Question

The total number of distinct $x\in [0,1]$ for which ${\int }_0^x\frac{t^2}{1+t^4}dt=2x-1$ is

Answer 1

Let $f\left(x\right)={\int }_0^x\frac{t^2}{1+t^4}dt$

Differentiating both sides, we get

$f^{\prime }\left(x\right)=\frac{x^2}{1+x^4}$

Observing $f^{\prime }\left(x\right)$, we can say that,

$f^{\prime }\left(x\right)>0$

$\Rightarrow f\left(x\right)$ is an increasing function.

Now at $x=0,f\left(0\right)=0$ and at $x=1,f\left(1\right)={\int }_0^1\frac{t^2}{1+t^4}dt$

Because, $0<\frac{x^2}{1+x^4}<1$

$\Rightarrow 0<{\int }_0^1\frac{t^2}{1+t^4}dt<{\int }_0^{1}1dt$

$\Rightarrow 0<{\int }_0^1\frac{t^2}{1+t^2}dt<1$

Plotting graph of $y=f\left(x\right)$ and $y=2x-1$, we find that they intersect at only one point.

Because $f\left(x\right)$ is increasing and less than $1$ for $x\in [0,1]$.

Therefore only one solution.