{\cos^3 \theta} = \theta \cdot \sec^3 \theta$

Hence, the correct answer is Option 3.

For the following two (02) items:

Consider the function $f(x)=1-\sqrt[3]{(x-1)^{2}}$

What is the domain of the function?

((a))
(1, ∞)
((b))
(- ∞, ∞)
((c))
(0, ∞)
((d))
(- ∞, ∞) \ {1}

Show Answer

Answer: ((b))

(- ∞, ∞)

Calculation:

Given the function:

$f(x) = 1 - \sqrt[3]{(x - 1)^2}$

$∴ (x - 1)^2 \geq 0 \Rightarrow (x - 1)^2$ is a real number

Since the cube root of any real number is also a real number, there are no restrictions on the values of $x$ .

∴ Domain $= (-\infty, \infty)$

Hence, the correct answer is Option 2.

The function has

((a))
a minimum at x = 1
((b))
a maximum at x = 1
((c))
neither maximum nor minimum at x = 1
((d))
no extremum

Show Answer

Answer: ((b))

a maximum at x = 1

Calculation:

Given the function:

$f(x)=1-\sqrt[3]{(x-1)^2}$

Note that:

⇒ $(x-1)^2 \ge 0 \text{ for all real } x$

Hence,

⇒ $\sqrt[3]{(x-1)^2} \ge 0 \Rightarrow f(x) \le 1$

Evaluate the function at $x=1$ :

⇒ $f(1)=1-\sqrt[3]{(1-1)^2}=1$

For any $x \neq 1$ :

⇒ $(x-1)^2>0 \Rightarrow \sqrt[3]{(x-1)^2}>0 \Rightarrow f(x)<1$

Also, as $|x| \to \infty$ :

⇒ $\sqrt[3]{(x-1)^2} \to \infty \Rightarrow f(x) \to -\infty$

Therefore, the function attains a maximum value at $x=1$ .

Hence, the correct answer is Option 2.

For the following two (02) items:

Consider the function

$f(x)=\begin{cases}4(5^{x}),&x<0\ 8k+x,&x\ge 0\end{cases}$

If the function is continuous, then what is the value of k?

((a))
0.5
((b))
1
((c))
1.5
((d))
2

Show Answer

Answer: ((a))

0.5

Calculation:

The function $f(x)$ is defined as:

$f(x) = \begin{cases} 4(5^x), & x < 0 \ 8k + x, & x \geq 0 \end{cases}$

Evaluating the left-hand limit as $x \to 0$ :

$\text{L.H.L.} = \lim_{x \to 0^-} 4(5^x) = 4(5^0) = 4$

Evaluating the right-hand limit as $x \to 0$ :

$\text{R.H.L.} = \lim_{x \to 0^+} (8k + x) = 8k$

Since $f(x)$ is continuous at $x = 0$ , we equate the limits:

$8k = 4 \Rightarrow k = \frac{1}{2} = 0.5$

Hence, the correct answer is Option 1.

What is f'(-1) equal to?

((a))
$\frac{2}{5}\ln 5$
((b))
$\frac{3}{5}\ln 5$
((c))
$\frac{4}{5}\ln 5$
((d))
$20\ln 5$

Show Answer

Answer: ((c))

$\frac{4}{5}\ln 5$

Calculation:

The derivative function $f'(x)$ is defined as:

$f'(x) = \begin{cases} 4(5^x) \cdot \log 5, & x < 0 \ 1, & x \geq 0 \end{cases}$

Evaluating the derivative at $x = -1$ :

$f'(-1) = 4(5^{-1}) \cdot \log 5 = \frac{4}{5} \cdot \log 5$

Hence, the correct answer is Option 3.

For the following two (02) items:

Let $u=\int e^{x}cosxdx $ and $v=\int e^{x}sinxdx.$

What is u + v equal to?

((a))
$-\frac{du}{dx}$
((b))
$-\frac{dv}{dx}$
((c))
$\frac{du}{dx}$
((d))
$\frac{dv}{dx}$

Show Answer

Answer: ((d))

$\frac{dv}{dx}$

Calculation:

$u+v=\int e^{x}\cos x,dx+\int e^{x}\sin x,dx$

$= \int e^x (\cos x + \sin x) , dx$

$= e^x \sin x \quad \left[ \therefore \frac{d \sin x}{dx} = \cos x \right]$

$= \frac{dv}{dx}$

Hence, the correct answer is Option 4.

Consider the following:

I. $\frac{du}{dx}=-v$

II. $\frac{dv}{dx}=-u$

Which of the above is/are correct?

