Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be a differentiable function such that its derivative \(f^{\prime}\) is continuous and \(f(\pi)=-6\). If \(F:[0, \pi] \rightarrow \mathbb{R}\) is defined by \(F(x)=\int_{0}^{x} f(t) dt\), and if \[\int_{0}^{\pi}\left(f^{\prime}(x)+F(x)\right) \cos x dx=2\] then the value of \(f(0)\) is ____

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Definite Integrals

Let

f : R \to R

be a differentiable function such that its derivative

f^{'}

is continuous and

f (π) = - 6

. If

F : [0, π] \to R

is defined by

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

, and if

\int_{0}^{π} (f^{'} (x) + F (x)) cos x d x = 2

then the value of

f (0)

is ____

Answer: 4

Since

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

, we have

F^{'} (x) = f (x)

. The integral becomes:

\int_{0}^{π} f^{'} (x) cos x d x + \int_{0}^{π} F (x) cos x d x = 2

For the first integral, integrate by parts:

\int_{0}^{π} f^{'} (x) cos x d x = [f (x) cos x]_{0}^{π} + \int_{0}^{π} f (x) sin x d x = f (π) cos π - f (0) cos 0 + \int_{0}^{π} f (x) sin x d x = - f (π) - f (0) + \int_{0}^{π} f (x) sin x d x = 6 - f (0) + \int_{0}^{π} F^{'} (x) sin x d x

.
For

\int_{0}^{π} F^{'} (x) sin x d x

, integrate by parts:

[F (x) sin x]_{0}^{π} - \int_{0}^{π} F (x) cos x d x = - \int_{0}^{π} F (x) cos x d x

.
So the original integral becomes:

6 - f (0) - \int_{0}^{π} F (x) cos x d x + \int_{0}^{π} F (x) cos x d x = 6 - f (0) = 2

, giving

f (0) = 4

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Numerical answer