JEE Advance - Mathematics (2020 - Paper 2 Offline - No. 17)
Let $$f:R \to R$$ be a differentiable function such that its derivative f' is continuous and f($$\pi $$) = $$-$$6.
If $$F:[0,\pi ] \to R$$ is defined by $$F(x) = \int_0^x {f(t)dt} $$, and if $$\int_0^\pi {(f'(x)} + F(x))\cos x\,dx$$ = 2
then the value of f(0) is ...........
If $$F:[0,\pi ] \to R$$ is defined by $$F(x) = \int_0^x {f(t)dt} $$, and if $$\int_0^\pi {(f'(x)} + F(x))\cos x\,dx$$ = 2
then the value of f(0) is ...........
Answer
4
Explanation
It is given that, for functions
$$f:R \to R$$
and $$F:[0,\pi ] \to R$$,
$$F(x) = \int_0^x {f(t)dt} $$, where $$f(\pi ) = - 6$$
$$ \Rightarrow F'(\pi ) = f(\pi ) = - 6$$ .... (i)
Now, $$\int_0^\pi {(f'(x)} + F(x))\cos x\,dx$$
$$ = \int_0^\pi {f'(x)\cos x\,dx + \int_0^\pi {F(x)\cos x\,dx} } $$
$$ = [\cos x\,f(x)]_0^\pi + \int_0^\pi {f(x)\sin x\,dx} + \int_0^\pi {F(x)\cos x\,dx} $$
{by integration by parts}
$$ = ( - 1)( - 6) - f(0) + [(\sin x)F(x)]_0^\pi - \int_0^\pi {F(x)\cos x\,dx + \int_0^\pi {F(x)\cos x\,dx} } $$
$$ = 6 - f(0) = 2$$ (given)
$$ \Rightarrow f(0) = 4$$
$$f:R \to R$$
and $$F:[0,\pi ] \to R$$,
$$F(x) = \int_0^x {f(t)dt} $$, where $$f(\pi ) = - 6$$
$$ \Rightarrow F'(\pi ) = f(\pi ) = - 6$$ .... (i)
Now, $$\int_0^\pi {(f'(x)} + F(x))\cos x\,dx$$
$$ = \int_0^\pi {f'(x)\cos x\,dx + \int_0^\pi {F(x)\cos x\,dx} } $$
$$ = [\cos x\,f(x)]_0^\pi + \int_0^\pi {f(x)\sin x\,dx} + \int_0^\pi {F(x)\cos x\,dx} $$
{by integration by parts}
$$ = ( - 1)( - 6) - f(0) + [(\sin x)F(x)]_0^\pi - \int_0^\pi {F(x)\cos x\,dx + \int_0^\pi {F(x)\cos x\,dx} } $$
$$ = 6 - f(0) = 2$$ (given)
$$ \Rightarrow f(0) = 4$$
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