((a))
I only
((b))
II only
((c))
Both I and II
((d))
Neither I nor II

Show Answer

Answer: ((d))

Neither I nor II

Calculation:

$u = \int e^x \cos x , dx$

$\frac{du}{dx} = e^x \cos x \ne -v$

$v = \int e^x \sin x , dx$

$\frac{dv}{dx} = e^x \sin x \ne -u$

$\Rightarrow e^x \sin x \ne -u$

Hence, the correct answer is Option 4.

For the following two (02) items:

Let the function f(x) = |x - 3| + |x - 4| be defined on the interval [0, 5].

What is $\frac{dy}{dx}$ at x = 3.5 equal to?

((a))
0
((b))
1
((c))
2
((d))
3.5

Show Answer

Answer: ((a))

Calculation:

The function $f(x)$ is defined as:

$f(x) = |x - 3| + |x - 4| = \begin{cases} -2x + 7, & x < 3 \ 1, & 3 \leq x < 4 \ 2x - 7, & x \geq 4 \end{cases}$

For $x = 3.5$

$y = f(x) = 1$

$\frac{dy}{dx} = \frac{d(1)}{dx} = 0$

Hence, the correct answer is Option 1.

Consider the following statements:

I. The function is differentiable at x = 3

II. The function is differentiable at x = 4

Which of the statements given above is/are correct?

((a))
I only
((b))
II only
((c))
Both I and II
((d))
Neither I nor II

Show Answer

Answer: ((d))

Neither I nor II

Calculation:

At $x = 3$ :

L. H. D = -2 and R. H. D = 0

Thus, L. H. D ≠ R. H. D

⇒ f(x) is not differentiable at x = 3

At $x = 4$ :

L. H. D = 0 and R. H. D = 2

Thus, L. H. D ≠ R. H. D

⇒ f(x) is not differentiable at x = 4

Hence, the correct answer is Option 4.

For the following two (02) items:

Consider the function $f(x)=\frac{10^{x}-10^{-x}}{10^{x}+10^{-x}}$

What is $f\circ f\circ f\circ f\circ f(0)$ equal to?

((a))
0
((b))
1
((c))
5
((d))
10

Show Answer

Answer: ((a))

Calculation:

$f(0) = \frac{10^0 - 10^0}{10^0 + 10^0} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0$

$\therefore f \circ f \circ f \circ f \circ f(0) = 0$

Hence, the correct answer is Option 1.

What is the inverse of the function?

((a))
$\log _{10}(2x-1)$
((b))
$\frac{1}{2}\log _{10}(2x-1)$
((c))
$\frac{1}{4}\log _{10}(\frac{2x}{2-x})$
((d))
$\frac{1}{2}\log _{10}(\frac{1+x}{1-x})$

Show Answer

Answer: ((d))

$\frac{1}{2}\log _{10}(\frac{1+x}{1-x})$

Calculation:

⇒ $\frac{y}{1} = \frac{10^x - 10^{-x}}{10^x + 10^{-x}} = \frac{10^{2x} - 1}{10^{2x} + 1}$

⇒ $\frac{y + 1}{y - 1} = \frac{10^{2x} - 1 + 10^{2x} + 1}{10^{2x} - 1 - 10^{2x} - 1} = \frac{10^{2x}}{-1}$

⇒ $\frac{1 + y}{1 - y} = 10^{2x} \Rightarrow 2x = \log_{10} \left( \frac{1 + y}{1 - y} \right)$

⇒ $x = \frac{1}{2} \log_{10} \left( \frac{1 + y}{1 - y} \right)$

⇒ $f^{-1}(x) = \frac{1}{2} \log_{10} \left( \frac{1 + x}{1 - x} \right)$

Hence, the correct answer is Option 4.

What is the degree of the differential equation

$(\frac{d^{2}y}{dx^{2}})^{\frac{3}{2}}=(\frac{dy}{dx})^{\frac{5}{2}}$

((a))
3
((b))
2
((c))
5/2
((d))
3/2

Show Answer

Answer: ((a))

Calculation:

$\left( \frac{d^2 y}{dx^2} \right)^{3/2} = \left( \frac{dy}{dx} \right)^{5/2}$

Squaring both sides:

$\left( \frac{d^2 y}{dx^2} \right)^3 = \left( \frac{dy}{dx} \right)^5$

Degree = 3

Hence, the correct answer is Option 1.

What is $\int _{n}^{n+1}(x-[x])dx$ , where [.] is the greatest integer function and n is natural number?

