Question

If $f\left(x\right)=\int \frac{dx}{{\left(1+{\displaystyle \frac{1}{x^2}}\right)}^{3/2}}$ and $f\left(0\right)=0$, then the value of$f\left(1\right)$ is

• A. $\frac{1}{\sqrt{2}}$
• B. $-\frac{1}{\sqrt{2}}$
• C. $\sqrt{2}$
• D. $-\sqrt{2}$

Answer 1

The correct option is A

$\frac{1}{\sqrt{2}}$

Explanation for the correct option:

Step 1: Simplifying the given function:

Given,

$f\left(x\right)=\int \frac{dx}{{\left(1+{\displaystyle \frac{1}{x^2}}\right)}^{3/2}}$

Let us consider ,

$\begin{array}{rcl}1+\frac{1}{x^2}& =& z^2...\left(1\right)\\ & \Rightarrow & 1+x^{-2}=z^2\\ & \Rightarrow & \frac{-2dx}{x^3}=2zdz\left[\mathrm{by}\mathrm{differentiation}\right]\end{array}$

Now we can write the upper equation as,

$\begin{array}{rcl}f\left(x\right)& =& -\int \frac{zdz}{{\left(z^2\right)}^{3/2}}\\ f\left(x\right)& =& -\int \frac{zdz}{z^3}\\ & =& -\int z^{-2}dz\\ & =& -\left[\frac{1}{z}\right]+c\\ & =& -\frac{1}{\sqrt{1+{\displaystyle \frac{1}{x^2}}}}+c\left[\mathrm{from}1\right]\end{array}$

Step 2: Find the value of $f\left(1\right)$:

Substitute $x=0$ in the above function,

$\begin{array}{rcl}f\left(0\right)& =& 0\\ & \Rightarrow & 0+c=0\\ & \Rightarrow & c=0\end{array}$

$f\left(x\right)=\frac{x}{\sqrt{1+x^2}}$

Now, substitute $x=1$ in the above function,

$\begin{array}{rcl}f\left(1\right)& =& \frac{1}{\sqrt{1+1^2}}\\ & =& \frac{1}{\sqrt{2}}\end{array}$

Hence, the correct option is A.