If a function y=f(x) such that f'(x) is continuous function and satisfies
`(f(x))^(2)=k+int_(0)^(x) [{f(t)}^(2)+{f'(t)}^(2)]dt,k in R^(+) `, then

Options

A. f(x) is an increasing function for all ` x in R`
B. f(x) is a bounded function
C. f(x) is neither even nor odd function
D. If k=100, then f(0)=10.

Answer

A, C

Explanation

We have
`(f(x))^(2)=k+underset(0)overset(x)int [{f(t)}^(2)+{f'(t)}^(2)]dt " "`.....(i)
Differentiating with respect of x, we obtain
`2f(x)f'(x)={f(x)}^(2)+{f'(x)}^(2)`
`rArr {f(x)=f'(x)}^(2)=0`
`rArr f(x)-f'(x)=0`
`rArr f'(x)=f(x)`
`(f'(x))/(f(x))=1`
`rArr log(f(x))=x+logC rArrf(x)=Ce^(x)" "`....(ii)
Putting x=0 in (i), we obtain
`f(0)=k+0 rArr f(0)=k rArr C=k`
Putting C=k in (ii), we obtain `f(x)=ke^(x)`.
We observe that f(x) is an increasing function for all `x in R` and is not bounded. Also, it is neither even nor odd.
If k=100, then f(x)`=ke^(x)`, gives
f(x)`=100e^(x) rArr f(0)=100=100`
Hence, options (a) and (c ) are correct.