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JEE Advanced 2020 · Paper 2, Mathematics question 17

Let f:R→Rf:\mathbb R\to\mathbb R be a differentiable function such that its derivative f′f' is continuous and f(π)=−6f(\pi) = -6. If F:[0,π]→RF:[0,\pi]\to\mathbb R is defined by F(x)=∫0xf(t) dtF(x) = \int_0^x f(t)\,dt, and if

∫0π(f′(x)+F(x))cos⁡x dx=2,\int_0^\pi\left(f'(x) + F(x)\right)\cos x\,dx = 2,

then the value of f(0)f(0) is ______

Answer and worked solution

Official answer: 4

First integral, by parts:
∫0πf′cos⁡x dx=[fcos⁡x]0π+∫0πfsin⁡x dx=−f(π)−f(0)+∫0πfsin⁡x dx.\int_0^\pi f'\cos x\,dx=\Big[f\cos x\Big]_0^\pi+\int_0^\pi f\sin x\,dx=-f(\pi)-f(0)+\int_0^\pi f\sin x\,dx.

Second integral, by parts with F′=fF'=f and F(0)=0F(0)=0:
∫0πFcos⁡x dx=[Fsin⁡x]0π−∫0πfsin⁡x dx=−∫0πfsin⁡x dx.\int_0^\pi F\cos x\,dx=\Big[F\sin x\Big]_0^\pi-\int_0^\pi f\sin x\,dx=-\int_0^\pi f\sin x\,dx.

Adding. The sine integrals cancel:
−f(π)−f(0)=2  ⟹  6−f(0)=2  ⟹  f(0)=4.-f(\pi)-f(0)=2\implies6-f(0)=2\implies f(0)=4.
Answer: 4

Topics:Properties of definite integralsIntegration by partsFundamental theorem of calculus
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