Q218 / 270 · NDA 2016 (Phase 1) · Mathematics

Consider the function $\mathrm{f}(\mathrm{x})=\left|\mathrm{x}^2-5 \mathrm{x}+6\right|$ What is $\mathrm{f}^{\prime}(4)$ equal to?

  • A
    –4
  • B
    –3
  • Correct answer
    C
    3
  • D
    2

Solution

$\begin{aligned} & \text { Here; } \mathrm{f}(\mathrm{x})=\left|x^2-5 x+6\right| \\ & f(x)=|(x-2)(x-3)| \\ & f(x)=\left\{\begin{array}{ccc} (x-2)(x-3) & ; & x \lt 2 \\ -(x-2)(x-3) & ; & 2 \leq x \lt 3 \\ (x-2)(x-3) & ; & x \geq 3 \end{array}\right. \\ & f^{\prime}(x)=\left\{\begin{array}{ccc} 2 x-5 & ; & x \lt 2 \\ -(2 x-5) & ; & 2 \leq x \lt 3 \\ (2 x-5) & ; & x \geq 3 \end{array}\right. \\ & f^{\prime \prime}(x)=\left\{\begin{array}{ccc} 2 & ; & x \lt 2 \\ -2 & ; & 2 \leq x \lt 3 \\ 2 & ; & x \geq 3 \end{array}\right. \\ & \Rightarrow \quad f^{\prime}(4)=2(4)-5=3 \end{aligned}$

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

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