JEE Advanced 2020 — Paper 2 — Question 50

Given $F(x)=\int_0^x f(t)\,dt$, so $F'(x)=f(x)$ and $F''(x)=f'(x)$. The integral is $\int_0^\pi [f'(x)+F(x)]\cos x\,dx = \int_0^\pi f'(x)\cos x\,dx + \int_0^\pi F(x)\cos x\,dx$. Evaluate $\int_0^\pi f'(x)\cos x\,dx$ by parts: $u=\cos x$, $dv=f'(x)dx$, so $du=-\sin x\,dx$, $v=f(x)$.

Then $\int_0^\pi f'(x)\cos x\,dx = [\cos x f(x)]_0^\pi + \int_0^\pi f(x)\sin x\,dx = (\cos\pi f(\pi)-\cos0 f(0)) + \int_0^\pi f(x)\sin x\,dx = (-1)(-6)-(1)f(0)+\int_0^\pi f(x)\sin x\,dx = 6-f(0)+\int_0^\pi f(x)\sin x\,dx$. Evaluate $\int_0^\pi F(x)\cos x\,dx$ by parts: $u=F(x)$, $dv=\cos x\,dx$, so $du=f(x)\,dx$, $v=\sin x$.

Then $\int_0^\pi F(x)\cos x\,dx = [F(x)\sin x]_0^\pi - \int_0^\pi f(x)\sin x\,dx = (F(\pi)\sin\pi - F(0)\sin0) - \int_0^\pi f(x)\sin x\,dx = 0-0-\int_0^\pi f(x)\sin x\,dx = -\int_0^\pi f(x)\sin x\,dx$. Adding: $\int_0^\pi [f'(x)+F(x)]\cos x\,dx = [6-f(0)+\int_0^\pi f(x)\sin x\,dx] + [-\int_0^\pi f(x)\sin x\,dx] = 6-f(0)$. Given equals 2, so $6-f(0)=2 \Rightarrow f(0)=4$.