Question

If ${\int }_0^{x}f\left(t\right)dt=x^2+2x+{\int }_x^{1}tf\left(t\right)dt$, then $f\left(x\right)$ is

• A. periodic
• B. $f\left(x\right)=3x$
• C. periodic and fundamental period exists
• D. $f\left(x\right)=4x$

Answer 1

The correct option is A periodic

Differentiating the LHS and RHS w.r.t x:-

$\frac{d}{dx}{\int }_0^{x}f\left(t\right)dt=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2x\right)+\frac{d}{dx}{\int }_x^{1}tf\left(t\right)dt$

According Leibniz integral rule-

$\frac{d}{dx}{\int }_{a\left(x\right)}^{b\left(x\right)}f\left(x\right)dx$

$=f\left(b\right(x\left)\right)\frac{d}{dx}b\left(x\right)-f\left(a\right(x\left)\right)\frac{d}{dx}a\left(x\right)$

Hence-

$f\left(x\right)=2x+2+(0-xf(x\left)\right)$

$⟹(x+1)f\left(x\right)=2(x+1)$
$⟹f\left(x\right)=2$

Hence,f(x) is constant function and hence a periodic function but no fundamental period.

Hence, answer is option-(A).