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Question: Let ѓ(x) =\(\left| \begin{matrix} x^{3} & \sin x & \cos x \\ 6 & - 1 & 0 \\ p & p^{2} & p^{3} \end{m...

Let ѓ(x) =x3sinxcosx610pp2p3\left| \begin{matrix} x^{3} & \sin x & \cos x \\ 6 & - 1 & 0 \\ p & p^{2} & p^{3} \end{matrix} \right|, where p is a constant. Then

d3dx3\frac{d^{3}}{dx^{3}}{ѓ(x)} at x = 0 is –

A

p

B

p + p2

C

p + p3

D

Independent of p

Answer

Independent of p

Explanation

Solution

ƒ(x)= – p3x3 – 6p3 . sin x + (6p2 + p) cos x

\ ƒข(x) = – 3p3x2 – 6p3 cos x – (6p2 + p) sin x

\ ƒขข(x) = – 6p3x + 6p3 sin x – (6p2 + p) cos x

\ ƒขขข(x) = – 6p3 + 6p3 cos x + (6p2 + p) sin x

\ ƒขขข(0) = – 6p3 + 6p3 . 1 = 0 = independent of p.