Question

If $f\left(x\right)={\left({\sin }^{2}x-1\right)}^n,\mathrm{then}x=\frac{\pi }{2}$ 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: (A), (D)

The correct options are
A local maximum, if n is odd
D local minimum, if n is even

$\mathrm{Given}\mathrm{that},f\left(x\right)={\left({\sin }^{2}x-1\right)}^{n}f\left(\frac{\pi }{2}\right)=0f\left({\frac{\pi }{2}}^{+}\right)={\left(\to 0^{-}\right)}^n\mathrm{and}f\left({\frac{\pi }{2}}^{-}\right)={\left(\to 0^{-}\right)}^n\mathrm{If}n\mathrm{is}\mathrm{even},f\left({\frac{\pi }{2}}^{+}\right)\mathrm{and}f\left({\frac{\pi }{2}}^{-}\right)>0.\mathrm{Then}x=\frac{\pi }{2}\mathrm{is}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{minima}.\mathrm{If}n\mathrm{is}\mathrm{odd},f\left({\frac{\pi }{2}}^{+}\right)\mathrm{and}f\left({\frac{\pi }{2}}^{-}\right)<0.\mathrm{Then}x=\frac{\pi }{2}\mathrm{is}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{maxima}.$