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The domain of the function
f(x)=\sin ^{-1}\left(\frac{3 x^{2}+x-1}{(x-1)^{2}}\right)+\cos ^{-1}\left(\frac{x-1}{x+1}\right) is :
Option: 1 \left[0, \frac{1}{2}\right]
Option: 2 \left[0, \frac{1}{4}\right]
Option: 3 \left[\frac{1}{4}, \frac{1}{2}\right] \cup\{0\}
Option: 4 [-2,0] \cup\left[\frac{1}{4}, \frac{1}{2}\right]

Answers (1)

best_answer

Domain

-1\leq \frac{3x^{2}+x-1}{\left ( x-1 \right )^{2}}\leq 1;x\neq1\\

\Rightarrow -\left ( x-1 \right )^{2}\leq 3x^{2}+x-1\leq \left ( x-1 \right )^{2}\\

\Rightarrow 4x^{2}-x\geq 0\; and\; 2x^{2}+3x-2\leq 0\\

\Rightarrow 4x\left ( x-\frac{1}{4} \right )\geq 0\; and\; 2x^{2}+4x-x-2\leq 0\\

\Rightarrow x\in\left ( -\infty,0 \right ]\cup \left [\frac{1}{4},\infty \right )\; and\; x\in\left [ -2,\frac{1}{2} \right ]\\

So\; x\in\; \left [ -2,0 \right ]\cup \left [ \frac{1}{4},\frac{1}{2} \right ]\\            ..........(1)

Also\; -1\leq \frac{x-1}{x+1}\leq 1\Rightarrow \frac{x-1}{x+1}+1\geq 0\; and\; \frac{x-1}{x+1}-1\leq 0\\

\Rightarrow \frac{2x}{x+1}\geq 0\; and\; \frac{-2}{x+1}\leq 0\Rightarrow \frac{1}{x+1}\geq 0,x\neq -1\\

\Rightarrow x \in\left ( -\infty,-1 \right ]\cup \left [ 0,\infty \right )\; and \; x\in\left ( -1,\infty \right )\\

\Rightarrow x\in\left [ 0,\infty \right )                          .......(2)

From (1) and (2)

x\in \{0\} \cup \left [ \frac{1}{4},\frac{1}{2} \right ]

Posted by

Kuldeep Maurya

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