Question

If $f\left(x\right)=({\sin }^{2}x-1{)}^n$, then $x=\frac{\pi }{2}$ is a point of

• A. Local Maximum, if $n$ is odd
• B. Local Minimum, if $n$ is odd
• C. Local Maximum, if $n$ is even
• D. Local Minimum, if $n$ is even

Answer 1

Answer: (B), (D)

The correct options are
B Local Minimum, if $n$ is odd
D Local Minimum, if $n$ is even

Given : $f\left(x\right)={\left({\sin }^{2}x-1\right)}^n$

Differentiate w.r.t $x$.

$f^{{}^{\prime }}\left(x\right)=n{\left({\sin }^{2}x-1\right)}^{n-1}\left(2\sin x\cos x\right)$

$Atx=\frac{\pi }{2}$

$f^{{}^{\prime }}\left(\frac{\pi }{2}\right)=n{\left({\sin }^2\left(\frac{\pi }{2}\right)-1\right)}^{n-1}2\sin \left(\frac{\pi }{2}\right)$

$f^{{}^{\prime }}\left(\frac{\pi }{2}\right)=0$

$f^{{}^{\prime }}\left(\frac{\pi }{2}\right)={\left(0^{-}\right)}^n$

$f^{{}^{\prime }}{\left(\frac{\pi }{2}\right)}^{}={\left(0^{-}\right)}^n$

$f^{{}^{\prime }}{\left(\frac{\pi }{2}\right)}^{}=0$

Which implies that $\Rightarrow$ If $n$ is odd, then the function is of local maximum.

$\Rightarrow$ If $n$ is even, then the function is of local minimum.

Hence, the correct answer is options B and D.