Question

The value of $\{si{n}^{-1}[x^2+\frac{1}{2}]+co{s}^{-1}[x^2-\frac{1}{2}]\}$ is :
where {.} and [.] denote fractional part function and greatest integer function respectively

• A. $\pi -3$
• B. $4-\pi$
• C. $2-\frac{\pi }{2}$
• D. 0

Answer 1

The correct option is A $\pi -3$

Given $\{\sin {}^{-1}[x^2+\frac{1}{2}]+\cos {}^{-1}[x^2-\frac{1}{2}]\}=?$

we know

$-1\le x^2+\frac{1}{2}\le 1\text{,}-1\le x^2-\frac{1}{2}\le 1\frac{-3}{2}\le x^2\le \frac{1}{2}\text{,}\frac{-1}{2}\le x^2\le \left(\frac{3}{2}\right)0\le x\le \frac{1}{\sqrt{2}}\text{,}0\le x\le \sqrt{\frac{3}{2}}$

combining both we get $0\le x\le \left(\frac{1}{\sqrt{2}}\right)$

for $x=0,{\sin }^{-1}\left[\frac{1}{2}\right]+{\cos }^{-1}\left[\frac{-1}{2}\right]=0+\pi =\pi$

for $x=\frac{1}{\sqrt{2}},{\sin }^{-1}\left(1\right)+{\cos }^{-1}\left(0\right)=\frac{\pi }{2}+\frac{\pi }{2}=\pi$

hence for all $0\le x\le \frac{1}{\sqrt{2}},$

${\sin }^{-1}[x^2+\frac{1}{2}]+{\cos }^{-1}[x^2-\frac{1}{2}]=\pi \{{\sin }^{-1}[x^2+\frac{1}{2}]+{\cos }^{-1}[x^2-\frac{1}{2}]\}=\left\{\pi \right\}=\pi -3$