Question

Suppose a continuous function $f:[0,\infty )\to R$ satisfies
$f\left(x\right)=2\underset{0}{\overset{x}{\int }}tf\left(t\right)dt+1$ for all $x\ge 0.$
Then $f\left(1\right)$ equals

• A. $e$
• B. $e^2$
• C. $e^4$
• D. $e^6$

Answer 1

The correct option is A $e$

$f\left(x\right)=2\underset{0}{\overset{x}{\int }}tf\left(t\right)dt+1\Rightarrow f^{\prime }\left(x\right)=2xf\left(x\right)\Rightarrow \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\int 2xdx\Rightarrow \ln f\left(x\right)=x^2+C$
We know $f\left(0\right)=1$
$\ln 1=0+C\Rightarrow C=0\Rightarrow f\left(x\right)=e^{x^2}\therefore f\left(1\right)=e$