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UPSC NDA I 2016 Mathematics Solved Paper

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For the next two (2) items that follow :

Consider the function

f(x)=\begin{vmatrix} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}

Where p is constant.

[NDA I 2016]

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Question : 34 of 120

Marks: +1, -0

\text{What is the value of } f'(0)?

$p^3$
$3p^3$
$6p^3$
$-6p^3$

Solution:

f(x)=\begin{vmatrix} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}

f'(x)=\begin{vmatrix} 3x^2 & \cos x & -\sin x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}+\begin{vmatrix} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ 0 & 0 & 0 \end{vmatrix}+\begin{vmatrix} x^3 & \sin x & \cos x \\ 0 & 0 & 0 \\ p & p^2 & p^3 \end{vmatrix}

f'(x)=\begin{vmatrix} 3x^2 & \cos x & -\sin x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}

f'(0)=\begin{vmatrix} 0 & \cos 0 & -\sin 0 \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}

=-6p^3

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