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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

  1. A 10120
  2. B 10118
  3. C 10122
  4. D 10116
Verified Solution

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…