For all $\mathrm{x} \in[0,2024]$ assume that $f(x)$ is differentiable, $f(0)=-2$ and $f^{\prime}(x) \geq 5$.…
For all $\mathrm{x} \in[0,2024]$ assume that $f(x)$ is differentiable, $f(0)=-2$ and $f^{\prime}(x) \geq 5$. Then the least possible value of $f$ (2024) is
10120
10118
10122
10116
Solution
Given $\frac{d(f(x))}{d x} \geq 5$
$\begin{aligned}
& \Rightarrow \int d(f(x)) \geq \int 5 d x \Rightarrow f(x) \geq 5 x+c \\
& f(0) \geq 0+c \Rightarrow-2 \geq c \text { So, } c=(-\infty,-2]
\end{aligned}$
Now $f(2024) \geq 5 \times 2024+c=10120+c$
For least value of $f(2024)$, we take $c=-2$
So, $f(2024)=10120-2 \quad$ (which is least)
$\Rightarrow$ Least value of $f(2024)=10118$