Let f - R - R and g - R - R be continuous functions- Then- the value of the integral- fx-f-x gx-g-x d x isA- NB- 1C- - 2D- - 4

The correct option is A nbsp;h(x)=f(x)+f( x)g(x) g( x)∴h( x)=f( x)+f(x)g( x) g(x)⇒h( x)= h(x)Hence, nbsp;h(x) nbsp;is nbsp;an nbsp;odd nbsp;functio ...

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Byju's Answer

Standard XII

Mathematics

Property 7

Let f : R → R...

Question

Let $f : R \to R a n d g : R \to R$ be continuous functions. Then, the value of the integral
$\int_{- π}^{π} (f (x) + f (- x)] (g (x) - g (- x)] dx$ is

0.0

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\frac{π}{2}

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\frac{π}{4}

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Solution

The correct option is A 0.0
$Let h (x) = (f (x) + f (- x)] (g (x) - g (- x)] ∴ h (- x) = (f (- x) + f (x)] (g (- x) - g (x)] \Rightarrow h (- x) = - h (x) Hence, h (x) is an odd function . ∴ \int_{- π}^{π} (f (x) + f (- x)] (g (x) - g (- x)] dx = 0$

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Let f:R→R and g:R→R be continuous functions. Then, the value of the integral ∫π−π (f(x)+f(−x)](g(x)−g(−x)] dx is

The correct option is A 0.0Let h(x)=(f(x)+f(−x)](g(x)−g(−x)]∴h(−x)=(f(−x)+f(x)](g(−x)−g(x)]⇒h(−x)=−h(x)Hence, h(x) is an odd function.∴∫π−π (f(x)+f(−x)](g(x)−g(−x)] dx =0

Let $f : R \to R a n d g : R \to R$ be continuous functions. Then, the value of the integral
$\int_{- π}^{π} (f (x) + f (- x)] (g (x) - g (- x)] dx$ is

The correct option is A 0.0
$Let h (x) = (f (x) + f (- x)] (g (x) - g (- x)] ∴ h (- x) = (f (- x) + f (x)] (g (- x) - g (x)] \Rightarrow h (- x) = - h (x) Hence, h (x) is an odd function . ∴ \int_{- π}^{π} (f (x) + f (- x)] (g (x) - g (- x)] dx = 0$