JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Differential Equations

Let for some function y = f(x), ∫₀ˣ t·f(t) dt = x²·f(x), x > 0 and f(2) = 3. Then f(6) is equal to:

Answer: (A)

Solution

Differentiate

\int_{0}^{x} t f (t) d t = x^{2} f (x)

with respect to

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}

Integrating,

ln f (x) = - ln x + c \Rightarrow f (x) = \frac{k}{x}

.

Using

f (2) = 3

k = 6

, so

f (x) = \frac{6}{x}

f (6) = \frac{6}{6} = 1

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