Question

The value of $$x$$at which the first derivative of the function${\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)}^2$w.r.t. $$x$$ is $${\displaystyle \frac{3}{4}}$$are

• A. $$\pm 2$$
• B. $$\pm {\displaystyle \frac{1}{2}}$$
• C. $$\pm {\displaystyle \frac{2}{\sqrt{3}}}$$
• D. $$\pm {\displaystyle \frac{\sqrt{3}}{2}}$$

Answer 1

The correct option is A

$$\pm 2$$

Explanation for the correct option:

Find the required value.

Given function: ${\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)}^2$
This function can be written as $$x+{\displaystyle \frac{1}{x}}+2$$
Now differentiating it with respect to $$x$$we get ,
$$\begin{gathered} \frac{d}{dx}{\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)}^2=\frac{d}{dx}\left(x+\frac{1}{x}+2\right) \\ =1-\frac{1}{x^2} \end{gathered}$$
We have given that $\frac{d}{dx}{\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)}^2=\frac{3}{4}$
Therefore we get ,
$$\begin{gathered} 1-\frac{1}{x^2}=\frac{3}{4} \\ \Rightarrow 1-\frac{3}{4}=\frac{1}{x^2} \\ \Rightarrow \frac{1}{4}=\frac{1}{x^2} \\ \Rightarrow x^2=4 \\ \Rightarrow x=\pm 2 \end{gathered}$$
Hence, option (A) is the correct answer.