Q72 / 200 · WBJEE 2014 · Mathematics

Let $f(x)$ be a differentiable function in $[2,7]$. If $f(2)=3$ and $f^{\prime}(x) \leq 5$ for all $x$ in $(2,7),$ then the maximum posaible value of $f(x)$ at $x=7$ is

  • A
    7
  • B
    15
  • Correct answer
    C
    28
  • D
    14

Solution

Given, $\quad f^{\prime}(x) \leq 5$ $\frac{f(7)-f(2)}{x-2} \leq 5$ (by Lagrange mean value theorem] $\begin{array}{ll}\Rightarrow & \frac{f(7)-3}{7-2} \leq 5 \\ \Rightarrow & f(7) \leq 5(7-2)+3 \\ \therefore & f(7) \leq 28\end{array}$

Hence, the correct answer is option: 'C'.

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