JEE Main 2020Inverse Trigonometric FunctionsInverse Trigonometric FunctionsHardQuestionThe domain of the function f(x)=sin−1(x2−3x+2x2+2x+7)f(x)=\sin ^{-1}\left(\frac{x^{2}-3 x+2}{x^{2}+2 x+7}\right)f(x)=sin−1(x2+2x+7x2−3x+2) is :OptionsA[1,∞)[1, \infty)[1,∞)B[−1,2][-1,2][−1,2]C[−1,∞)[-1, \infty)[−1,∞)D(−∞,2](-\infty, 2](−∞,2]Check AnswerHide SolutionSolutionf(x)=sin−1(x2−3x+2x2+2x+7)f(x)=\sin ^{-1}\left(\frac{x^{2}-3 x+2}{x^{2}+2 x+7}\right)f(x)=sin−1(x2+2x+7x2−3x+2) −1≤x2−3x+2x2+2x+7≤1-1 \leq \frac{x^{2}-3 x+2}{x^{2}+2 x+7} \leq 1−1≤x2+2x+7x2−3x+2≤1 x2−3x+2x2+2x+7≤1x2−3x+2≤x2+2x+75x≥−5x≥−1\begin{aligned} & \frac{x^{2}-3 x+2}{x^{2}+2 x+7} \leq 1 \\\\ & x^{2}-3 x+2 \leq x^{2}+2 x+7 \\\\ & 5 x \geq-5 \\\\ & x \geq-1 \end{aligned}x2+2x+7x2−3x+2≤1x2−3x+2≤x2+2x+75x≥−5x≥−1 And x2−3x+2x2+2x+7≥−1\frac{x^{2}-3 x+2}{x^{2}+2 x+7} \geq-1x2+2x+7x2−3x+2≥−1 x2−3x+2≥−x2−2x−7x^{2}-3 x+2 \geq-x^{2}-2 x-7x2−3x+2≥−x2−2x−7 2x2−x+9≥02 x^{2}-x+9 \geq 02x2−x+9≥0 x∈Rx \in Rx∈R (i) ∩\cap∩ (ii) Domain ∈[−1,∞)\in[-1, \infty)∈[−1,∞)