If $f(0)=0$ and $f'(x)\in(0,1]\;\;\forall x\in[0,1]$ then $\dfrac{\left(\int_0^1f'(x)dx\right)^2}{\int_0^1(f(x))^3}$ can not take values

Question

The following question is taken from the practice set of JEE exam.

If $f(0)=0$ and $f'(x)\in(0,1]\;\;\forall x\in[0,1]\;$ then $\dfrac{\left(\int_0^1f'(x)dx\right)^2}{\int_0^1(f(x))^3}$ can not take values

• A) $\dfrac12$
• B) $2$
• C) $\dfrac14$
• D) $1$

Numerator can be written as $f^2(1)=\int_0^1f^2(1)dx$

Also, for $x\in(0,1), f'(x)>0\implies f(x)$ is increasing. So, we can say $\int_0^1f^2(1)dx>\int_0^1f^2(x)dx$

But don't know how to compare it to $f^3(x)$.

Can we say $\int_0^1f^2(1)dx>\int_0^1f^3(x)dx$?

Can we use Cauchy Schwarz inequality here? Not able to figure out how.

Answer 1

Can we say $\int_0^1f^2(1)dx>\int_0^1f^3(x)dx$?

Yes, we can.

Note that by MVT, we have $$|f(1) - f(0)| = f'(\xi)|1-0| \in (0,1] \implies f(1) \in (0,1] \tag{$\star$}$$ for some $\xi \in (0,1]$. Therefore $$\int_0^1 f^3(x)dx < f^3(1) \overset{(\star)}\le f^2(1) = \int_0^1 f^2(1) dx$$ as you wish.

Answer 2

The quantity is

$$ Q(f) = \frac{f(1)^2}{\int_0^1 f^3(x)dx}$$

First observation: for each $\alpha \in (0,1)$,

$$Q(\alpha f) = \alpha^{-1} Q(f).$$

Thus the quantity can get as large as we please. So we need to minimize $Q(f)$.

For now assume that $c = f(1)$ is fixed. Since $f(x) \le x$ and $f(x) \le f(1) = c$, we have

$$ f(x) \le \min \{x, c\} = \begin{cases} x & x\in [0,c] \\ c & x\in [c, 1].\end{cases}$$

and thus

$$\int_0^1 f^3(x) dx < \frac{c^4}{4}+ c^3(1-c) = c^3\left( 1- \frac 34 c\right)$$

(we have strictly inequality here, since $f$ is differentiable and $f$ cannot be $\min\{x, c\}$). This implies

$$Q(f) > \frac{1}{c (1-3c/4)}.$$

The function $c\mapsto c(1-3c/4)$ is maximized at $c = 2/3$, thus

$$ Q(f) >3.$$

So it seems that all four options are correct...