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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Determinants

NDA 2016 · Mathematics · Determinants

Consider the function f(x)=| array ccc x^3 & x & x 6 & -1 & 0 p &...

Consider the function f(x)=\left|x3sin⁡xcos⁡x6−10pp2p3\begin{array}{ccc}x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3\end{array}\right|, where pp is a constant. What is the value of pp for which f′′(0)=0f^{\prime \prime}(0)=0 ?

Options

  1. A−16-\frac{1}{6} or 0
  2. B-1 or 0
  3. C−16-\frac{1}{6} or 1
  4. D-1 or 1

Correct answer

A. −16-\frac{1}{6} or 0

Step-by-step solution

f′′(x)=∣6x−sin⁡x−cos⁡x6−10pp2p3∣f′′(0)=∣00−cos⁡06−10pp2p3∣=−1(6p2+p)f′′(0)=0−(6p2+p)=0p(6p+1)=0p=0 or p=−16\begin{aligned} & f^{\prime \prime}(x)=\left|\begin{array}{ccc}6 x & -\sin x & -\cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3\end{array}\right| \\ & f^{\prime \prime}(0)=\left|\begin{array}{ccc}0 & 0 & -\cos 0 \\ 6 & -1 & 0 \\ p & p^2 & p^3\end{array}\right| \\ & \quad=-1\left(6 p^2+p\right) \\ & f^{\prime \prime}(0)=0 \\ & \begin{array}{l}-\left(6 p^2+p\right)=0 \\ p(6 p+1)=0\end{array} \\ & p=0 \text{ or } p=-\frac{1}{6}\end{aligned}
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