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TS EAMCET 5-Aug-2021 Shift 1 Question Paper

Section: Mathematics

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Question : 26 of 160

Marks: +1, -0

Let

k > 0

t

\operatorname{Sech}^{-1}\left(\frac{1}{2}\right)-\operatorname{Cosech}^{-1}\left(\frac{3}{k}\right)

3e^{t}

2+\sqrt{3}

k

=

$2$
$4$
$3\sqrt{3}$
$3\sqrt{2}$

Solution: 👈: Video Solution

t=\operatorname{sech}^{-1}\left(\frac{1}{2}\right)-\operatorname{cosech}^{-1}\left(\frac{3}{k}\right)

\operatorname{sech}^{-1}(x)=\ln\left(\frac{1+\sqrt{1-x^{2}}}{x}\right)

\operatorname{sech}^{-1}\left(\frac{1}{2}\right)=\ln\left(\frac{1+\frac{\sqrt{3}}{2}}{\frac{1}{2}}\right)=\ln(\sqrt{3}+2)

\operatorname{cosech}^{-1}(x)=\ln\left(\frac{1+\sqrt{1+x^{2}}}{x}\right)

\operatorname{cosech}^{-1}\left(\frac{3}{k}\right)=\operatorname{In}\left(\frac{k+\sqrt{k^{2}+9}}{3}\right)

\ln(\sqrt{3}+2)-\operatorname{In}\left(\frac{k+\sqrt{k^{2}+9}}{3}\right)=t

\ln\left[\frac{(\sqrt{3}+2)3}{k+\sqrt{k^{2}+9}}\right]=t

3\sqrt{3}+6=(k+\sqrt{k^{2}+9})e^{t}

3\sqrt{3}+6=(k+\sqrt{k^{2}+9})\left(\frac{2+\sqrt{3}}{3}\right)

If

k=4

, then

\Rightarrow 3\sqrt{3}+6=(4+5)\left(\frac{2+\sqrt{3}}{3}\right)=3(\sqrt{3}+2)

k=4

satifies the equation.

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