JEE Main 2024Inverse Trigonometric FunctionsInverse Trigonometric FunctionsEasyQuestionThe domain of the function f(x)=sin−1(3x2+x−1(x−1)2)+cos−1(x−1x+1)f(x) = {\sin ^{ - 1}}\left( {{{3{x^2} + x - 1} \over {{{(x - 1)}^2}}}} \right) + {\cos ^{ - 1}}\left( {{{x - 1} \over {x + 1}}} \right)f(x)=sin−1((x−1)23x2+x−1)+cos−1(x+1x−1) is :OptionsA[0,14]\left[ {0,{1 \over 4}} \right][0,41]B[−2,0]∪[14,12][ - 2,0] \cup \left[ {{1 \over 4},{1 \over 2}} \right][−2,0]∪[41,21]C[14,12]∪{0}\left[ {{1 \over 4},{1 \over 2}} \right] \cup \{ 0\} [41,21]∪{0}D[0,12]\left[ {0,{1 \over 2}} \right][0,21]Check AnswerHide SolutionSolutionf(x)=sin−1(3x2+x−1(x−1)2)+cos−1(x−1x+1)f(x) = {\sin ^{ - 1}}\left( {{{3{x^2} + x - 1} \over {{{(x - 1)}^2}}}} \right) + {\cos ^{ - 1}}\left( {{{x - 1} \over {x + 1}}} \right)f(x)=sin−1((x−1)23x2+x−1)+cos−1(x+1x−1) −1≤x−1x+1≤1⇒0≤x<∞- 1 \le {{x - 1} \over {x + 1}} \le 1 \Rightarrow 0 \le x < \infty−1≤x+1x−1≤1⇒0≤x<∞ .... (1) −1≤3x2+x−1(x−1)2≤1⇒x∈[−14,12]∪{0}- 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\}−1≤(x−1)23x2+x−1≤1⇒x∈[4−1,21]∪{0} .... (2) (1) & (2) ⇒\Rightarrow⇒ Domain = [14,12]∪{0}\left[ {{1 \over 4},{1 \over 2}} \right] \cup \{ 0\} [41,21]∪{0}