Q65 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Mathematics
Assertion $(\mathrm{A}) \operatorname{cosech}^{-1}(\mathrm{~B})=\log \left(\frac{1+\sqrt{10}}{3}\right)$
Reason (R) $e^{\operatorname{cosech}^{-1} x}$ is a root of the quadratic equation $x p^2-2 p-x=0$
The correct option among the following is
- Correct answerA(A) is true, (R) is true and (R) is the correct explanation for (A)
- B(A) is true, (R) is true but (R) is not the correct explanation for (A)
- C(A) is true but (R) is false
- D(A) is false but (R) is true
Solution
Reason (R) $e^{\operatorname{cosech}^{-1} x}$ is a root of the quadratic equation $x p^2-2 p-x=0$
We have, $x p^2-2 p-x=0$
$\begin{aligned}
\Rightarrow \quad p & =\frac{2 \pm \sqrt{4+4 x^2}}{2 x}=\frac{1 \pm \sqrt{1+x^2}}{x} \\
& =\frac{1+\sqrt{1+x^2}}{x}, \frac{1-\sqrt{1+x^2}}{x} \\
& =e^{\ln \left(\frac{1+\sqrt{1+x^2}}{x}\right)}, e^{\ln \left(\frac{1-\sqrt{1+x^2}}{x}\right)}
\end{aligned}$
Since, $\operatorname{cosech}^{-1} x=\ln \left(\frac{1+\sqrt{1+x^2}}{x}\right)$
$\therefore e^{\operatorname{cosech}^{-1} x}$ is a solution of given equation.
Now, $\operatorname{cosech}{ }^{-1}(3)=\ln \left(\frac{1+\sqrt{1+3^2}}{3}\right)=\ln \left(\frac{1+\sqrt{10}}{3}\right)$
Hence, the correct answer is option: 'A'.