Question

If $m=\tan \left(x\right)$is the slope of the tangent to the curve $e^y=1+x^2$ then

• A. $\left|\tan \left(x\right)\right|>1$
• B. $\left|\tan \left(x\right)\right|<1$
• C. $-1\le \tan \left(x\right)\le 1$
• D. $-1\le \left|\tan \left(x\right)\right|\le 1$

Answer 1

The correct option is C

$-1\le \tan \left(x\right)\le 1$

Explanation for correct option

We know that slope of a tangent is $\frac{dy}{dx}=m$

Given $m=\tan \left(x\right)$

Therefore, $\frac{dy}{dx}=m=\tan \left(x\right)$

Now, given equation of the curve is $e^y=1+x^2$

Now differentiating with respect to $x$ we have

$$\begin{gathered} \Rightarrow e^y\frac{dy}{dx}=2x \\ \Rightarrow \frac{dy}{dx}=\frac{2x}{e^y} \\ \Rightarrow \frac{dy}{dx}=\frac{2x}{1+x^2} \end{gathered}$$

$$\begin{gathered} \therefore {\left(x-1\right)}^2\ge 0\forall x\in R \\ \Rightarrow x^2-2x+1\ge 0 \\ \Rightarrow x^2+1\ge 2x \\ \Rightarrow \frac{1}{x^2+1}\le \frac{1}{2x} \\ \Rightarrow \frac{2x}{x^2+1}\le 1 \end{gathered}$$

$$\begin{gathered} \therefore \left|\frac{dy}{dx}\right|\le 1 \\ \Rightarrow \left|\tan \left(x\right)\right|\le 1 \end{gathered}$$

Hence, option(C) i.e. $-1\le \tan \left(x\right)\le 1$ is correct