Question

Let $f\left(x\right)={\sin }^{-1}x+{\cos }^{-1}x,$ then $\frac{\pi }{2}$ is equal to

• A. $f(-\frac{1}{2})$
• B. $f(k^2-2k+3),k\in R$
• C. $f\left(\frac{1}{1+k^2}\right),k\in R$
• D. $f(-2)$

Answer 1

Answer: (A), (C)

$f(-\frac{1}{2})$
$f\left(\frac{1}{1+k^2}\right),k\in R$

$f\left(x\right)={\sin }^{-1}x+{\cos }^{-1}x.$
We know that
${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}x\in [-1,1]$
Now, option A, $f(-\frac{1}{2})=\frac{\pi }{2}$ as $-\frac{1}{2}\in [-1,1]$
For option B, $k^2\ge 0$
$\Rightarrow k^2+3\ge 3$
$\Rightarrow k^2-2k+3\notin [-1,1]$
So, option B does not holds.
Option C, $k^2+1\ge 1$
$\Rightarrow \frac{1}{k^2+1}\le 1$
Hence, $f\left(\frac{1}{k^2+1}\right)=\frac{\pi }{2}$

Option D, $-2\notin [-1,1]$
So, does not hold.