Let g(x)=f(x)sinx ,w h e r ef(x) is a tw...

Let `g(x)=f(x)sinx ,w h e r ef(x)` is a twice differentiable function on `(-oo,oo)` such that `f'(-pi)=1.` The value of `|g^(-pi)|` equals __________

A

1

B

2

C

`-2`

D

0

Text Solution

Verified by Experts

The correct Answer is:

C

We have `g(x) = f(x) sin x" (1)"`
On differentiating equation (1) w.r.t. x, we get
`g'(x)=f(x)cos x+f'(x) sinx" (2)"`
Again differentiating equation (2) w.r.t. x, we get
`g''(x)=f(x)(-sinx)+f'(x) cosx+f'(x)cosx+f''(x)sinx" (3)"`
`rArr" "g''(-pi)=2f'(-pi)cos(-pi)=2xx1xx-1=-2`
Hence `g''(-pi)=-2`

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