Question

If $f\left(x\right)$ is a differentiable function satisfying $\left|f^{{}^{\prime }}\left(x\right)\right|\le 4\forall x\in \left[0,4\right]$ and $f\left(0\right)=0,$ then

• A. $f\left(x\right)=18$ has no solution in $x\in \left[0,4\right]$
• B. $f\left(x\right)=18$ has more than $2$ solutions in $x\in \left[0,4\right]$
• C. $f\left(x\right)=14$ has no solution in $x\in \left[0,4\right]$
• D. $f\left(x\right)=20$ has $2$ solution in $x\in \left[0,4\right]$

Answer 1

Answer: (A)

Applying LMVT in $x\in \left[0,t\right]$ for $f\left(x\right),$ we get
$f^{{}^{\prime }}\left(c\right)=\frac{f\left(t\right)-f\left(0\right)}{t-0}$
$\Rightarrow \left|f^{{}^{\prime }}\left(c\right)\right|=\left|\frac{f\left(t\right)}{t}\right|\le 4$
$\Rightarrow \left|f\left(t\right)\right|\le 4t$
As $t\in \left[0,4\right]$
$\therefore \left|f\left(t\right)\right|\le 16$