NIMCET 2008 Question Paper with solutions

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Question : 3 of 120

Marks: +1, -0

If

f : R \to R

and

g : R \to R

and continuous functions, then the value of the integral

∫↙{-π / 2}↖{π / 2}[f(x)+f(-x)][g(x)-g(-x)] d x

is

π
1
-1
0

Solution:

Let

h(x)= [f(x)+f(-x)\][g(x)-g(-x)]

⇒ h(-x)=[f(-x)+f(x)][g(-x)-g(x)]

=-[f(x)+f(-x)][g(x)-g(-x)]

=-h(x)

⇒ h(x)

is an odd function.

⇒ ∫↙{-π / 2}↖{π / 2} h(x) d x=0

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