f(x)= {vmatrix} x³ & sinx & cosx \\ 6 & -1 & 0 \\ p & p² & p³ {vmatrix} , where p is a constant What is the value of f’(0)?

Correct Answer is : −6p3-6p³−6p3

Correct Answer is : −6p3-6p³−6p3

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UPSC NDA I 2016 Mathematics Solved Paper

Question Numbers: 34-35

Consider the function

Question : 34 of 120

Marks: +1, -0

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

What is the value of f’(0)?

Solution:

Concept:

The derivative of a determinant with only one varying row is found by differentiating that row and keeping the others unchanged.

Explanation:

We are given

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

Since only the first row depends on

x

, the derivative is:

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

Compute the derivatives:

\frac{d}{dx}(x^{3}) = 3x^{2}

\frac{d}{dx}(\sin x) = \cos x

\frac{d}{dx}(\cos x) = -\sin x

Thus

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

Now set

x = 0

\cos 0 = 1

\sin 0 = 0

, so the first row becomes

(0,\;1,\;0)

Hence

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

Expand along the first row:

0 \cdot \text{minor} - 1 \cdot (6 \cdot p^{3} - 0 \cdot p) + 0 \cdot \text{minor} = -6p^{3}

Answer:

-6p^{3}

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