If $f\left( x \right) = \mathop \smallint \nolimits_0^x \,\,({\sin ^4}t\,\, + \,\,{\cos ^4}\,\,t)\,dt$, then \[f\,\,\left( {x\,\, + \,\,{\rm{\pi }}} \right)\] will be equal to

$f\left( x \right)\,\, + \,\,f\left( {\rm{\pi }} \right)$ or $f\left( x \right)\,\, + \,\,2f\left( {\frac{{\rm{\pi }}}{2}} \right)$

$f\left( x \right)\,\, + \,\,f\left( {\frac{{\rm{\pi }}}{2}} \right)$

$f\left( x \right)\,\, - \,\,f\left( {\rm{\pi }} \right)$

$f\left( x \right)\,\, - \,\,2\,f\left( {\frac{{\rm{\pi }}}{2}} \right)$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Hard

1 of 1

Application +4 | -1

f\left( x \right) = \int \nolimits_0^x \,\,({\sin ^4}t\,\, + \,\,{\cos ^4}\,\,t)\,dt

, then

f\,\,\left( {x\,\, + \,\,\mathrm{\pi }} \right)

will be equal to

View answer and explanation

Correct option: B

f\left( x \right)\,\, + \,\,f\left( \mathrm{\pi } \right)

f\left( x \right)\,\, + \,\,2f\left( {\frac{\mathrm{\pi }}{2}} \right)

Option B (

f(x) + f(\pi)

f(x) + 2f(\frac{{\pi }}{2})

) is correct because the function is periodic and the integral can be split into parts.