((a))
$\frac{4n+1}{2}$
((b))
$\frac{2n+1}{2}$
((c))
1/2
((d))
1

Show Answer

Answer: ((c))

1/2

Calculation:

$\int_n^{n+1} (x - [x]) , dx = \int_n^{n+1} (x - n) , dx$

$= \left[ \frac{x^2}{2} - nx \right]_n^{n+1}$

$= \frac{(n+1)^2}{2} - n(n+1) - \frac{n^2}{2} + n^2$

$= \frac{1 - n^2}{2} + \frac{n^2}{2} = \frac{1}{2}$

Hence, the correct answer is Option 3.

Consider the following statements:

I. $y=xe^{2x}$ is the solution of $\frac{dy}{dx}=y(2+\frac{1}{x})$

II. $y=x\ln |x|+cx$ is the solution of $\frac{dy}{dx}=\frac{x+y}{x}$

Which of the statements given above is/are correct?

((a))
I only
((b))
II only
((c))
Both I and II
((d))
Neither I nor II

Show Answer

Answer: ((c))

Both I and II

Calculation:

Given:

$y = xe^{2x} \Rightarrow \frac{dy}{dx} = e^{2x} + 2xe^{2x}$

$\frac{dy}{dx} = \frac{y}{x} + 2y = y\left(2 + \frac{1}{x}\right)$

So, statement I is correct.

$\frac{dy}{dx} = \frac{x + y}{x} = 1 + \frac{y}{x}$

$\Rightarrow \frac{dy}{dx} - \frac{1}{x} \cdot y = 1$

Integrating factor:

$\text{I.F.} = e^{\int -\frac{1}{x} dx} = e^{-\log x} = \frac{1}{x}$

$\therefore y \cdot \frac{1}{x} = \int \frac{1}{x} dx$

$\frac{y}{x} = \ln|x| + c \Rightarrow y = x \ln|x| + cx$

So, statement II is correct.

Hence, the correct answer is option 3.

If k is an arbitrary constant, then what is the general solution of the equation $(x+y)^{2}\frac{dy}{dx}=k^{2}$

((a))
y + x = tan(x + c) + k
((b))
$x+y=k\tan \left(\frac{y-c}{k}\right)$
((c))
$x-y=k\tan \left(\frac{y-c}{k}\right)$
((d))
y - x = tan(x + c) + k

Show Answer

Answer: ((b))

$x+y=k\tan \left(\frac{y-c}{k}\right)$

Calculation:

$\frac{dy}{dx} = \frac{k^2}{(x+y)^2}$

Let $x + y = t \Rightarrow \frac{dt}{dx} = 1 + \frac{dy}{dx}$

⇒ $\frac{dt}{dx} - 1 = \frac{k^2}{t^2} \Rightarrow \frac{dt}{dx} = \frac{k^2 + t^2}{t^2}$

⇒ $\int \frac{t^2}{k^2 + t^2} , dt = \int dx$

⇒ $\int \left(1 - \frac{k^2}{t^2 + k^2} \right) dt = x$

⇒ $t - k^2 \cdot \frac{1}{k} \tan^{-1} \left( \frac{t}{k} \right) = x + c$

$\Rightarrow x + y - k \tan^{-1} \left( \frac{x + y}{k} \right) = x + c$

$\Rightarrow \frac{y - c}{k} = \tan^{-1} \left( \frac{x + y}{k} \right) \Rightarrow \frac{x + y}{k} = \tan \left( \frac{y - c}{k} \right) \Rightarrow x + y = k \tan \left( \frac{y - c}{k} \right)$

Hence, the correct answer is Option 2.

What is $\int \frac{dx}{10^{x}+10^{-x}}$ equal to?

((a))
$\tan ^{-1}(10^{x})+c$
((b))
$(\ln 10)\tan ^{-1}(10^{x})+c$
((c))
$\frac{1}{\ln 10}\tan ^{-1}(10^{x})+c$
((d))
$\ln (10^{x}+10^{-x})+c$

Show Answer

Answer: ((c))

$\frac{1}{\ln 10}\tan ^{-1}(10^{x})+c$

Calculation:

$\int \frac{dx}{10^x + 10^{-x}} = \int \frac{10^x}{(10^x)^2 + 1} , dx$

Let $10^x = t \Rightarrow 10^x \cdot \log_{10} e , dx = dt$

$= \int \frac{1}{1 + t^2} \cdot \log_{10} e , dt$

$= \log_{10} e \cdot \tan^{-1}(t) + c = \log_{10} e \cdot \tan^{-1}(10^x) + c$

$= \frac{1}{\log_e 10} \cdot \tan^{-1}(10^x) + c$

Hence, the correct answer is Option 3.

A wire of length 20 cm is to be bent into a rectangle. Which of the following statements is/are correct?

