Question

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

Answer 1

Answer: 1

Let $g(x)=\underset{0}{\overset{x}{\int }}\frac{t^2}{1+t^4}dt-(2x)+1$
Let $I(x)=\underset{0}{\overset{x}{\int }}\frac{t^2}{1+t^4}dt$
$=\frac{1}{2}\underset{0}{\overset{x}{\int }}\frac{2t^2}{1+t^4}dt$
$=\frac{1}{2}\underset{0}{\overset{x}{\int }}\left(\frac{t^2+1}{1+t^4}+\frac{t^2-1}{1+t^4}\right)dt$
$=\frac{1}{2}\underset{0}{\overset{x}{\int }}\frac{1+\frac{1}{t^2}}{{\left(t-\frac{1}{t}\right)}^2+2}dt+\frac{1}{2}\underset{0}{\overset{x}{\int }}\frac{\left(1-\frac{1}{t^2}\right)dt}{{\left(t+\frac{1}{t}\right)}^2-2}$
$I(x)=\frac{1}{2}\frac{1}{\sqrt{2}}{\left[{\tan }^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)\right]}_0^x+\frac{1}{2}\times \frac{1}{2\sqrt{2}}{\left[\ell n\left|\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right|\right]}_0^x$
Now, $g^{\prime }(x)=\frac{x^2}{1+x^4}-2$
$g^{\prime }(x)=\frac{1}{\frac{1}{x^2}+x^2}-2$
since $\frac{1}{x^2}+x^2\ge 2$
$\therefore g^{\prime }(x)<0$
So $\frac{1}{x^2+\frac{1}{x^2}}\le \frac{1}{2}$
Now, $g(0)=1$
$\&g(1)=\underset{0}{\overset{1}{\int }}\frac{t^2}{1+t^4}dt-2+1$
$=\underset{0}{\overset{1}{\int }}\frac{t^2}{1+t^4}dt-1$
$=\left[0+\frac{1}{4\sqrt{2}}\ell n\left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)\right]-\left[\frac{1}{2\sqrt{2}}\left(-\frac{\pi }{2}\right)+\frac{1}{4\sqrt{2}}\ell n\left(\frac{1}{1}\right)\right]-1$
$=\frac{1}{4\sqrt{2}}\ell n\left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)+\frac{\pi }{4\sqrt{2}}-1$
$g(1)=-ve$
$\therefore g(0)>0=\&g(1)<0$
$\&g^{\prime }(x)<0$
$\therefore g(x)$ has one root in $[0,1]$