Question

If $y=\sqrt{\frac{1-x}{1+x}}$, then $\left(1-x^2\right)\frac{dy}{dx}+y$ is equal to

• A. $1$
• B. $-1$
• C. $2$
• D. $0$

Answer 1

The correct option is D

$0$

Explanation for the correct option.

Step 1. Simplify the given equation.

In the right hand side of the given equation $y=\sqrt{\frac{1-x}{1+x}}$ multiply and divide $\sqrt{1-x}$ and simplify.

$$\begin{gathered} y=\sqrt{\frac{1-x}{1+x}}\times \frac{\sqrt{1-x}}{\sqrt{1-x}} \\ \Rightarrow y=\frac{\sqrt{\left(1-x\right)\left(1-x\right)}}{\sqrt{\left(1+x\right)\left(1-x\right)}} \\ \Rightarrow y=\frac{\sqrt{{\left(1-x\right)}^2}}{\sqrt{\left(1-x^2\right)}}\left[\left(a-b\right)\left(a+b\right)=a^2-b^2\right] \\ \Rightarrow y=\frac{1-x}{\sqrt{1-x^2}}...\left(1\right) \end{gathered}$$

Step 2. Find the value of $\frac{dy}{dx}$ and form an equation.

Now differentiate $y=\frac{1-x}{\sqrt{1-x^2}}$ with respect to $x$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}\left(\frac{1-x}{\sqrt{1-x^2}}\right)\\ & \Rightarrow & \frac{dy}{dx}=\frac{\left(-1\right)\sqrt{1-x^2}-\left(1-x\right)\times {\displaystyle \frac{1}{2\sqrt{1-x^2}}}\times \left(-2x\right)}{{\left(\sqrt{1-x^2}\right)}^2}\\ & \Rightarrow & \left(1-x^2\right)\frac{dy}{dx}=-\sqrt{1-x^2}+\frac{x\left(1-x\right)}{\sqrt{1-x^2}}\\ & \Rightarrow & \left(1-x^2\right)\frac{dy}{dx}=\frac{-{\left(\sqrt{1-x^2}\right)}^2+x\left(1-x\right)}{\sqrt{1-x^2}}\\ & \Rightarrow & \left(1-x^2\right)\frac{dy}{dx}=\frac{-1+x^2+x-x^2}{\sqrt{1-x^2}}\\ & \Rightarrow & \left(1-x^2\right)\frac{dy}{dx}=\frac{-\left(1-x\right)}{\sqrt{1-x^2}}\\ & \Rightarrow & \left(1-x^2\right)\frac{dy}{dx}=-y\left[\mathit{U}\mathit{s}\mathit{i}\mathit{n}\mathit{g}\mathbf{}\mathit{e}\mathit{q}\mathit{u}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{}\mathbf{1}\right]\\ & \Rightarrow & \left(1-x^2\right)\frac{dy}{dx}+y=0\end{array}$

So the value of the expression $\left(1-x^2\right)\frac{dy}{dx}+y$ is $0$.

Hence, the correct option is D.