Question

$f\left(x\right)$ $=\{\begin{array}{c}{\sin }^{2}x;x\le 0\\ {\cos }^{2}x;x>0\end{array}$
The value of $f(\frac{\pi }{3})+f(\frac{-\pi }{2})$ is

• A. $\frac{1}{4}$
• B. $\frac{5}{4}$
• C. $1$
• D. $0$

Answer 1

The correct option is B $\frac{5}{4}$

$\Rightarrow$ In function $f\left(\frac{\pi }{3}\right)$, $x$ if greater than $0$

$\Rightarrow$ $f\left(x\right)={\cos }^{2}x$

$\Rightarrow$ $f\left(\frac{\pi }{3}\right)={\cos }^2\frac{\pi }{3}$

$\therefore$ $f\left(\frac{\pi }{3}\right)=(\frac{1}{2}{)}^2=\frac{1}{4}$ ----- ( 1 ).

$\Rightarrow$ In function $f\left(\frac{-\pi }{2}\right)$, $x$ is less than $0$.

$\Rightarrow$ So, $f\left(x\right)={\sin }^{2}x$

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

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

From ( 1 ) and ( 2 ),

$\Rightarrow$ $f\left(\frac{\pi }{3}\right)+f\left(\frac{-\pi }{2}\right)=\frac{1}{4}+1=\frac{5}{4}$