AP EAMCET · Maths · Functions
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
- A 10120
- B 10118
- C 10122
- D 10116
Answer & Solution
Correct Answer
(B) 10118
Step-by-step Solution
Detailed explanation
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…
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