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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 $30$.

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.

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.

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.

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.

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.

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.

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.

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.

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.