JEE Main 2026 Jan 24 Shift 1, Mathematics Q21: Differentiation Under the Integral Sign

JEE Main2026Jan 24, Shift 1Mathematics

Let a differentiable function $f$ satisfy the equation

$\int_{0}^{36} f (\frac{t x}{36}) d t = 4 α f (x)$ .

If $y = f (x)$ is a standard parabola passing through the points $(2, 1)$ and $(- 4, β)$ , then $β^{α}$ is equal to _____ .

Your answer

Solution

Substitute $u = \frac{t x}{36}$ (treating $x > 0$ ): $d u = \frac{x}{36} d t$ , and as $t$ goes from $0$ to $36$ , $u$ goes from $0$ to $x$ .

$\int_{0}^{x} f (u) \cdot \frac{36}{x} d u = 4 α f (x) \Rightarrow \int_{0}^{x} f (u) d u = \frac{α x}{9} f (x)$ .

Differentiate both sides w.r.t. $x$ :

$f (x) = \frac{α}{9} [f (x) + x f^{'} (x)] \Rightarrow (9 - α) f (x) = α x f^{'} (x)$ .

$\frac{f ^{'} ( x )}{f ( x )} = \frac{9 - α}{α x} \Rightarrow ln f (x) = (\frac{9}{α} - 1) ln x + C$ .

$\Rightarrow f (x) = c x^{9/ α - 1}$ .

For $f$ to be a standard parabola $y = c x^{2}$ , the exponent must equal $2$ : $\frac{9}{α} - 1 = 2 \Rightarrow α = 3$ .

$(2, 1)$ on the curve: $1 = 4 c \Rightarrow c = \frac{1}{4}$ , so $y = \frac{x ^{2}}{4}$ .

At $x = - 4$ : $β = \frac{16}{4} = 4$ .

$β^{α} = 4^{3} = 64$ .

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