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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

If y(x)= ^ -1 ( 1+a^2 x^2 -1 a x ) and (1+a^2 x^2) y^ +g(x) y^ =0...

If y(x)=tan⁡−1(1+a2x2−1ax)and(1+a2x2)y′′+g(x)y′=0y(x)=\tan ^{-1} \left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right) and \left(1+a^2 x^2\right) y^{\prime \prime}+g(x) y^{\prime}=0 then, the sum of the roots of the equation 1+a2x2+g(x)=01+a^2 x^2+g(x)=0 is

Options

  1. A2a2 a
  2. B−2a2-2 a^2
  3. C2
  4. D-2

Correct answer

D. -2

Step-by-step solution

Given, y=tan⁡−1(1+a2x2−1ax)y=\tan ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right) put ax=tan⁡θ⇒θ=tan⁡−1(ax)y=tan⁡−1(sec⁡θ−1tan⁡θ)⇒y=tan⁡−1(1−cos⁡θsin⁡θ)=tan⁡−1tan⁡θ2=θ2⇒y=12tan⁡−1(ax)⇒y′=12a1+a2x2⇒(1+a2x2)y′=a2⇒(1+a2x2)y′′+2a2xy′=0∴g(x)=2a2x⇒1+a2x2+g(x)=0a2x2+2a2x+1=0 Sum of roots =−2a2a2=−2\begin{aligned} & \text { put } a x=\tan \theta \Rightarrow \theta=\tan ^{-1}(a x) \\ & y=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right) \\ & \Rightarrow \quad y=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)=\tan ^{-1} \tan \frac{\theta}{2}=\frac{\theta}{2} \\ & \Rightarrow \quad y=\frac{1}{2} \tan ^{-1}(a x) \Rightarrow y^{\prime}=\frac{1}{2} \frac{a}{1+a^2 x^2} \\ & \Rightarrow \quad\left(1+a^2 x^2\right) y^{\prime}=\frac{a}{2} \Rightarrow\left(1+a^2 x^2\right) y^{\prime \prime}+2 a^2 x y^{\prime}=0 \\ & \therefore \quad g(x)=2 a^2 x \Rightarrow 1+a^2 x^2+g(x)=0 \\ & a^2 x^2+2 a^2 x+1=0 \\ & \text { Sum of roots }=\frac{-2 a^2}{a^2}=-2\end{aligned}$$
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