Q104 / 160 · TS EAMCET 2023 (13 May Shift 1) · Mathematics

If $f(x)=|x-1|+|x-2|$, then $f^{\prime}(-2023)+$ $\mathrm{f}^{\prime}\left(\frac{2024}{2023}\right)+\mathrm{f}^{\prime}(2023)=$

  • A
    $1$
  • B
    $-1$
  • Correct answer
    C
    $0$
  • D
    $3$

Solution

$\begin{aligned} & \text { When } x=-2023 \Rightarrow f(x)=-(x-1)-(x-2)=-2 x+3 \\ & \Rightarrow f^{\prime}(x)=-2 \Rightarrow f^{\prime}(-2023)=-2 \\ & \text { When } x=\frac{2024}{2023} \Rightarrow f(x)=(x-1)-(x-2)=1 \\ & \Rightarrow f^{\prime}(x)=0 \Rightarrow f^{\prime}\left(\frac{2024}{2023}\right)=0 \\ & \text { When } x=2023 \Rightarrow f(x)=(x-1)+(x-2)=2 x-3 \\ & \Rightarrow f^{\prime}(x)=2 \Rightarrow f^{\prime}(2023)=2 \\ & \text { Therefore } f^{\prime}(-2023)+f^{\prime}\left(\frac{2024}{2023}\right)+f^{\prime}(-2023)=0\end{aligned}$

Hence, the correct answer is option: 'C'.

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