Let f be a twice differentiable function on

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 :

  1. (1) $f(5)+f^{\prime}(5) \leq 26$

  2. (2) $f(5)+f^{\prime}(5) \geq 28$

  3. (3) $f^{\prime}(5)+f^{\prime \prime}(5) \leq 20$

  4. (4) $f(5) \leq 10$


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, $f(5)+f^{\prime}(5) \geq 28$

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