If a function $y=f(x)$ such that $f'(x)$ is continuous function and satisfies
$(f(x))^2=k+\int\limits_0^x\left[\{f(t)\}^2+\left\{f'(t)\right\}^2\right] d t, k \in R^{+}$, then
(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 & explanation
Correct answer: option 2
We have
$(f(x))^2=k+\int\limits_0^x\left[f^2(t)+f'^2(t)\right] dt$ ......(1)
Differentiating with respect to $x$
$2 f(x) f'(x)=f^2(x)+f'^2(x)$
so $\left(f'(x)-f(x)\right)^2=0$
so $f'(x)=f(x)⇒f'(x)/f(x)=1$
so $\log f(x)=x+\log c$ ....(2)
at $x = 0$ from (1)
$f^2(0)=k⇒f(0)=\sqrt{k}$
from (2) $\log f(0)=0=0+\log c$
so $\log \sqrt{k}=\log c⇒c=\sqrt{k}$
from (2) $\log f(x)=x+\log c$
so $f(x)=\sqrt{k}e^x$
f(x) is increasing for all $x∈R$ and not Bounded and neither odd/even
$f(0) = \sqrt{100}e^0=10$ (at k = 100)