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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Differentiation

NDA 2015 · Mathematics · Differentiation

Consider ^ (x)= x^2 2 -kx+1 such that f(0)=0 and f ( 3)=15 ^...

Consider \mathrmf′(x)=x22−kx+1\mathrmf^{\prime}(\mathrm{x})=\frac{\mathrm{x}^2}{2}-\mathrm{kx}+1 such that f(0)=0\mathrm{f}(0)=0 and f\mathrm{f}(3)=15\mathrmf′′(−23)isequalto3)=15\mathrmf^{\prime \prime} \left(-\frac{2}{3}\right)is equal to

Options

  1. A-1
  2. B1/3
  3. C1/2
  4. D1

Correct answer

D. 1

Step-by-step solution

∵f′(x)=x22+53x+1⇒\mathrmf′′(x)=x+53\mathrmf′′(−23)=−23+53=1\begin{aligned} & \because f^{\prime}(\mathrm{x})=\frac{\mathrm{x}^2}{2}+\frac{5}{3} \mathrm{x}+1 \\ & \Rightarrow \mathrmf^{\prime \prime}(\mathrm{x})=\mathrm{x}+\frac{5}{3} \\ & \mathrmf^{\prime \prime}\left(-\frac{2}{3}\right)=\frac{-2}{3}+\frac{5}{3} \\ & =1 \end{aligned} So option (4) is correct.
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