Question

The difference between the greatest and least possible values of $y=(si{n}^{-1}x{)}^2+(co{s}^{-1}x{)}^2$ is

• A. $\frac{5}{8}{\pi }^2$
• B. $\frac{3}{4}{\pi }^2$
• C. $\frac{9}{8}{\pi }^2$
• D. $\frac{7}{8}{\pi }^2$

Answer 1

The correct option is C

$\frac{9}{8}{\pi }^2$

Let $si{n}^{-1}x=t.$ Then $y=t^2+(\frac{\pi }{2}-t{)}^2,-\frac{\pi }{2}\le t\le \frac{\pi }{2}$

We need to check extremum of $y=f\left(t\right)$ and its value at boundary points.

Now $y=f\left(t\right)=2t^2-\pi t+\frac{{\pi }^2}{4}$

$f^{\prime }\left(t\right)=0\Rightarrow 4t-\pi =0$ at $t=\frac{\pi }{4}$

$f\left(\frac{\pi }{4}\right)=\frac{{\pi }^2}{8}.$

Boundary points are

$f(-\frac{\pi }{2})=\frac{5{\pi }^2}{4},f\left(\frac{\pi }{2}\right)=\frac{{\pi }^2}{4}$

$\Rightarrow f(t{)}_{max}=\frac{5{\pi }^2}{4},f(t{)}_{min}=\frac{{\pi }^2}{8}.$ Required difference $=\frac{9}{8}{\pi }^2$