Question

JEE MAINS MATHS QUESTION SEPTEMBER 2020

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$\displaystyle \: y_1 = {tan}^{ - 1} \bigg( \: \dfrac{ \sqrt{1 + {x}^{2} } - 1}{x} \bigg)$

And

$\displaystyle \: y_2 = {tan}^{ - 1} \bigg( \: \dfrac{2x \sqrt{1 - {x}^{2} } }{1 - 2 {x}^{2} } \bigg)$

TO DETERMINE

$\displaystyle \: \dfrac{dy_1}{dy_2}$

EVALUATION

$\displaystyle \: y_1 = {tan}^{ - 1} \bigg( \: \dfrac{ \sqrt{1 + {x}^{2} } - 1}{x} \bigg)$

Let

$x = \tan \theta$

Then

$\displaystyle \: y_1 = {tan}^{ - 1} \bigg( \: \dfrac{ \sqrt{1 + { \tan }^{2} \theta} - 1}{ \tan \theta } \bigg)$

$\implies \: \displaystyle \: y_1 = {tan}^{ - 1} \bigg( \: \dfrac{ { \sec } \theta - 1}{ \tan \theta } \bigg)$

$\implies \: \displaystyle \: y_1 = {tan}^{ - 1} \bigg( \: \dfrac{ { 1 - \cos } \theta }{ \sin \theta } \bigg)$

$\implies \: \displaystyle \: y_1 = {tan}^{ - 1} \bigg( \: \tan \frac{ \theta}{2} \bigg)$

$\implies \: \displaystyle \: y_1 = \frac{ \theta}{2}$

$\implies \: \displaystyle \: y_1 = \frac{1}{2} \: {tan}^{ - 1}x$

So

$\displaystyle \: \dfrac{dy_1}{dx} = \frac{1}{2} \times \frac{1}{1 + {x}^{2} }$

Now

$\displaystyle \: y_2 = {tan}^{ - 1} \bigg( \: \dfrac{2x \sqrt{1 - {x}^{2} } }{1 - 2 {x}^{2} } \bigg)$

Let

$x = \sin \alpha$

So

$\displaystyle \: y_2 = {tan}^{ - 1} \bigg( \: \dfrac{2 \sin \alpha \sqrt{1 - {\sin }^{2} \alpha } }{1 - 2 { \sin}^{2}\alpha } \bigg)$

$\implies \: \displaystyle \: y_2 = {tan}^{ - 1} \bigg( \: \dfrac{2 \sin \alpha \cos \alpha }{ \cos2\alpha } \bigg)$

$\implies \: \displaystyle \: y_2 = {tan}^{ - 1} \bigg( \: \dfrac{ \sin2 \alpha }{ \cos2\alpha } \bigg)$

$\implies \: \displaystyle \: y_2 = {tan}^{ - 1} \bigg( { \tan2 \alpha } \bigg)$

$\implies \: \displaystyle \: y_2 = 2 \alpha$

$\implies \: \displaystyle \: y_2 = 2 { \sin}^{ - 1} x$

So

$\displaystyle \: \dfrac{dy_2}{dx} = \frac{2}{ \sqrt{1 - {x}^{2} } }$

Hence

$\displaystyle \: \dfrac{dy_1}{dy_2} = \frac{1}{4} \frac{ \sqrt{ 1 - {x}^{2} } }{1 + {x}^{2} }$