Mathematics · Application of Derivatives

JEE Main 2026 — 6 April, Evening Shift — Question 36

Let $A = \begin{vmatrix}1 & 3 & -1\\ 2 & 1 & \alpha \\ 0 & 1 & -1\end{vmatrix}$ be a singular matrix. Let $f(x) = \int_{0}^{x}(t^{2} + 2t + 3)dt$ , $x∈[1,α].$ If M and m are respectively the maximum and the minimum values of f in $[1,α]$ then $3(M-m)$ is equal to:

Option A:
$64$
Option B:
$68$
Correct
Option C:
$72$
Option D:
$76$

Answer: B

Step-by-step solution

$|A|=0$ $\Rightarrow 1 .(-1-\alpha)-3(-2-0)-1(2-0)=0$ $\Rightarrow \alpha=3$ $f(x)=\int_{0}^{x}\left(t^{2}+2 t+3\right) d t ; x \in[1,3]$ $f(x)=\frac{x^{3}}{3}+x^{2}+3 x$ $\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^{2}+2 \mathrm{x}+3 \quad$ where $\mathrm{D}<0$ $\Rightarrow \mathrm{f}(\mathrm{x})$ is strictly increasing $M=f(3)=9+9+9=27$ $\mathrm{m}=\mathrm{f}(1)=\frac{1}{3}+1+3=\frac{13}{3}$ $3(M-m)=3\left(27-\frac{13}{3}\right)=68$

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Exam: JEE Main 2026
Paper: 6 April, Evening Shift
Subject: Mathematics
Chapter: Application of Derivatives
Topic: Maxima and Minima