Question

Let $m$ be the slope of tangent to the curve $e^{2y}=1+x^2$ then set of all values of $m$ is :

• A. $[-\frac{1}{2},\frac{1}{2}]$
• B. $[-\infty ,-\frac{1}{2}]\cup [\frac{1}{2},0]$
• C. $[-\frac{1}{2},\frac{1}{2}]-\left\{0\right\}$
• D. $[-2,2]-\left\{0\right\}$

Answer 1

Answer: (A)

$[-\frac{1}{2},\frac{1}{2}]$

$m$ be the slope of tangent.

Given equation of curve is

$e^{2y}=1+x^2$

Taking log both side and we get,

$\log e^{2y}=\log (1+x^2)$

$2y=\log (1+x^2)$

On differentiating and we get,

$2\frac{dy}{dx}=\frac{1}{1+x^2}\frac{d}{dx}(1+x^2)$

$\Rightarrow 2\frac{dy}{dx}=\frac{1}{1+x^2}\frac{d}{dx}(1+x^2)$

$\Rightarrow 2\frac{dy}{dx}=\frac{1}{1+x^2}\frac{d}{dx}2x$

$\Rightarrow \frac{dy}{dx}=\frac{x}{1+x^2}$

$m=\frac{dy}{dx}=\frac{x}{1+x^2}$

On put $x=(-1,1)$

So,

$m=\frac{1}{1+1}=\frac{1}{2}$

$m=\frac{-1}{1+1}=\frac{-1}{2}$

Hence, the value of $m$ is $[-\frac{1}{2},\frac{1}{2}]$

Hence, this is the answer.