Q243 / 270 · NDA 2015 (Phase 2) · Mathematics

Consider $\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{x}^2}{2}-\mathrm{kx}+1$ such that $\mathrm{f}(0)=0$ and $\mathrm{f}(3)=15$ The value of $k$ is

  • A
    5/3
  • B
    3/5
  • Correct answer
    C
    -5/3
  • D
    -3/5

Solution

$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{x}^2}{2}-\mathrm{kx}+1 \\ & \mathrm{f}(0)=0 ; \mathrm{f}(3)=15 \\ & \mathrm{f}(\mathrm{x})=\frac{1}{6} \mathrm{x}^3-\frac{\mathrm{k}}{2} \mathrm{x}^2+\mathrm{x}+\mathrm{c} \end{aligned}$ Putting $\mathrm{x}=0$ $\begin{aligned} & \mathrm{f}(0)=\mathrm{c}=0 \\ & \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^3}{6}-\frac{k}{2} \mathrm{x}^2+x \end{aligned}$ Putting $x=3$ $\begin{aligned} & f(3)=\frac{(3)^3}{6}-\frac{k}{2}(3)^2+3 \\ & 15=\frac{9}{2}-\frac{9}{2} k+3 \\ & \Rightarrow \mathrm{k}=-\frac{5}{3} \end{aligned}$

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

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