(B) Given $f^{\prime}(x) \geq 5$. By the Mean Value Theorem,for $x \in [0, 2024]$,there exists some $c \in (0, 2024)$ such that $\frac{f(2024) - f(0)}

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(B) Given $f^{\prime}(x) \geq 5$. By the Mean Value Theorem,for $x \in [0, 2024]$,there exists some $c \in (0, 2024)$ such that $\frac{f(2024) - f(0)}{2024 - 0} = f^{\prime}(c)$. Since $f^{\prime}(c) \geq 5$,we have $\frac{f(2024) - (-2)}{2024} \geq 5$. $f(2024) + 2 \geq 5 \times 2024$. $f(2024) + 2 \geq 10120$. $f(2024) \geq 10118$. Thus,the least possible value of $f(2024)$ is $10118$.

Class 12 Mathematics Continuity and Differentiability Rolle’s theorem, Lagrange's mean value theorem

Easy

For all $x \in [0, 2024]$,assume that $f(x)$ is differentiable,$f(0) = -2$,and $f^{\prime}(x) \geq 5$. Then the least possible value of $f(2024)$ is:

AP EAMCET 2024 All AP EAMCET

A
$10120$
B
$10118$
C
$10122$
D
$10116$

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Continuity and Differentiability

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Rolle’s theorem, Lagrange's mean value theorem

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For all $x \in [0, 2024]$,assume that $f(x)$ is differentiable,$f(0) = -2$,and $f^{\prime}(x) \geq 5$. Then the least possible value of $f(2024)$ is:

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