A) 0
B) 1
C) -1
D) 2
Correct Answer: D
Solution :
[d] Let \[f(x)=p\,\,\sin x+\frac{\sin 3x}{3}\] Differentiate both side w.r.t(x). \[\Rightarrow f'(x)=p\,\,\cos \,\,x+\frac{3\cos 3x}{3}=p\cos x+\cos 3x\] It is given that \[f(x)\] has extreme value at \[x=\pi /3\] \[\therefore f'\left( \frac{\pi }{3} \right)=0\Rightarrow p\cos \frac{\pi }{3}+\cos \pi =0\] \[\Rightarrow \frac{p}{2}-1=0\Rightarrow p=2\]You need to login to perform this action.
You will be redirected in
3 sec