Question

If $f:R\to R$ be a continuous function such that $f\left(x\right)={\int }_1^{x}2tf\left(t\right)dt$, then which of the following does not hold(s) good?

• A. $f\left(\pi \right)=e^{{\pi }^2}$
• B. $f\left(1\right)=e$
• C. $f\left(0\right)=1$
• D. $f\left(2\right)=-2$

Answer 1

The correct option is D $f\left(2\right)=-2$

$f\left(x\right)={\int }_1^{x}2tf\left(t\right)dt$

$f^{\prime }\left(x\right)=\frac{d}{dx}{\int }_1^{x}2tf\left(t\right)dt$ $\to$

$\therefore f^{\prime }\left(x\right)=2xf\left(x\right)$

$\therefore \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=2x$

Now integrating,

$\int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\int 2x$

$\therefore lnf\left(x\right)=x^2$

$\therefore f\left(x\right)=e^{x^2}$

$\therefore f\left(x\right)=e^{2^2}=e^4$

$e^4\ne -2$.