Question:
Let $f$ be a function which is continuous and differentiable for all $x$. If $f(1) = 1$ and $f'(x) \leq 5$ for all $x$ in $[1, 5]$, then the maximum value of $f(5)$ is
Let $f$ be a function which is continuous and differentiable for all $x$. If $f(1) = 1$ and $f'(x) \leq 5$ for all $x$ in $[1, 5]$, then the maximum value of $f(5)$ is
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$f(b) \leq f(a) + f'(max)(b-a)$.
Updated On: Apr 26, 2026
- $5$
- $20$
- $6$
- $21$
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The Correct Option is D
Solution and Explanation
Step 1: Mean Value Theorem
$\frac{f(5) - f(1)}{5 - 1} = f'(c)$ for some $c \in (1, 5)$.
Step 2: Inequality
Since $f'(x) \leq 5$, then $\frac{f(5) - 1}{4} \leq 5$.
Step 3: Solve
$f(5) - 1 \leq 20 \implies f(5) \leq 21$.
The maximum value is 21.
Final Answer: (D)
$\frac{f(5) - f(1)}{5 - 1} = f'(c)$ for some $c \in (1, 5)$.
Step 2: Inequality
Since $f'(x) \leq 5$, then $\frac{f(5) - 1}{4} \leq 5$.
Step 3: Solve
$f(5) - 1 \leq 20 \implies f(5) \leq 21$.
The maximum value is 21.
Final Answer: (D)
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