The maximum value of function f(x) = _0^1 t , , ( x + t ) dt, ,x … - Vidyadip

MCQ

The maximum value of function $f(x) = \int\limits_0^1 {t\,\sin \,\left( {x + \pi t} \right)} dt,\,x \in \,R$ is

A
$\frac{1}{\pi }\sqrt {{\pi ^2} + 4} $
✓
$\frac{1}{{{\pi ^2}}}\sqrt {{\pi ^2} + 4} $
C
$\sqrt {{\pi ^2} + 4} $
D
$\frac{1}{{2{\pi ^2}}}\sqrt {{\pi ^2} + 4} $

✓

Answer

Correct option: B.

$\frac{1}{{{\pi ^2}}}\sqrt {{\pi ^2} + 4} $

b
$f(x) = \left[ { - \frac{t}{\pi }\cos (x + \pi t)} \right]_0^1 + \frac{1}{\pi }\int\limits_0^1 {\cos } (x + \pi t)dt$

$=\frac{1}{\pi} \cos x-\frac{2}{\pi^{2}} \sin x$

$\therefore \mathrm{f}(\mathrm{x})_{\max }=\sqrt{\frac{1}{\pi^{2}}+\frac{4}{\pi^{4}}}=\frac{\sqrt{\pi^{2}+4}}{\pi^{2}}$

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