JEE Mathematics 2025 Question with Solution

Given:

$$f(x) = \left| \begin{matrix} a + \frac{\sin x}{x} & 1 & b\\ a & 1 + \frac{\sin x}{x} & b\\ a & 1 & b + \frac{\sin x}{x} \end{matrix} \right|$$

and

$$\lim_{x \to 0} \frac{\sin x}{x} = 1$$

Find: $(\lambda + \mu + \nu)^2$ where

$$\lim_{x \to 0} f(x) = \lambda + \mu a + \nu b$$

As $x \to 0$, the determinant becomes

$$\left| \begin{matrix} a + 1 & 1 & b\\ a & 2 & b\\ a & 1 & b + 1 \end{matrix} \right|$$

Now apply row operations using the structure visible in the matrix:

$$R_1 \to R_1 - R_2, \qquad R_3 \to R_3 - R_2$$

Then the determinant becomes

$$\left| \begin{matrix} 1 & -1 & 0\\ a & 2 & b\\ 0 & -1 & 1 \end{matrix} \right|$$

Expanding this determinant,

$$\begin{aligned} \det &= 1\cdot \begin{vmatrix} 2 & b\\ -1 & 1 \end{vmatrix} - (-1)\cdot \begin{vmatrix} a & b\\ 0 & 1 \end{vmatrix} \\ &= (2\cdot 1 - b(-1)) + a \\ &= 2 + b + a \end{aligned}$$

So,

$$\lambda + \mu a + \nu b = a + b + 2$$

Comparing coefficients,

$$\lambda = 2, \qquad \mu = 1, \qquad \nu = 1$$

Hence,

$$\lambda + \mu + \nu = 2 + 1 + 1 = 4$$

and therefore

$$(\lambda + \mu + \nu)^2 = 16$$

The determinant-based working gives $16$. The solution marks option D, but that contradicts the extracted determinant calculation. The most defensible option from the working is B.

Given:

$$\left| \begin{matrix} a + 1 & 1 & b\\ a & 2 & b\\ a & 1 & b + 1 \end{matrix} \right|$$

after taking the limit $x \to 0$.

Find: $(\lambda + \mu + \nu)^2$.

Expand along the first row:

$$\begin{aligned} f(0) &= (a+1)\begin{vmatrix} 2 & b\\ 1 & b+1 \end{vmatrix} - 1\begin{vmatrix} a & b\\ a & b+1 \end{vmatrix} + b\begin{vmatrix} a & 2\\ a & 1 \end{vmatrix} \end{aligned}$$

Compute the minors:

$$\begin{aligned} \begin{vmatrix} 2 & b\\ 1 & b+1 \end{vmatrix} &= 2(b+1) - b = b+2 \\ \begin{vmatrix} a & b\\ a & b+1 \end{vmatrix} &= a(b+1) - ab = a \\ \begin{vmatrix} a & 2\\ a & 1 \end{vmatrix} &= a - 2a = -a \end{aligned}$$

Substitute back:

$$\begin{aligned} f(0) &= (a+1)(b+2) - a + b(-a) \\ &= ab + 2a + b + 2 - a - ab \\ &= a + b + 2 \end{aligned}$$

Therefore,

$$\lambda = 2, \quad \mu = 1, \quad \nu = 1$$

So,

$$(\lambda + \mu + \nu)^2 = (2+1+1)^2 = 16$$

Therefore, the correct option according to the actual determinant evaluation is B.