"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 :"

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Accepted Answer

"$$f(x) = {\sin ^{ - 1}}\left( {{{3{x^2} + x - 1} \over {{{(x - 1)}^2}}}} \right) + {\cos ^{ - 1}}\left( {{{x - 1} \over {x + 1}}} \right)$$

$- 1 \le {{x - 1} \over {x + 1}} \le 1 \Rightarrow 0 \le x < \infty$ .... (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 = $\left[ {{1 \over 4},{1 \over 2}} \right] \cup \{ 0\}$"

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 :

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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 :

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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 :