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Question

If f(x) is a function such that f′′(x)+f(x)=0 and g(x)=[f(x)]2+[f′(x)]2 and g(3)=8, then g(8)=_____

A
0
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B
3
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C
5
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D
8
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Solution

The correct option is D 8
We have, g(x)=[f(x)]2+[f(x)]2
Differentiate the function g(x)
g(x)=2f(x)f(x)+2f(x)f′′(x), use chain rule
=2f(x)[f(x)+f′′(x)]=2f(x)(0)=0, use the given condition
Hence g(x) is a constant function
g(x)=c, constant
But g(3)=8, so g(x)=8, for all real x
Hence g(8)=8

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