JEE Main 2025 Jan 28 Shift 1, Mathematics Q16: Differentiation Under the Integral Sign

JEE Main2025Jan 28, Shift 1Mathematics

Let for some function $y = f (x)$ , $\int_{0}^{x} t f (t) d t = x^{2} f (x)$ , $x > 0$ and $f (2) = 3$ . Then $f (6)$ is equal to :

A
$1$
B
$2$
C
$6$
D
$3$

Solution

Differentiate both sides of $\int_{0}^{x} t f (t) d t = x^{2} f (x)$ with respect to $x$ . By the fundamental theorem of calculus the LHS becomes $x f (x)$ ; the RHS by product rule becomes $2 x f (x) + x^{2} f^{'} (x)$ .

$x f (x) = 2 x f (x) + x^{2} f^{'} (x)$

$- x f (x) = x^{2} f^{'} (x) \Rightarrow \frac{f ^{'} ( x )}{f ( x )} = - \frac{1}{x}$ (for $x > 0$ and $f \neq = 0$ ).

Integrate: $ln ∣ f (x) ∣ = - ln x + C \Rightarrow f (x) = \frac{K}{x}$ .

Use $f (2) = 3$ : $K = 6$ . So $f (x) = \frac{6}{x}$ , and $f (6) = 1$ .

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JEE Main 2025 Jan 28 Shift 1, Mathematics Q16: Differentiation Under the Integral Sign

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