JEE MAIN - Mathematics (2021 - 31st August Evening Shift - No. 2)
The domain of the function
$$f(x) = {\sin ^{ - 1}}\left( {{{3{x^2} + x - 1} \over {{{(x - 1)}^2}}}} \right) + {\cos ^{ - 1}}\left( {{{x - 1} \over {x + 1}}} \right)$$ is :
$$f(x) = {\sin ^{ - 1}}\left( {{{3{x^2} + x - 1} \over {{{(x - 1)}^2}}}} \right) + {\cos ^{ - 1}}\left( {{{x - 1} \over {x + 1}}} \right)$$ is :
$$\left[ {0,{1 \over 4}} \right]$$
$$[ - 2,0] \cup \left[ {{1 \over 4},{1 \over 2}} \right]$$
$$\left[ {{1 \over 4},{1 \over 2}} \right] \cup \{ 0\} $$
$$\left[ {0,{1 \over 2}} \right]$$
Explanation
$$f(x) = {\sin ^{ - 1}}\left( {{{3{x^2} + x - 1} \over {{{(x - 1)}^2}}}} \right) + {\cos ^{ - 1}}\left( {{{x - 1} \over {x + 1}}} \right)$$
$$ - 1 \le {{x - 1} \over {x + 1}} \le 1 \Rightarrow 0 \le x < \infty $$ .... (1)
$$ - 1 \le {{3{x^2} + x - 1} \over {{{(x - 1)}^2}}} \le 1 \Rightarrow x \in \left[ {{{ - 1} \over 4},{1 \over 2}} \right] \cup \{ 0\} $$ .... (2)
(1) & (2)
$$\Rightarrow$$ Domain = $$\left[ {{1 \over 4},{1 \over 2}} \right] \cup \{ 0\} $$
$$ - 1 \le {{x - 1} \over {x + 1}} \le 1 \Rightarrow 0 \le x < \infty $$ .... (1)
$$ - 1 \le {{3{x^2} + x - 1} \over {{{(x - 1)}^2}}} \le 1 \Rightarrow x \in \left[ {{{ - 1} \over 4},{1 \over 2}} \right] \cup \{ 0\} $$ .... (2)
(1) & (2)
$$\Rightarrow$$ Domain = $$\left[ {{1 \over 4},{1 \over 2}} \right] \cup \{ 0\} $$
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