I. The rectangle of the largest area is the square.

II. It is possible to form a rectangle of an area of $27c m ^ 2$

Select the answer using the code given below.

((a))
I only
((b))
II only
((c))
Both I and II
((d))
Neither I nor II

Show Answer

Answer: ((a))

I only

Calculation:

Given:

Perimeter of rectangle:

$2(x + y) = 20 \Rightarrow x + y = 10 \Rightarrow y = 10 - x$

Area:

$A = xy = x(10 - x) = 10x - x^2$

First derivative:

$\frac{dA}{dx} = 10 - 2x$

Setting derivative to zero for critical point:

$10 - 2x = 0 \Rightarrow x = 5$

Second derivative:

$\frac{d^2A}{dx^2} = -2 < 0$

∴ Area of maximum at =5

y = 10 - 5 = 5 = x

Statement I is correct.

Now testing if area 27 cm² is possible:

$A = 10x - x^2 = 27 \Rightarrow x^2 - 10x + 27 = 0$

Discriminant:

$D = (-10)^2 - 4 \cdot 27 = 100 - 108 = -8 < 0$

∴ Area 27 cm² is not possible So, statement II is not correct.

Hence, the correct answer is Option 1.

If $I_{1}=\int {e}^{e^{2}}\frac{dx}{ln~x}$ and $I{2}=\int _{1}^{2}\frac{e^{x}}{x}dx$ then which one of the following is correct?

((a))
$I_{1}-I_{2}=0$
((b))
$I_{1}+I_{2}=0$
((c))
$I_{1}-2I_{2}=0$
((d))
$2I_{1}-I_{2}=0$

Show Answer

Answer: ((a))

$I_{1}-I_{2}=0$

Calculation:

Given:

$I_1 = \int_e^{e^2} \frac{dx}{\ln x}$

Let $\ln x = t \Rightarrow x = e^t$ , then $dx = e^t dt$

Limits change:

When $x = e \Rightarrow t = 1$

When $x = e^2 \Rightarrow t = 2$

Substituting:

$I_1 = \int_1^2 \frac{e^t}{t} dt = I_2$

$\therefore I_1 - I_2 = 0$

Hence, the correct answer is Option 1.

What is the area of the region bounded by |x| <= 2k and |y| <= k where k is a positive real number?

((a))
$2k ^ 2$
((b))
$4k ^ 2$
((c))
$5k ^ 2$
((d))
$8k ^ 2$

Show Answer

Answer: ((d))

$8k ^ 2$

Calculation:

$|x|\le 2k ;\Rightarrow; -2k \le x \le 2k$

$|y|\le k ;\Rightarrow; -k \le y \le k$

$AB = CD = 2k - (-2k) = 4k$

$AD = BC = k - (-k) = 2k$

Area of rectangle:

$\text{Area} = 4k \cdot 2k = 8k^2$

Hence, the correct answer is Option 4.

Consider the following statements regarding the function f(x) = 1/(x - 5)

Statement-I: f(x) is decreasing on the intervals x < 5 and x > 5

Statement-II: f''(x) > 0 for all x ≠ 5.

Which one of the following is correct in respect of the above statements?

((a))
Both Statement-I and Statement-II are correct and Statement-II explains Statement-I
((b))
Both Statement-I and Statement-II are correct but Statement-II does not explain Statement-I
((c))
Statement-I is correct but Statement-II is not correct
((d))
Statement-I is not correct but Statement-ll is correct

Show Answer

Answer: ((c))

Statement-I is correct but Statement-II is not correct

Concept:

Monotonicity and Concavity of Function:

The first derivative f′(x) determines whether a function is increasing or decreasing.
If f′(x) < 0, the function is decreasing.
If f′(x) > 0, the function is increasing.
The second derivative f″(x) determines the concavity of the function.
If f″(x) > 0, the function is concave upward.
If f″(x) < 0, the function is concave downward.
For rational functions like f(x) = 1/(x − a), domain excludes x = a.

Calculation:

Given, f(x) = 1/(x − 5)

First derivative:

⇒ f′(x) = −1/(x − 5)²

⇒ (x − 5)² > 0 for x ≠ 5

⇒ f′(x) < 0 for all x ≠ 5

⇒ Function decreasing for x < 5 and x > 5

Second derivative:

⇒ f″(x) = 2/(x − 5)³

⇒ For x > 5, (x − 5)³ > 0 ⇒ f″(x) > 0

⇒ For x < 5, (x − 5)³ < 0 ⇒ f″(x) < 0

⇒ f″(x) not positive for all x ≠ 5

∴ Only Statement I is correct.

