JEE Advance - Mathematics (2016 - Paper 1 Offline - No. 4)
The total number of distinct $$x \in \left[ {0,1} \right]$$ for which
$$\int\limits_0^x {{{{t^2}} \over {1 + {t^4}}}} dt = 2x - 1$$
$$\int\limits_0^x {{{{t^2}} \over {1 + {t^4}}}} dt = 2x - 1$$
Answer
1
Explanation
Let $f(x)=\int_0^x \frac{t^2}{1+t^4} d t-(2 x-1)$
Differentiating both sides w.r.t. $x$
$$ \begin{aligned} & \therefore f^{\prime}(x)=\frac{x^2}{1+x^4}-2 \\\\ & \therefore f^{\prime}(x)<0(\because x \in[0,1]) \end{aligned} $$ $\Rightarrow f$ is strictly decreasing function
$$ \begin{aligned} \text { Let } \mathrm{I} & =\frac{1}{2} \int_0^x \frac{2 t^2}{t^4+1} d t=\frac{1}{2} \int_0^x \frac{\left(t^2+1\right)+\left(t^2-1\right)}{t^4+1} \\\\ & =\frac{1}{2} \int_0^x \frac{t^2+1}{t^4+1} d t+\frac{1}{2} \int_0^x \frac{t^2-1}{t^4+1} d t \end{aligned} $$
$\begin{aligned} & =\frac{1}{2} \int_0^x \frac{1+\frac{1}{t^2}}{\left(t-\frac{1}{t}\right)^2+2} d t+\frac{1}{2} \int_0^x \frac{1-\frac{1}{t^2}}{\left(t+\frac{1}{t}\right)^2-2} d t \\\\ & =\frac{1}{2} \frac{1}{\sqrt{2}}\left[\tan ^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)\right]_0^x+\frac{1}{4 \sqrt{2}}\left[\ln \left(\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right)\right]_0^x \\\\ & =\frac{1}{2 \sqrt{2}}\left[\tan ^{-1}\left(\frac{t^2-1}{t \sqrt{2}}\right)\right]_0^x+\frac{1}{4 \sqrt{2}}\left[\ln \left(\frac{t^2-\sqrt{2} t+1}{t^2+\sqrt{2} t+1}\right)\right]_0^x \\\\ & =\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{x \sqrt{2}}\right)-\tan ^{-1}(-\infty) \\\\ & +\frac{1}{4 \sqrt{2}}\left[\ln \left(\frac{x^2-\sqrt{2} x+1}{x^2+\sqrt{2} x+1}\right)\right]-0 \\\\ & =\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{x \sqrt{2}}\right)+\frac{1}{4 \sqrt{2}}\ln\left(\frac{x^2-\sqrt{2} x+1}{x^2+\sqrt{2} x+1}\right)+\frac{1}{2 \sqrt{2}}\left(\frac{\pi}{2}\right)\end{aligned}$
$$ \begin{aligned} & \because f(x)=I-(2 x-1) \\\\ & \therefore f(0)=\frac{1}{2 \sqrt{2}} \tan ^{-1}(-\infty)+\frac{1}{4 \sqrt{2}} \ln (1)+\frac{1}{2 \sqrt{2}}\left(\frac{\pi}{2}\right)+1 \\\\ & \Rightarrow f(0)=\frac{-\pi}{4 \sqrt{2}}+0+\frac{\pi}{4 \sqrt{2}}+1 \\\\ & \Rightarrow f(0)=1 \\\\ & f(1)=\frac{1}{2 \sqrt{2}} \tan ^{-1}(-0)+\frac{1}{4 \sqrt{2}} \ln \left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)+\frac{\pi}{4 \sqrt{2}}-1 \\\\ & \Rightarrow f(1)=\frac{1}{4 \sqrt{2}}\left[\ln \left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right) \pi\right]-1 \end{aligned} $$
Here, $f(1)<0$ and $f(0)=1>0$
$$ \Rightarrow f(0) \cdot f(1)<0 $$
By intermediate value theorem
$\Rightarrow$ There exists at least one $c \in(0,1)$ such that $f(c)=0$
Since, $f(x)$ is a decreasing function such that $f(x)=0$ has exactly one root.
