JEE Main 2026 — Jan 28, Shift 2 — Mathematics solution

JEE Main2026Jan 28, Shift 2Mathematics

Let $f$ be a differentiable function satisfying $f (x) = 1 - 2 x + \int_{0}^{x} e^{(x - t)} f (t) d t, x \in R$ and let $g (x) = \int_{0}^{x} (f (t) + 2)^{15} (t - 4)^{6} (t + 12)^{17} d t, x \in R$ . If $p$ and $q$ are respectively the points of local minima and local maxima of $g$ , then the value of $∣ p + q ∣$ is equal to ______ .

Your answer

Solution

Find $f$ . Rewrite $f (x) = 1 - 2 x + e^{x} \int_{0}^{x} e^{- t} f (t) d t$ . Multiplying by $e^{- x}$ : $e^{- x} f (x) = (1 - 2 x) e^{- x} + \int_{0}^{x} e^{- t} f (t) d t$ .

Differentiate both sides: $e^{- x} f^{'} (x) - e^{- x} f (x) = - 2 e^{- x} - (1 - 2 x) e^{- x} + e^{- x} f (x)$ , which simplifies to $f^{'} (x) - 2 f (x) = 2 x - 3$ .

Solve this linear ODE (integrating factor $e^{- 2 x}$ ): $y \cdot e^{- 2 x} = \int e^{- 2 x} (2 x - 3) d x$ . After integration and applying $f (0) = 1$ , the result simplifies to $f (x) = 1 - x$ .

Analyse $g$ . $g^{'} (x) = (f (x) + 2)^{15} (x - 4)^{6} (x + 12)^{17} = (3 - x)^{15} (x - 4)^{6} (x + 12)^{17}$ . Equivalently $g^{'} (x) = - (x - 3)^{15} (x - 4)^{6} (x + 12)^{17}$ .

Sign of $g^{'}$ across critical points $x = - 12, 3, 4$ (with the factor $(x - 4)^{6} \geq 0$ always):

$x < - 12$ : $- (-)^{15} (+) (-)^{17} = - (-) (-) = -$ . At $x = - 12$ sign changes from $-$ to $+$ : local minimum.

$- 12 < x < 3$ : positive. At $x = 3$ sign flips to negative (odd power of $(x - 3)$ ): local maximum.

So $p = - 12$ (local min) and $q = 3$ (local max). $∣ p + q ∣ = ∣ - 12 + 3∣ = 9$ .

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