Question

If $f\left(x\right)$ be a function such that $\left(f\right(x){)}^{2007}=\underset{0}{\overset{x}{\int }}\frac{\left(f\right(t){)}^{2006}}{2+t^2}dt,$ then which of the following is/are true?

• A. There are only two such functions are possible.
• B. There are infinite number of such functions are possible.
• C. $f\left(x\right)=\frac{1}{2007}{\tan }^{-1}\left(\frac{x}{\sqrt{2}}\right)$
• D. $f\left(x\right)=\frac{1}{2007\sqrt{2}}{\tan }^{-1}\left(\frac{x}{\sqrt{2}}\right)$

Answer 1

Answer: (A), (D)

$f\left(x\right)=\frac{1}{2007\sqrt{2}}{\tan }^{-1}\left(\frac{x}{\sqrt{2}}\right)$

Given:
$\left(f\right(x){)}^{2007}=\underset{0}{\overset{x}{\int }}\frac{\left(f\right(t){)}^{2006}}{2+t^2}dt$

Using Newton-Leibniz theorem, we get
$2007\left(f\right(x){)}^{2006}\times f^{\prime }(x)=1\times \frac{\left(f\right(x){)}^{2006}}{2+x^2}\Rightarrow (f\left(x\right){)}^{2006}\left(2007f^{\prime }\right(x)-\frac{1}{2+x^2})=0\Rightarrow f\left(x\right)=0\text{or}2007f^{\prime }\left(x\right)=\frac{1}{2+x^2}$

Now,
$2007f^{\prime }\left(x\right)=\frac{1}{2+x^2}\Rightarrow 2007\int f^{\prime }\left(x\right)dx=\int \frac{dx}{2+x^2}\Rightarrow 2007f\left(x\right)=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x}{\sqrt{2}}\right)+C\cdots \left(1\right)$

From the given integral, we get
$\left(f\right(0){)}^{2007}=\underset{0}{\overset{0}{\int }}\frac{\left(f\right(t){)}^{2006}}{2+t^2}dt\Rightarrow f(0)=0$

Using equation $\left(1\right)$, we get
$2007f\left(0\right)=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{0}{\sqrt{2}}\right)+C\Rightarrow C=0$

Therefore,
$f\left(x\right)=0$
OR
$f\left(x\right)=\frac{1}{2007\sqrt{2}}{\tan }^{-1}\left(\frac{x}{\sqrt{2}}\right)$