Mathematics · Definite Integration

JEE Main 2025 — 24 January, Morning Shift — Question 21

Let f be a differentiable function such that $2(x+2)^{2} f(x)-3(x+2)^{2}=10 \int_{0}^{x}(t+2) f(t) d t$ ,

$x \geq 0$ . Then $f(2)$ is equal to $\_\_\_\_\_$ .

Answer: 19

Numerical answer — enter this value.

Differentiate both sides

$4(x+2) f(x)+2(x+2)^{2} f^{\prime}(x)-6(x+2)=10(x+2) f(x)$

$2(x+2)^{2} f^{\prime}(x)-6(x+2) f(x)=6(x+2)$

$(x+2) \frac{d y}{d x}-3 y=3$

$\int \frac{d y}{y+1}=3 \int \frac{d x}{x+2}$

$\ln (y+1)=3 \ln (x+2)+C$

$(y+1)=C(x+2)^{3}$

$f(0)=\frac{3}{2}$

$f(2)=19$

Answer key and solution verified before publishing.

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