Question

If $m$ is the slope of a tangent to the curve $e^y=1+x^2$, then

• A. $\left|m\right|<1$
• B. $\left|m\right|\le 1$
• C. $\left|m\right|>1$
• D. $m<1$

Answer 1

The correct option is B

$\left|m\right|\le 1$

The explanation for the correct option:

The given equation of the curve $e^y=1+x^2$.

Differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \therefore \frac{d}{dx}\left(e^y\right)=\frac{d}{dx}\left(1+x^2\right) \\ \Rightarrow e^y\frac{dy}{dx}=2x \\ \Rightarrow \frac{dy}{dx}=\frac{2x}{e^y} \\ \Rightarrow \frac{dy}{dx}=\frac{2x}{1+x^2}\left[\because e^y=1+x^2\right] \end{gathered}$$

Thus, the slope of the curve can be given by, $m=\frac{dy}{dx}=\frac{2x}{1+x^2}$.

Now

$$\begin{gathered} x+\frac{1}{x}\ge 2(x>0)andx+\frac{1}{x}\le -2(x<0) \\ \Rightarrow \frac{1}{x+{\displaystyle \frac{1}{x}}}\le \frac{1}{2}and\frac{1}{x+{\displaystyle \frac{1}{x}}}\ge \frac{-1}{2} \\ \Rightarrow \frac{2x}{1+x^2}\le 1and\frac{2x}{1+x^2}\ge -1 \\ \Rightarrow m\le 1andm\ge -1 \\ \Rightarrow \left|m\right|\le 1 \end{gathered}$$

Hence, (B) is the correct option.