Question

The abscissa of the point on the curve $y=a\left(e^{\frac{x}{a}}+e^{–\frac{x}{a}}\right)$ where the tangent is parallel to the $x$ - axis is

• A. $0$
• B. $a$
• C. $2a$
• D. $-2a$

Answer 1

The correct option is A

$0$

Explanation For The Correct Option:

Finding the abscissa of the point on the curve where the tangent is parallel to the x-axis

The given curve,

$y=a\left(e^{\frac{x}{a}}+e^{–\frac{x}{a}}\right)...\left(1\right)$

We know that the coordinates of any point of the curve be in the form of $\left(x,y\right)$

And in the point $\left(x,y\right)$, $x$ is called abscissa.

Since given that the tangent is parallel to the x-axis

Therefore slope will be zero at that point.

i.e. $\frac{dy}{dx}=0$

Now differentiating $\left(i\right)$ with respect to $x$ and equation to zero

$$\begin{gathered} \Rightarrow a\left[e^{\frac{x}{a}}\left(\frac{1}{a}\right)+e^{-\frac{x}{a}}\left(-\frac{1}{a}\right)\right]=0 \\ \Rightarrow e^{\frac{x}{a}}+e^{-\frac{x}{a}}=0 \\ \Rightarrow e^{\frac{x}{a}}=e^{-\frac{x}{a}} \\ \Rightarrow e^{\frac{x}{a}}=\frac{1}{e^{\frac{x}{a}}}\left[\because a^{-1}=\frac{1}{a}\right] \\ \Rightarrow e^{2\left(\frac{x}{a}\right)}=1 \\ \Rightarrow e^{2\left(x/a\right)}=e^0[\because \text{Anything to the power zero is}1] \\ \Rightarrow 2\left(\frac{x}{a}\right)=0\left[\text{equating exponent}\right] \\ \Rightarrow x=0 \end{gathered}$$

Hence, the correct answer is option (A)