Question
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) =x36psinx−1p2cosx0p3, where p is a constant. Then
dx3d3{ѓ(x)} at x = 0 is –
A
p
B
p + p2
C
p + p3
D
Independent of p
Answer
Independent of p
Explanation
Solution
(x)= – p3x3 – 6p3 . sin x + (6p2 + p) cos x
\ ข(x) = – 3p3x2 – 6p3 cos x – (6p2 + p) sin x
\ ขข(x) = – 6p3x + 6p3 sin x – (6p2 + p) cos x
\ ขขข(x) = – 6p3 + 6p3 cos x + (6p2 + p) sin x
\ ขขข(0) = – 6p3 + 6p3 . 1 = 0 = independent of p.