100

Consider the following statements:

Statement-I: The function $f(x)=\frac{x^{3}+128}{x}$ has a minimum value 48 at x = 4

Statement-II: As x increases through 4, f'(x) changes sign from positive to negative.

Which one of the following is correct in respect of the above statements?

((a))
Both Statement-I and Statement-II are correct and Statement-II explains Statement-I
((b))
Both Statement-I and Statement-II are correct but Statement-II does not explain Statement-I
((c))
Statement-I is correct but Statement-II is not correct
((d))
Statement-I is not correct but Statement-II is correct

Show Answer

Answer: ((c))

Statement-I is correct but Statement-II is not correct

Calculation:

Given function:

$f(x) = \frac{x^3 + 128}{x}$

⇒ $f'(x) = \frac{3x^2 \cdot x - (x^3 + 128)}{x^2} = \frac{2x^3 - 128}{x^2}$

Setting $f'(x) = 0$ :

⇒ $2x^3 - 128 = 0 \Rightarrow x = 4$

⇒ $f''(x) = \frac{6x^2 \cdot x^2 - (2x^3 - 128)(2x)}{x^4}$

At $x = 4$ :

⇒ $f''(4) = \frac{6 \cdot 16 \cdot 16 - 0}{16 \cdot 16} = 6 > 0$

Minimum occurs at $x = 4$

Minimum value: $f(4) = \frac{64 + 128}{4} = 48$

So, statement I is correct.

Behavior of function:

⇒ $f(x)$ is decreasing for $x < 4$

⇒ $f(x)$ is increasing for $x > 4$

Sign of $f'(x)$ changes from negative to positive as $x$ increases through 4.

Hence, the correct answer is Option 3.

101

What is the harmonic mean of the numbers C(10,3) C(10,4) C(10,5) C(10,6) and C(10,7) ?

((a))
3150/19
((b))
4000/19
((c))
252
((d))
225

Show Answer

Answer: ((a))

3150/19

Calculation:

Combinations:

${}^{10}C_3 = 120$

${}^{10}C_4 = 210$

${}^{10}C_5 = 252$

${}^{10}C_6 = 210$

${}^{10}C_7 = 120$

Sum of reciprocals:

$\frac{1}{120} + \frac{1}{210} + \frac{1}{252} + \frac{1}{210} + \frac{1}{120} = 2\left(\frac{1}{120} + \frac{1}{210}\right) + \frac{1}{252}$

$= 2\left(\frac{7 + 4}{840}\right) + \frac{1}{252} = \frac{11}{420} + \frac{1}{252} = \frac{33 + 5}{1260} = \frac{38}{1260} = \frac{19}{630}$

Harmonic Mean:

$\text{H.M.} = \frac{5}{\frac{19}{630}} = \frac{3150}{19}$

Hence, the correct answer is Option 1.

102

In a sample survey of a village, the probability that a farmer is in debt is 0.60. What is the probability that three randomly selected farmers are all in debt (assume independence of events)?

((a))
0.000216
((b))
0.064
((c))
0.216
((d))
0.512

Show Answer

Answer: ((c))

0.216

Calculation:

The probability that one farmer is in debt is given as P = 0.60

Since the events are independent, the probability that three randomly selected farmers are all in debt is:

P³ = (0.60)³= 0.216

Hence, the correct answer is Option 3.

103

The probability that a family owns a laptop is 0.68; that it also owns a desktop is 0.56. If the probability that it owns both is 0.48, then what is the probability that a randomly selected family owns a laptop or a desktop?

((a))
0.80
((b))
0.76
((c))
0.36
((d))
0.28

Show Answer

Answer: ((b))

0.76

Calculation:

Given:

$P(L) = 0.68$

$P(D) = 0.56$

$P(L \cap D) = 0.48$

Using the formula for union of two events:

⇒ $P(L \cup D) = P(L) + P(D) - P(L \cap D)$

Substituting values:

⇒ $P(L \cup D) = 0.68 + 0.56 - 0.48 = 0.76$

Hence, the correct answer is Option 2.

104

An urn contains 10 white and 5 red balls. If two balls are drawn at random, then what is the probability that both the balls are red?

((a))
2/21
((b))
1/7
((c))
4/21
((d))
3/7

Show Answer

Answer: ((a))

2/21

Calculation:

Given:

White balls = 10,Red balls= 5 ,Total balls= 15

Total ways to draw 2 balls from 15:

$\binom{15}{2}=\frac{15\times 14}{2}=105$

Favourable ways to draw 2 red balls from 5 red balls:

$\binom{5}{2}=\frac{5\times 4}{2}=10$

Therefore, required probability:

$P=\frac{\binom{5}{2}}{\binom{15}{2}}=\frac{10}{105}=\frac{2}{21}$

Hence, the correct answert is Option 1.

