Question

Suppose that $f$ is differentiable for all x such that $f^{\prime }\left(x\right)$ $\le$ $2$ for all $x$, if $f\left(1\right)=2$ and $f\left(4\right)=8$ then $f\left(2\right)$ has the value equal to

• A. $3$
• B. $4$
• C. $6$
• D. $8$

Answer 1

Answer: (B)

$4$

$xϵ(1,4)$
Divide this interval in two parts $xϵ(1,2)$ and $xϵ(2,4)$ and apply LMVT individually to both parts.

$\frac{f\left(2\right)-f\left(1\right)}{2-1}=f‘\left(c_1\right)\forall c_1ϵ(1,2)$

$f\left(2\right)-2$ $\le 2$ $\because f^{\prime }\left(x\right)\le 2\Rightarrow f\left(2\right)\le 4$ ..........(1)

Similarly applying LMVT in $[2,4]$

$\frac{f\left(4\right)-f\left(2\right)}{4-2}=f‘\left(c_2\right)\forall c_2ϵ(2,4)$

$\frac{8-f\left(2\right)}{2}\le 2\Rightarrow f\left(2\right)\ge 4$

From (1) & (2) $f\left(2\right)=4$