Question

Let $\{\ast \}$ and $[\ast ]$ be the fractional part function and greatest integer function repectively. If the range of the function $f\left(x\right)={\sin }^{-1}{x}^2+\left[\left\{\ln \sqrt{x-\left[x\right]}\right\}\right]+{\cot }^{-1}\left(\frac{1}{1+\sqrt{2}{x}^2}\right)$ is $(\frac{\pi }{c},\frac{a\pi }{b}),$ then

• A. $a=7$
• B. $b=8$
• C. $c=5$
• D. $c=4$

Answer 1

Answer: (A), (B), (D)

$c=4$

Domain of function is $(-1,1)-\left\{0\right\}.$
because $x-\left[x\right]=0$ for integral value of $x$, hence, middle term will not be defined.
Also, $\left[\left\{f\right\}\right]=0,$ whenever $f$ is meaningful.

$\therefore f\left(x\right)={\sin }^{-1}{x}^2+{\tan }^{-1}(1+\sqrt{2}{x}^2)\left(\begin{array}{c}\because {\cot }^{-1}x={\tan }^{-1}\frac{1}{x},x>0\end{array}\right)$
Function is continuous and is even.
$f^{\prime }\left(x\right)=\frac{2x}{\sqrt{1-x^4}}+\frac{2\sqrt{2}x}{1+(1+\sqrt{2}{x}^2{)}^2}$
From $f^{\prime }\left(x\right)=0\Rightarrow x=0$
$x=0$ is the point of local minima

When $x\to 0,$ $f\to \frac{\pi }{4}$
Check $f\left(x\right)$ at boundary values
$\underset{x\to \pm 1}{lim}={\sin }^{-1}1+{\tan }^{-1}(1+\sqrt{2})$
$=\frac{\pi }{2}+\frac{3\pi }{8}=\frac{7\pi }{8}$
$\therefore$ Range of $f\left(x\right)$ is$(\frac{\pi }{4},\frac{7\pi }{8})$
$\Rightarrow a=7,b=8,c=4$