Question

Let $S$ be the set of real numbers $p$ such that there is no nonzero continuous function $f:R\to R$ satisfying $\underset{0}{\overset{x}{\int }}f\left(t\right)dt=pf\left(x\right)$ for all $x\in R$. Then $S$ is

• A. the empty set
• B. the set of all rational numbers
• C. the set of all irrational numbers
• D. the whole set $R$

Answer 1

The correct option is D the whole set $R$

Given :
$\underset{0}{\overset{x}{\int }}f\left(t\right)dt=pf\left(x\right)$
Putting $x=0\Rightarrow pf\left(0\right)=0$.......(i)

differentiating w.r.t. x
$f\left(x\right)=p{f}^{\prime }\left(x\right)\Rightarrow \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\frac{1}{p}$
Integrating w.r.t. x
$\ln f\left(x\right)=\frac{x}{p}+c$
$\Rightarrow f\left(x\right)=k.e^{\frac{x}{p}}$ ........(ii)

From equation (i) we say that
Case 1: when $p\ne 0$ then $\Rightarrow f\left(0\right)=0$
i.e. $p\ne 0$ then there is no non zero continuous $f\left(x\right)$

Case 2: when $p=0$
from equation (ii) we can say that if function needs to exist then $k=0$ so,
$f\left(x\right)=0$

Hence for $p\in R$ there is no non zero continous function satisfying the condition given $S=R$.