Question

Let $f\left(x\right)=cose{c}^{-1}[1+{\sin }^{2}x]$, where $[.]$ denotes the greatest integer function. Then $f\left(x\right)$ equals;

• A. $\left\{\frac{\pi }{2}\right\}$
• B. $\{\frac{\pi }{2},cose{c}^{-1}2\}$
• C. $\left\{cose{c}^{-1}2\right\}$
• D. none of these

Answer 1

Answer: (B)

$\{\frac{\pi }{2},cose{c}^{-1}2\}$

$-1\le \sin x\le 1\Rightarrow 0\le {\sin }^{2}x\le 1$

For ${\sin }^{2}x=0f\left(x\right)=cose{c}^{-1}\{1+0\}=cose{c}^{-1}1=\frac{\pi }{2}$

And for ${\sin }^{2}x=1f\left(x\right)=cose{c}^{-1}\{1+1\}=cose{c}^{-1}2$