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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 __________

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