Question

Let f(x) be a continuous function and $f(2x^2-1)=2xf\left(x\right)\forall x$ then ${\int }_{-1}^1\left|f\right(x\left)\right|dx$ id equal to ............

• A. 0
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is A 0

Put $x=\cos \theta \Rightarrow f(2{\cos }^2\theta -1)=2\cos \theta f\left(\cos \theta \right)$
or $g\left(2\theta \right)=\frac{f\left(\cos 2\theta \right)}{\sin 2\theta }=\frac{f\left(\cos \theta \right)}{\sin \theta }=g\left(\theta \right)$ (say) ..........(1)
also $g(\theta +\pi )=g\left(\theta \right)$ (as f is odd) ...........(2)
From equation (1) and (2) we get
$g\left(1\right)=g\left(2^k\right)=g(2^k+n\pi )=g(1+\frac{n}{2^k}\pi )$
We take $\underset{x\to a}{lim}g\left(x\right)=L=g\left(a\right)$ since g is continuous wherever defined
If $L\ne g\left(1\right)$ then we may adjust n and sufficiently large k to obtain points indefinitely close to a such than $g\left(x\right)=g\left(1\right)$
Hence $g\left(x\right)$ is a constant function
$\Rightarrow f\left(x\right)$ is a constant function $x\in (-1,1)$ also $f(-1)=0$ hence $f\left(x\right)=0$