Q66 / 160 · TS EAMCET 2021 (06 Aug Shift 2) · Mathematics

If the range of $\operatorname{sech}^{-1} x+\operatorname{cosech}^{-1} x$ is $[a, b]$, then

  • A
    $a=0, b=1$
  • B
    $a=\sqrt{2}, b=\infty$
  • Correct answer
    C
    $a=\log (1+\sqrt{2}), b=\infty$
  • D
    $a=0, b=\log (1+\sqrt{2})$

Solution

$\begin{aligned} & \text { Range of } \operatorname{sech}^{-1} x=\operatorname{cosech}^{-1} x \text { is }[a, b] \\ & \because \operatorname{sech}^{-1} x=\cosh ^{-1}\left(\frac{1}{x}\right)=\ln \left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right) \\ & \therefore \text { Domain } x \in(0,1] \\ & \operatorname{cosech}^{-1}(x)=\sinh ^{-1}\left(\frac{1}{x}\right)=\ln \left(\frac{1}{x}+\sqrt{\frac{1}{x^2}+1}\right) \\ & \therefore \text { Domain } x \in R-\{0\} \\ & \therefore \text { Domain of } \operatorname{sech}^{-1}(x)+\operatorname{cosech}^{-1}(x) \text { is Eqs. (i) and (ii) } \\ & \Rightarrow \quad x \in(0,1] \\ & \therefore \quad f(0)=\operatorname{sech}^{-1}(0)+\operatorname{cosech}^{-1}(0)=\infty \\ & \quad f(1)=\operatorname{sech}^{-1}(1)+\operatorname{cosech}^{-1}(1)=0+\ln (1+\sqrt{2}) \\ & \therefore \text { Range }=[(\ln (1+\sqrt{2}), \infty) \\ & \text { Hence, } a=\ln (1+\sqrt{2}), b=\infty\end{aligned}$

Hence, the correct answer is option: 'C'.

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