Question

If $f\left(x\right)=\underset{n\to \infty }{lim}(2x+4x^3+......+2^n{x}^{2n-1})(0<x<\frac{1}{\sqrt{2}})$, then the value of $\int f\left(x\right)dx$ is equal to

$\text{Note}$: $c$ is the constant of integration.

• A. $\log \left(\frac{1}{\sqrt{1-x^2}}\right)+c$
• B. $\log \sqrt{1-2x^2+x}+c$
• C. $\log \left(\frac{1}{\sqrt{1-2x^2}}\right)+c$
• D. None of these

Answer 1

The correct option is C $\log \left(\frac{1}{\sqrt{1-2x^2}}\right)+c$

When $n\to \infty$, $f\left(x\right)$ becomes an infinites series of G.P. with $a=2x$, $r=2x^2$

Hence, $f\left(x\right)=\frac{2x}{1-2x^2}$

Since, $0<x<\frac{1}{\sqrt{2}}$
So, the integral $I=\int f\left(x\right)dx$ becomes $\int \frac{2x}{1-2x^2}dx$

Substitute, $1-2x^2=t\Rightarrow -4xdx=dt$
After substitution, the integral becomes

$-\frac{1}{2}\int \frac{dt}{t}=-\frac{1}{2}\log \left(t\right)+c$
$=-\frac{1}{2}\log (1-2x^2)+c$
$I=\log \left(\frac{1}{\sqrt{1-2x^2}}\right)+c$

Hence, option C.