Question

Let $f$ be a twice differentiable function on $(1,6)$. If $f\left(2\right)=8,f^{\prime }\left(2\right)=5,f^{\prime }\left(x\right)\ge 1$ and $f^{\prime \prime }\left(x\right)\ge 4,$ for all $x\in (1,6),$ then

• A. $f\left(5\right)+$$f^{\prime }\left(5\right)\ge 28$
• B. $f^{\prime }\left(5\right)+f^{\prime \prime }\left(5\right)\le 20$
• C. $f\left(5\right)\le 10$
• D. $f\left(5\right)+f^{\prime }\left(5\right)\le 26$

Answer 1

The correct option is A $f\left(5\right)+$$f^{\prime }\left(5\right)\ge 28$

$f\left(2\right)=8,f^{\prime }\left(2\right)=5,f^{\prime }\left(x\right)\ge 1,f^{\prime \prime }\left(x\right)\ge 4\forall x\in (1,6)$
Using LMVT,
$f^{\prime \prime }\left(c_1\right)=\frac{f^{\prime }\left(5\right)-f^{\prime }\left(2\right)}{5-2}\ge 4\forall x\in (1,6)$
$\Rightarrow f^{\prime }\left(5\right)\ge 17\cdots \left(1\right)$
and
$f^{\prime }\left(c_2\right)=\frac{f\left(5\right)-f\left(2\right)}{5-2}\ge 1\forall x\in (1,6)$
$\Rightarrow f\left(5\right)\ge 11\cdots \left(2\right)$

Adding $\left(1\right)$ and $\left(2\right),$ we get
$f\left(5\right)+f^{\prime }\left(5\right)\ge 28$