105

An urn contains 5 white, 6 red and 4 blue balls. Three balls are drawn at random. What is the probability that a white ball, a red ball and a blue ball are drawn?

((a))
28/91
((b))
2/7
((c))
24/91
((d))
23/91

Show Answer

Answer: ((c))

24/91

Calculation:

Given:

White balls=5, Red balls = 6, Blue balls = 4

$\text{Total balls}=5+6+4=15$

Total ways to draw 3 balls from 15:

$\binom{15}{3}=\frac{15\times14\times13}{6}=455$

Favourable ways to draw one white, one red and one blue ball:

$\binom{5}{1}\times\binom{6}{1}\times\binom{4}{1} =5\times6\times4=120$

Therefore, required probability:

$P=\frac{120}{455}=\frac{24}{91}$

Hence, the correct answer is Option 3.

106

Under which of the following conditions may binomial distribution be used?

I. The number of trials is infinite and not fixed.

II. The trials are independent.

III. Each trial has two possible outcomes.

Select the correct answer using the code given below.

((a))
II only
((b))
III only
((c))
I and II
((d))
II and III

Show Answer

Answer: ((d))

II and III

Calculation:

For a binomial distribution, the required conditions are:

The number of trials must be finite and fixed
The trials must be independent
Each trial must have exactly two possible outcomes (success/failure)
Probability of success remains constant

Hence, the correct answer is Option 4.

107

A person X speaks the truth 4 out of 5 times and person Y speaks the truth 5 out of 6 times. What is the probability that they will contradict each other in stating the fact?

((a))
3/10
((b))
1/15
((c))
1/6
((d))
7/10

Show Answer

Answer: ((a))

3/10

Calculation:

We are asked to compute:

$P(X \cap Y') + P(X' \cap Y)$

Using the formula:

$= P(X)[1 - P(Y)] + [1 - P(X)]P(Y)$

Substituting values:

$= \frac{4}{5} \left(1 - \frac{5}{6}\right) + \left(1 - \frac{4}{5}\right) \cdot \frac{5}{6}$

$= \frac{4}{5} \cdot \frac{1}{6} + \frac{1}{5} \cdot \frac{5}{6} = \frac{4}{30} + \frac{5}{30} = \frac{9}{30} = \frac{3}{10}$

Hence, the correct answer is Option 1.

108

The probability that a student passes Physics test is 2/3 and the probability that he passes both Physics test and English test is 11/15. The probability that he passes at least one test is 4/5 What is the probability that he passes English test?

((a))
11/15
((b))
13/15
((c))
14/15
((d))
1

Show Answer

Answer: ((b))

13/15

Calculation:

Let:

$P(P)=\frac{2}{3},\quad P(P\cap E)=\frac{11}{15},\quad P(P\cup E)=\frac{4}{5}$

Using the formula:

⇒ $P(P\cup E)=P(P)+P(E)-P(P\cap E)$

Substitute the given values:

⇒ $\frac{4}{5}=\frac{2}{3}+P(E)-\frac{11}{15}$

Convert fractions to a common denominator:

⇒ $\frac{4}{5}=\frac{12}{15},\quad \frac{2}{3}=\frac{10}{15}$

Simplifying:

⇒ $\frac{12}{15}=\frac{10}{15}+P(E)-\frac{11}{15}$

⇒ $\frac{12}{15}=P(E)-\frac{1}{15}$

Therefore,

$P(E)=\frac{12}{15}+\frac{1}{15}=\frac{13}{15}$

Hence, the correct answer is Option 2.

109

An event X can happen with probability pand event Y can happen with probability q. Further, X and Y are independent events. Which of the following statements is/are correct?

The probability that exactly one of the events happens is p + q - pq

II. The probability that at least one of the events happens is p + q - 2pq

Select the answer using the code given below.

((a))
I only
((b))
II only
((c))
Both I and II
((d))
Neither I nor II

Show Answer

Answer: ((d))

Neither I nor II

Calculation:

Let:

$P(X) = p$

$P(Y) = q$

X and Y are independent ⇒ $P(X \cap Y) = pq$

Probability of exactly one event occurring:

$P(\text{exactly one}) = P(X) + P(Y) - 2P(X \cap Y) = p + q - 2pq$

So, statement I is incorrect.

Probability of at least one event occurring:

$P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = p + q - pq$

So, statement II is also incorrect.

