Question:
Let $f$ be a twice differentiable function on $(1,6)$. If $f(2)=8$, $f^{\prime}(2)=5, f^{\prime}(x) \geq 1$ and $f^{\prime \prime}(x) \geq 4$, for all $x \in(1,6)$, then :
Correct Option: , 2
Solution:
Let $f$ be twice differentiable function
$\because f^{\prime}(x) \geq 1$
$\Rightarrow \frac{f(5)-f(2)}{3} \geq 1$
$\Rightarrow f(5) \geq 3+f(2)$
$\Rightarrow f(5) \geq 3+8 \Rightarrow f(5) \geq 11$
and also $f^{\prime \prime}(x) \geq 4$
$\Rightarrow \frac{f^{\prime}(5)-f^{\prime}(2)}{5-2} \geq 4 \Rightarrow f^{\prime}(5) \geq 12+f^{\prime}(2)$
$\Rightarrow f^{\prime}(5) \geq 17$
Hence, $f(5)+f^{\prime}(5) \geq 28$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.