Let $f: \Bbb R \to (0, \infty)$ be a twice differentiable function such that $f(0)=1$ and $\int_a^b f(x)dx=\int_a^b$$f^{'}(x)dx$ for all a,b $\in \Bbb R$, with $a \le b$. Which of the following statements is/are true?
$1$. $f$ is one to one.
$2$. The image of $f$ is compact.
$3$. $f$ is unbounded.
$4.$ There is only one such function.
first I tired to rule out the options. Taking $f(x)= e^{x}$ which satisfy all the required conditions but image of $f(x)= e^{x}$ is not bounded so $2.$ option is ruled out.
$$and\int_a^b f(x)dx=\int_a^bf^{'}(x)dx $$
implies that $\int_a^b f(x)dx=f(b)-f(a)$ all a,b $\in \Bbb R$…..$(A)$
For $1.$ let $c\neq d$ such that $f(c)=f(d)$ . Now using these $c,d$ we get $\int_c^d f(x)dx=0$ which is contradiction becuase $f(x) \gt 0$ for all x$\in \Bbb R$ so $f$ is one to one.
For $3.$ Let $f$ is bounded i.e there exists $M \ge 0$ st $ \vert f(x) \vert \le M$ for all x $ \in \Bbb R $. then $(A)$ implies $$ \vert f(b)-f(a) \vert \le M(b-a) \vert$$
Taking limit $ b \to a$ we get $ \vert f^{'}(a) \vert \le M$ for all $a \in \Bbb R$ i.e derivative of $f$ must be bounded on $\Bbb R$ but as taking $f(x)= e^{x}$ which satisfy all required condition but derivative of $e^{x}$ is not bounded. so $f$ is not bounded.
For $4$. i need explaination…and please check the proof of above options also
Best Answer
Just differentiate the equation with respect to $b$ to get $f(b)=f'(b)$ for all $b$. This implies $f(x)=ce^{x}$ and the condition $f(0)=1$ gives $c=1$. Hence $f(x)=e^{x}$ is the only solution. [ $f(b)=f'(b)$ implies $\frac d {db} \log f(b)=1$ so $\log f(b)=b+d$ where $d$ is a constant. Hence $f(b)=e^{b+d}=ce^{b}$ where $c=e^{d}$].