Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be such that $f(0)=0$ and $\left|f^{\prime}(x)\right| \leq 5$ for all $x$. Then $f(1)$ is in

(A) $(5,6)$

(B) $[-5,5]$

(C) $(-\infty,-5) \cup(5, \infty)$

(D) $[-4,4]$

Here from the hypothesis we know the functional value at a point and we know the bound of its derivatives. We have to find the functional value at an other point. This suggest us that we have to apply Mean Value Theorem.

Mean Value Theorem:

This theorem states that  if $f$ is a continuous function on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists a point $c$ in $(a, b)$ such that
$$f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$$

For this problem we are going to take $a=0$ and $b=1$. Also note that by given hypothesis $|f'(x)|\leq 5$ for all $x$. Therefore,  $$|f'(x)|\leq 5,\qquad\text{for all } x\in(0,1)\tag{Eq. 1}$$

Hence, by applying MVT, we get

\begin{aligned}f(1)-f(0)&=(1-0)f'(c) &&\text{for some } c\in(0,1)\\ |f(1)-f(0)|&=|f'(c)| && \text{By taking both side modulus}\\ |f(1)-0|&\leq 5 && \text{By (Eq. 1) and f(0)=0}\\ |f(1)|&\leq 5\end{aligned}

Hence, the above expression suggest that $-5\leq f(1)\leq 5$, that means, $f(1)\leq [-5,5]$.

The correct option is (B).

by Expert (2.6k points)