Hence, the Correct answer is Option 4.

110

Three faces of a die are black, two faces are white and one face is red. The die is tossed three times. What is the probability that the colours black, white and red appear in the first, second and third tosses respectively?

((a))
1/36
((b))
1/6
((c))
7/36
((d))
5/36

Show Answer

Answer: ((a))

1/36

Calculation:

Given:

Number of black faces = 3

Number of white faces = 2

Number of red faces = 1

Probability that black, white, and red appear in first, second, and third positions respectively:

$P = \frac{3}{6} \cdot \frac{2}{6} \cdot \frac{1}{6} = \frac{6}{216} = \frac{1}{36}$

Hence, the correct answer is Option 1.

111

A fair coin is tossed 4 times. What is the probability that two heads do not occur consecutively?

((a))
1/8
((b))
3/8
((c))
7/16
((d))
1/2

Show Answer

Answer: ((d))

1/2

Calculation:

Total number of outcomes for 4 coin tosses:

$n(S) = 2^4 = 16$

Favorable outcomes with two consecutive heads:

⇒ $E = { \text{HHTT, THHT, TTHH, HHHT, HTHH, HHTH, THHH, HHHH} }$

⇒ $n(E) = 8 \Rightarrow n(E') = 16 - 8 = 8$

Probability of not getting two consecutive heads:

⇒ $P(E') = \frac{8}{16} = \frac{1}{2}$

Hence, the correct answer is Option 4.

112

In a throw of three dice, what is the probability of getting one prime number, one composite number and one number which is neither prime nor composite?

((a))
1/2
((b))
1/3
((c))
1/4
((d))
1/6

Show Answer

Answer: ((d))

1/6

Calculation:

Given:

Prime numbers = {2, 3, 5} ⇒ 3 choices

Composite numbers = {4, 6} ⇒ 2 choices

Neither prime nor composite = {1} ⇒ 1 choice

Probability of selecting one prime, one composite, and one neither prime nor composite:

$P = \frac{3}{6} \cdot \frac{2}{6} \cdot \frac{1}{6} \cdot 3!$

Calculating:

$= \frac{6}{216} \cdot 6 = \frac{36}{216} = \frac{1}{6}$

Hence, the correct answer is Option 4.

113

An integer is chosen at random from the first 50 integers. What is the probability that the integer is neither divisible by 5 nor 9?

((a))
7/10
((b))
18/25
((c))
37/50
((d))
19/25

Show Answer

Answer: ((b))

18/25

Calculation:

Given:

Numbers divisible by 5 = 10

Numbers divisible by 9 = 5

Numbers divisible by both 5 and 9 = 1

Using the inclusion-exclusion principle:

Divisible by 5 or 9 $= 10 + 5 - 1 = 14$

Total numbers = 50

Not divisible by 5 or 9 $= 50 - 14 = 36$

Required probability:

$P = \frac{36}{50} = \frac{18}{25}$

Hence, the correct answer is Option 2.

114

Out of 50 consecutive natural numbers, two integers are chosen at random. What is the probability that their sum is odd?

((a))
1/2
((b))
24/49
((c))
1/4
((d))
25/49

Show Answer

Answer: ((d))

25/49

Calculation:

Out of 50 consecutive natural numbers:

Even numbers = 25

Odd numbers = 25

Sum of one even and one odd number is always odd.

Number of ways to choose one even and one odd:

$= 25 \cdot 25 = 625$

Total number of ways to choose any two numbers from 50:

$= {}^{50}C_2 = \frac{50 \cdot 49}{2} = 1225$

Required probability:

$P(\text{sum is odd}) = \frac{625}{1225} = \frac{25}{49}$

Hence, the Correct answer is Option 4.

115

The standard deviation of 100 observations is 10. If 20 is added to each observation, then what will be the new standard deviation?

((a))
10
((b))
15
((c))
20
((d))
25

Show Answer

Answer: ((a))

Calculation:

Given:

Standard deviation of 100 observations = 10

A constant value $20$ is added to each observation.

Property of standard deviation:

Adding or subtracting a constant from each observation does not change the standard deviation

Hence,

New standard deviation = Old standard deviation = 10

Hence, the correct answer is Option 1.

116

Let X be a random variable following binomial distribution with parameters n = 5 and p = k Further, P(X = 1) = 0.4096 and P(X = 2) = 0.2048 . What is the value of k?

