Q42 / 75 · JEE Main 2026 (06 April Shift 2) · Mathematics
Let $A = \begin{bmatrix} 1 & 3 & -1 \\ 2 & 1 & \alpha \\ 0 & 1 & -1 \end{bmatrix}$ be a singular matrix. Let $f(x) = \int\limits_0^x (t^2 + 2t + 3)\,dt$, $x \in [1, \alpha]$. If $M$ and $m$ are respectively the maximum and the minimum values of $f$ in $[1, \alpha]$, then $3(M - m)$ is equal to :
- A$64$
- Correct answerB$68$
- C$72$
- D$76$
Solution
Since $A$ is a singular matrix, $|A| = 0$.
$|A| = 1(-1 - \alpha) - 3(-2 - 0) - 1(2 - 0) = 0$
$-1 - \alpha + 6 - 2 = 0 \Rightarrow \alpha = 3$
The function is $f(x) = \int\limits_0^x (t^2 + 2t + 3)\,dt = \dfrac{x^3}{3} + x^2 + 3x$
Differentiating with respect to $x$, we get $f'(x) = x^2 + 2x + 3 = (x+1)^2 + 2 > 0$ for all $x$.
Thus, $f(x)$ is a strictly increasing function on the interval $[1, 3]$.
The maximum value $M$ occurs at $x = 3$:
$M = f(3) = \dfrac{3^3}{3} + 3^2 + 3(3) = 9 + 9 + 9 = 27$
The minimum value $m$ occurs at $x = 1$:
$m = f(1) = \dfrac{1^3}{3} + 1^2 + 3(1) = \dfrac{13}{3}$
We need to find $3(M - m)$:
$3(M - m) = 3\left(27 - \dfrac{13}{3}\right) = 81 - 13 = 68$
Answer: $68$
$|A| = 1(-1 - \alpha) - 3(-2 - 0) - 1(2 - 0) = 0$
$-1 - \alpha + 6 - 2 = 0 \Rightarrow \alpha = 3$
The function is $f(x) = \int\limits_0^x (t^2 + 2t + 3)\,dt = \dfrac{x^3}{3} + x^2 + 3x$
Differentiating with respect to $x$, we get $f'(x) = x^2 + 2x + 3 = (x+1)^2 + 2 > 0$ for all $x$.
Thus, $f(x)$ is a strictly increasing function on the interval $[1, 3]$.
The maximum value $M$ occurs at $x = 3$:
$M = f(3) = \dfrac{3^3}{3} + 3^2 + 3(3) = 9 + 9 + 9 = 27$
The minimum value $m$ occurs at $x = 1$:
$m = f(1) = \dfrac{1^3}{3} + 1^2 + 3(1) = \dfrac{13}{3}$
We need to find $3(M - m)$:
$3(M - m) = 3\left(27 - \dfrac{13}{3}\right) = 81 - 13 = 68$
Answer: $68$
Hence, the correct answer is option: 'B'.