Q65 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Mathematics
If $\operatorname{sech}^{-1}\left(\frac{1}{2}\right)-\operatorname{cosec~h}^{-1}\left(\frac{3}{4}\right)=\log _e k$, then
- A$3 k^2-12 k-1=0$
- B$3 k^2-12 k+1=0$
- Correct answerC$9 k^2-12 k+1=0$
- D$9 k^2-12 k-1=0$
Solution
Given, $\operatorname{sech}^{-1}\left(\frac{1}{2}\right)-\operatorname{cosech}^{-1}\left(\frac{3}{4}\right)=\log _e k$
$\Rightarrow \log _e\left(\sqrt{2^2-1}+2\right)-\log _e\left(\sqrt{\left(\frac{4}{3}\right)^2}+1+\frac{4}{3}\right)=\log _e k$
$\Rightarrow \log _e(\sqrt{3}+2)-\log \left(\frac{5}{3}+\frac{4}{3}\right)=\log _e k$
$\Rightarrow \quad \log _e\left(\frac{\sqrt{3}+2}{3}\right)=\log _e k \Rightarrow 3 k-2=\sqrt{3}$
$\Rightarrow 9 k^2-12 k+4=3 \Rightarrow 9 k^2-12 k+1=0$
Hence, the correct answer is option: 'C'.