Applications-Of-Derivatives Question 436
Question: What is the value of p for which the function $ f(x)=p\sin x+\frac{\sin 3x}{3} $ has an extremum at $ x=\frac{\pi }{3} $ ?
Options:
A) 0
B) 1
C) -1
D) 2
Show Answer
Answer:
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’( \frac{\pi }{3} )=0\Rightarrow p\cos \frac{\pi }{3}+\cos \pi =0 $
$ \Rightarrow \frac{p}{2}-1=0\Rightarrow p=2 $
BETA
