2(x+2)2f(x)−3(x+2)2=10∫0x(t+2)f(t)dt Differentiating both side $\begin{aligned}
& 4(x+2) f(x)+2 f(x)(x+2)^2-6(x+2)=10(x+2) f(x) \
& =(x+2) \frac{d y}{d x}-3 y=3 \
& \frac{1}{3} \int \frac{d y}{y+1}=\int \frac{d x}{x+2} \
& \ln |y+1=3 \ln | x+2 \mid+\ln c \
& y+1=(x+2)^3 c \
& \because y(0)=\frac{3}{2} \
& \Rightarrow \frac{5}{16}=c \
& \therefore \quad y=\frac{5}{16}(x+2)^3-1 \
& y(2)=\frac{5}{16} \times 64-1=19
\end{aligned}$
Let f be a differentiable function such that 2(x+2)^2 f(x)-3(x+2)^2=10 ∫_0^x(t+2) f(t) d t, x ≥ 0. Then f(2) is equal to .
JEE Main 2025 — Mathematics Calculus
Let f be a differentiable function such that 2(x+2)2f(x)−3(x+2)2=10∫0x(t+2)f(t)dt,x≥0. Then f(2) is equal to ______.
Held on 24 Jan 2025 · Verified 6 Jul 2026.
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
JEE Main Calculus in other years
More JEE Main Calculus questions
If $\int\left(\frac{1-5 \cos ^{2} x}{\sin ^{5} x \cos ^{2} x}\right) d x=f(x)+\mathrm{C}$, where C is the constant of integration, then $f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right)$ is equal to
If the solution curve $y=f(x)$ of the differential equation $\left(x^{2}-4\right) y^{\prime}-2 x y+2 x\left(4-x^{2}\right)^{2}=0, x>2$, passes through the point $(3,15)$, then the local maximum value of $f$ is $\_\_\_\_$.
Let $[\cdot]$ be the greatest integer function. If $\alpha=\int_{0}^{64}\left(x^{1 / 3}-\left[x^{1 / 3}\right]\right) \mathrm{d} x$, then $\frac{1}{\pi} \int_{0}^{\alpha \pi}\left(\frac{\sin ^{2} \theta}{\sin ^{6} \theta+\cos ^{6} \theta}\right) \mathrm{d} \theta$ is equal to $\_\_\_\_$ .
Let $P_{1}: y=4 x^{2}$ and $P_{2}: y=x^{2}+27$ be two parabolas. If the area of the bounded region enclosed between $P_{1}$ and $P_{2}$ is six times the area of the bounded region enclosed between the line $y=\alpha x, \alpha>0$ and $P_{1}$, then $\alpha$ is equal to :
If $y=y(x)$ satisfies the differential equation $16(\sqrt{x+9 \sqrt{x}})(4+\sqrt{9+\sqrt{x}}) \cos y \mathrm{~d} y=(1+2 \sin y) \mathrm{d} x, x>0$ and $y(256)=\frac{\pi}{2}, y(49)=\alpha$, then $2 \sin \alpha$ is equal to :
Other Mathematics topics for JEE Main
Keep practicing Calculus
Work through every JEE Main Calculus PYQ, year by year.