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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 \mathrmf′(0)\mathrmf^{\prime}(0) ?

Options

  1. Ap3\mathrm{p}^3
  2. B3p33 p^3
  3. C6p36 \mathrm{p}^3
  4. D−6p3-6 p^3

Correct answer

D. −6p3-6 p^3

Step-by-step solution

f(x)=∣x3sin⁡xcos⁡x6−10pp2p3∣f′(x)=∣3x2cos⁡x−sin⁡x6−10pp2p3∣+∣x3sin⁡xcos⁡x6−10000∣+∣x3sin⁡xcos⁡x000pp2p3∣\begin{aligned} f(x) & =\left|\begin{array}{ccc}x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3\end{array}\right| \\ f^{\prime}(x)= & \left|\begin{array}{ccc}3 x^2 & \cos x & -\sin x \\ 6 & -1 & 0 \\ p & p^2 & p^3\end{array}\right|+\left|\begin{array}{ccc}x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ 0 & 0 & 0\end{array}\right| \\ & +\left|\begin{array}{ccc}x^3 & \sin x & \cos x \\ 0 & 0 & 0 \\ p & p^2 & p^3\end{array}\right|\end{aligned}f′(x)=∣3x2cos⁡x−sin⁡x6−10pp2p3∣f′(0)=∣0cos⁡0−sin⁡06−10pp2p3∣=−6p3\begin{aligned} f^{\prime}(x) & =\left|\begin{array}{ccc}3 x^2 & \cos x & -\sin x \\ 6 & -1 & 0 \\ p & p^2 & p^3\end{array}\right| \\ f^{\prime}(0) & =\left|\begin{array}{ccc}0 & \cos 0 & -\sin 0 \\ 6 & -1 & 0 \\ p & p^2 & p^3\end{array}\right| \\ & =-6 p^3\end{aligned}
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