((a))
0.2
((b))
0.25
((c))
0.3
((d))
0.35

Show Answer

Answer: ((a))

0.2

Calculation:

Given:

$n = 5$

$p = k$ , $q = 1 - k$

Binomial probabilities:

⇒ $P(X = 1) = {}^5C_1 \cdot k \cdot (1 - k)^4 = 0.4096$

⇒ $P(X = 2) = {}^5C_2 \cdot k^2 \cdot (1 - k)^3 = 0.2048$

Taking ratio:

⇒ $\frac{{}^5C_2 \cdot k^2 \cdot (1 - k)^3}{{}^5C_1 \cdot k \cdot (1 - k)^4} = \frac{0.2048}{0.4096}$

Simplifying:

⇒ $\frac{10 \cdot k}{5 \cdot (1 - k)} = \frac{1}{2} \Rightarrow 4k = 1 - k \Rightarrow 5k = 1 \Rightarrow k = \frac{1}{5} = 0.2$

Hence, the correct answer is Option 1.

117

The frequency distribution of the marks obtained by students in a Science examination is given below:

Marks	5-15	15-25	25-35	35-45
Number of students	20	30	30	20

What is the arithmetic mean?

((a))
20
((b))
25
((c))
30
((d))
35

Show Answer

Answer: ((b))

Calculation:

The given data is a grouped frequency distribution.

Class Interval	f_i	x_i	x_if_i
5 – 15	20	10	200
15 – 25	30	20	600
25 – 35	30	30	900
35 – 45	20	40	800
Total	∑f_i=100		2500

Compute the totals:

⇒ $\sum f_i = 20+30+30+20 = 100$

⇒ $\sum x_i f_i = 200+600+900+800 = 2500$

Mean of the distribution is given by:

⇒ $\text{Mean} = \frac{\sum x_i f_i}{\sum f_i}$

Substituting the values:

⇒ $\text{Mean} = \frac{2500}{100} = 25$

Hence, the correct answer is Option 2.

118

If P(A) = 0.3 , P(B) = 0.4 and P( A |B)=0.5, then what is the value P( B |A) ?

((a))
0.325
((b))
0.333
((c))
0.375
((d))
0.667

Show Answer

Answer: ((d))

0.667

Calculation:

Given:

P(A) = 0.3 , P(B) = 0.4 and P( A |B)=0.5

Using the formula of conditional probability:

$P!\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}$

Substitute the given values:

$0.5=\frac{P(A\cap B)}{0.4}$

Solving for $P(A\cap B)$ :

$P(A\cap B)=0.5\times 0.4=0.2$

Now find $P!\left(\frac{B}{A}\right)$ :

$P!\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P(A)}$

Substitute the values:

$P!\left(\frac{B}{A}\right)=\frac{0.2}{0.3}=\frac{2}{3}\approx 0.667$

Hence, the correct answer is Option 4.

119

If P(A) = 1/3 , P(B) = 1/2 and $P(A\cap B)=1/4$ , then what is the value of $P(\overline{A}\cup B)$ ?

((a))
7/12
((b))
2/3
((c))
3/4
((d))
11/12

Show Answer

Answer: ((d))

11/12

Calculation:

Given:

$P(A)=\frac{1}{3},\quad P(B)=\frac{1}{2},\quad P(A\cap B)=\frac{1}{4}$

We know that:

⇒ $P(\bar{A})=1-P(A)$

Using the formula:

⇒ $P(\bar{A}\cup B)=P(\bar{A})+P(B)-P(\bar{A}\cap B)$

Substitute $P(\bar{A})=1-P(A)$ :

⇒ $P(\bar{A}\cup B)=1-P(A)+P(B)-[P(B)-P(A\cap B)]$

Substitute the given values:

$=1-\frac{1}{3}+\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{4}\right)$

Simplifying:

$=\frac{2}{3}+\frac{1}{2}-\frac{1}{4}$

$=\frac{8+6-3}{12}=\frac{11}{12}$

Hence, the correct answer is Option 4.

120

Consider the following statements:

I. Mean and variance have the same unit of measurement.

II. Mean deviation and standard deviation have the same unit of measurement.

Which of the statements given above is/are correct?

((a))
I only
((b))
II only
((c))
Both I and II
((d))
Neither I nor II

Show Answer

Answer: ((b))

II only

Calculation:

Statement I: Mean and variance have the same unit of measurement.

Mean has the same unit as the original data.
Variance is the square of the unit of the data (since deviations are squared).
Therefore, mean and variance do NOT have the same unit.
Statement I is incorrect.

Statement II: Mean deviation and standard deviation have the same unit of measurement.

Mean deviation is based on absolute deviations, so it has the same unit as the data.
Standard deviation is the square root of variance, hence it also has the same unit as the data.
Therefore, Statement II is correct.

Hence, the correct answer is Option 2.

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