CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If $$\displaystyle f(x) = \int_{0}^{x} (cos(sin t) + cos (cos t)) dt$$, then $$f(x + \pi)$$ is


A
f(x)+f(π)
loader
B
f(x)+2f(π)
loader
C
f(x)+f(π2)
loader
D
f(x)+2f(π2)
loader

Solution

The correct options are
B $$\displaystyle f(x) + 2f(\frac {\pi}{2})$$
C $$\displaystyle f(x) + f(\pi)$$
$$\displaystyle f(x+\pi) = \int_{0}^{x+\pi} (cos(sin \: t) + cos(cos \: t)) dt$$
$$\displaystyle = \int_{0}^{\pi} (cos(sin \: t) + cos(cos \: t)) dt$$
$$\displaystyle +\int_{\pi}^{x+\pi} (cos(sin \: t) + cos(cos \: t)) dt$$
$$\displaystyle = f(\pi) + \int_{0}^{x} (cos(sin \: t) + cos(cos \: t)) dt$$
$$\displaystyle [\therefore $$ for $$g(x) = cos(sin \: x) + cos(cos \: x), f(x + \pi) = f(x)]$$
$$\displaystyle = f(\pi) + f(x)$$
$$\displaystyle = f(\pi) + 2f(\frac {\pi}{2})$$
$$\displaystyle [\therefore g(x) \:$$ has period $$\: \pi/2]$$
Ans: A,D

Maths

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image