JEE Main 2026 Apr 6 Shift 2, Mathematics Q17: Definite Integration And Fundamental Theorem

JEE Main2026Apr 6, Shift 2Mathematics

Let $A = 13 - 1 21 α 01 - 1$ be a singular matrix. Let $f (x) = \int_{0}^{x} (t^{2} + 2 t + 3) d t$ , $x \in [1, α]$ . If $M$ and $m$ are respectively the maximum and the minimum values of $f$ in $[1, α]$ , then $3 (M - m)$ is equal to:

A
64
B
68
C
72
D
76

Solution

$∣ A ∣ = 1 (- 1 - α) - 3 (- 2 - 0) - 1 (2 - 0) = - 1 - α + 6 - 2 = 3 - α = 0$ $\Rightarrow$ $α = 3$ .

$f (x) = \frac{x ^{3}}{3} + x^{2} + 3 x$ . Derivative $f^{'} (x) = x^{2} + 2 x + 3$ has discriminant $4 - 12 < 0$ , so $f^{'} (x) > 0$ . Hence $f$ is strictly increasing on $[1, 3]$ .

$M = f (3) = 9 + 9 + 9 = 27$ . $m = f (1) = \frac{1}{3} + 1 + 3 = \frac{13}{3}$ .

$3 (M - m) = 3 (27 - \frac{13}{3}) = 81 - 13 = 68$ .

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JEE Main 2026 Apr 6 Shift 2, Mathematics Q17: Definite Integration And Fundamental Theorem

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