Let S be the set of real numbers p such that there is no nonzero continuous function `f: R to R` satisfying `int_(0)^(x) f(t) dt= p f(x)` for all `x in R`. Then S is

Options

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

Answer

D

Explanation

`int_(0)^(x)f(t)dtphi(f(x))`...(1)
Differentiable both side with respect to x
f(x)=pf'(x)
`(f'(x))/(f(x))=(1)/(p)`
Now, integrating both side w.r.t x
`In f(x) =(x)/(p)+C`
`f(x)=k.e^(x//p)`….(2)
patting x=0 in original equation (1)
0=pf(0)
`implies` either p=0 or {f(0)=0 & p`ne`0}
Let cse I where f(0)=0 & `phi ne 0 implies=0` then from equation (2)
`f(x) =0" " implies f(x)=0`
`implies` i.e, `p ne0` then there is no zero continuous to f(x)
Case II p=0
`int_(0)^(x) f(t)dt=o AA x in R`
It is only possible when f(x)=0
Hence `AA p in R`: There is no nonzero continuous function, satisfying the given condition.
Hence `S in R`