Question

Let $f$ be differentiable for all $x$. If $f\left(1\right)=-2$ and $f’\left(x\right)\ge 2$for all $x\in [1,6]$ then

• A. $f\left(6\right)<8$
• B. $f\left(6\right)\ge 8$
• C. $f\left(6\right)\ge 5$
• D. $f\left(6\right)\le 5$

Answer 1

The correct option is B

$f\left(6\right)\ge 8$

Explanation for correct option:

Lagrange's theorem:

We know that, $f$ be differentiable for all $x$.This means that $f$ will also be continuous for all $x$, where $x\in \left[1,6\right]$.

According to Lagrange's theorem, there is a value $C$ between $1$ and $6$, such that $1<C<6$.

Now,

$$\begin{gathered} \Rightarrow f\text{'}\left(C\right)=\frac{f\left(6\right)-f\left(1\right)}{6-1} \\ \Rightarrow f\text{'}\left(C\right)=\frac{f\left(6\right)-(-2)}{5}\mathbf{\left[}\mathbf{f}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{\right]} \\ \Rightarrow \frac{f\left(6\right)+\left(2\right)}{5}\ge 2\mathbf{\left[}\mathbf{\because }\mathbf{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\ge }\mathbf{2}\mathbf{\right]} \\ \Rightarrow f\left(6\right)+\left(2\right)\ge 10 \\ \Rightarrow f\left(6\right)\ge 10-2 \\ \Rightarrow f\left(6\right)\ge 8 \end{gathered}$$

Hence, option(B) is correct.