Differentiating both sides w.r.t. $x$
$$ \begin{aligned} & \therefore f^{\prime}(x)=\frac{x^2}{1+x^4}-2 \\\\ & \therefore f^{\prime}(x)<0(\because x \in[0,1]) \end{aligned} $$ $\Rightarrow f$ is strictly decreasing function
$$ \begin{aligned} \text { Let } \mathrm{I} & =\frac{1}{2} \int_0^x \frac{2 t^2}{t^4+1} d t=\frac{1}{2} \int_0^x \frac{\left(t^2+1\right)+\left(t^2-1\right)}{t^4+1} \\\\ & =\frac{1}{2} \int_0^x \frac{t^2+1}{t^4+1} d t+\frac{1}{2} \int_0^x \frac{t^2-1}{t^4+1} d t \end{aligned} $$
$\begin{aligned} & =\frac{1}{2} \int_0^x \frac{1+\frac{1}{t^2}}{\left(t-\frac{1}{t}\right)^2+2} d t+\frac{1}{2} \int_0^x \frac{1-\frac{1}{t^2}}{\left(t+\frac{1}{t}\right)^2-2} d t \\\\ & =\frac{1}{2} \frac{1}{\sqrt{2}}\left[\tan ^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)\right]_0^x+\frac{1}{4 \sqrt{2}}\left[\ln \left(\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right)\right]_0^x \\\\ & =\frac{1}{2 \sqrt{2}}\left[\tan ^{-1}\left(\frac{t^2-1}{t \sqrt{2}}\right)\right]_0^x+\frac{1}{4 \sqrt{2}}\left[\ln \left(\frac{t^2-\sqrt{2} t+1}{t^2+\sqrt{2} t+1}\right)\right]_0^x \\\\ & =\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{x \sqrt{2}}\right)-\tan ^{-1}(-\infty) \\\\ & +\frac{1}{4 \sqrt{2}}\left[\ln \left(\frac{x^2-\sqrt{2} x+1}{x^2+\sqrt{2} x+1}\right)\right]-0 \\\\ & =\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{x \sqrt{2}}\right)+\frac{1}{4 \sqrt{2}}\ln\left(\frac{x^2-\sqrt{2} x+1}{x^2+\sqrt{2} x+1}\right)+\frac{1}{2 \sqrt{2}}\left(\frac{\pi}{2}\right)\end{aligned}$
$$ \begin{aligned} & \because f(x)=I-(2 x-1) \\\\ & \therefore f(0)=\frac{1}{2 \sqrt{2}} \tan ^{-1}(-\infty)+\frac{1}{4 \sqrt{2}} \ln (1)+\frac{1}{2 \sqrt{2}}\left(\frac{\pi}{2}\right)+1 \\\\ & \Rightarrow f(0)=\frac{-\pi}{4 \sqrt{2}}+0+\frac{\pi}{4 \sqrt{2}}+1 \\\\ & \Rightarrow f(0)=1 \\\\ & f(1)=\frac{1}{2 \sqrt{2}} \tan ^{-1}(-0)+\frac{1}{4 \sqrt{2}} \ln \left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)+\frac{\pi}{4 \sqrt{2}}-1 \\\\ & \Rightarrow f(1)=\frac{1}{4 \sqrt{2}}\left[\ln \left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right) \pi\right]-1 \end{aligned} $$
Here, $f(1)<0$ and $f(0)=1>0$
$$ \Rightarrow f(0) \cdot f(1)<0 $$
By intermediate value theorem
$\Rightarrow$ There exists at least one $c \in(0,1)$ such that $f(c)=0$
Since, $f(x)$ is a decreasing function such that $f(x)=0$ has exactly one root.
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