Q65 / 160 · AP EAMCET 2018 (24 Apr Shift 1) · Mathematics

If $x=-\frac{1}{2}, \sinh ^{-1} x+\operatorname{cosech}^{-1} x=$

  • Correct answer
    A
    $\log _e\left(\frac{7-3 \sqrt{5}}{2}\right)$
  • B
    $\log _6\left(\frac{3+\sqrt{5}}{2}\right)$
  • C
    $\log _e\left[\frac{(\sqrt{5}-1)(2+\sqrt{3})}{2}\right]$
  • D
    $\log _e\left[\frac{(\sqrt{5}+1)(2+\sqrt{3})}{2}\right]$

Solution

$\begin{aligned} & \text { Given, } x=-\frac{1}{2} \\ & \because \\ & \text { and } \quad \sinh ^{-1} x=\ln \left(x+\sqrt{x^2+1}\right) \\ & \therefore \sinh ^{-1} x+\operatorname{cosech}^{-1} x=\ln \left(\frac{1}{x}+\sqrt{\frac{1}{x^2}+1}\right) \\ & =\ln \left(x+\sqrt{x^2+1}\right) \\ & =\ln \left(-\frac{1}{2}+\sqrt{\frac{1}{4}+1}\right)+\ln \left(-2+\sqrt{\frac{1}{x^2}+1}\right) \\ & =\ln \left(\frac{\sqrt{4}+1)}{2}\right)+\ln (-2+\sqrt{5}) \\ & =\ln \left(\frac{(\sqrt{5}-1)}{2} \times(\sqrt{5}-2)\right] \ln (\sqrt{5}-2) \\ & =\ln \left(\frac{5-2 \sqrt{5}-\sqrt{5}+2}{2}\right)=\ln \left(\frac{7-3 \sqrt{5}}{2}\right)\end{aligned}$

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

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