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Let f(x)=|x^3sinxcosx6-1 0p p^2p^3|,w h...

Let `f(x)=|x^3sinxcosx6-1 0p p^2p^3|,w h e r ep` is a constant. Then `(d^3)/(dx^3)(f(x))` at `x=0` is `p` (b) `p-p^3` `p+p^3` (d) independent of `p`

A

p

B

`p + p^(2)`

C

`p + p^(3) `

D

independent of p

Text Solution

Verified by Experts

The correct Answer is:
D

Given, `f(x) = |{:(x^(3)" "sin x" "cos x),(6" "-1" "0),(p" "p^(2)" "p^(3)):}|`
On differentiating w.r.t. x, we get
`f'(x) =|{:(3x^(2)" "cos x" "-sin x),(6" "-1" "0),(p" "p^(2) " "p^(3)):}|+|{:(x^(3)" "sin x" "cos x),(0" "0" "0),(p" "p^(2)" "p^(3)):}|+|{:(x^(3) " "sin x" "cos x),(6" "-1" "0),(0" "0" "0):}|`
` rArr" "f'(x)=|{:(3x^(2)" "cos x" "-sin x),(6" "-1" "0),(p" "p^(2)" "p^(3)):}|`
`rArr" "f''(x)=|{:(6x" "-sinx" "-cos x),(6" "-1" "0),(p" "p^(2)" "p^(3)):}|+0+0`
`and f'''(x) =|{:(6" "-cos x" "sin x),(6" "-1" "0),(p" "p^(2)" "p^(3)):}|+0+0`
`:. f'''(0)=|{:(6" "-1" "0),(6" "-1" "0),(p" "p^(2)" "p^(3)):}|=0="independent of p"`
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