Let f x - g x - h x be continous in -0-2 a - and satisfies f 2 a - x - f x - g 2 a - x - g x - h x - h 2 a - x -3- f 2 a - x g 2 a - x -f x g x then -02 a f x g x h x dx -A- 3 -0 a f x g x dxB- -02 a f x g x dxC- -0 a f x g x dxD- 2 -0 a f x g x dx

The correct option is A 3 ∫_0^a f(x)g(x)dxI = ∫_0^2a f(x)g(x)h(x)dx = ∫_0^2a f(2a x)g(2a x)h(2a x)dx = ∫_0^2a f(x)g(x) [3 h(x)]dx I = 3/2∫_0^2a f(x)g(x)dx = ...

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Byju's Answer

Standard XII

Mathematics

Inequalities of Integrals

Let f x , g x...

Question

Let f(x), g(x), h(x) be continous in [0, 2a] and satisfies $f (2 a - x) = f (x), g (2 a - x) = g (x), h (x) + h (2 a - x) = 3, f (2 a - x) g (2 a - x) = f (x) g (x) t h e n \int_{0}^{2 a} f (x) g (x) h (x) d x =$

\int_{0}^{2 a} f (x) g (x) d x

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3 \int_{0}^{a} f (x) g (x) d x

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2 \int_{0}^{a} f (x) g (x) d x

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\int_{0}^{a} f (x) g (x) d x

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Solution

The correct option is B $3 \int_{0}^{a} f (x) g (x) d x$
$I = \int_{0}^{2 a} f (x) g (x) h (x) d x = \int_{0}^{2 a} f (2 a - x) g (2 a - x) h (2 a - x) d x = \int_{0}^{2 a} f (x) g (x) [3 - h (x)] d x I = \frac{3}{2} \int_{0}^{2 a} f (x) g (x) d x = 3 \int_{0}^{a} f (x) g (x) d x$

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Let f(x), g(x), h(x) be continous in [0, 2a] and satisfies f(2a−x)=f(x),g(2a−x)=g(x),h(x)+h(2a−x)=3,f(2a−x)g(2a−x)=f(x)g(x)then∫2a0f(x)g(x)h(x)dx=

The correct option is B 3∫a0f(x)g(x)dxI=∫2a0f(x)g(x)h(x)dx=∫2a0f(2a−x)g(2a−x)h(2a−x)dx=∫2a0f(x)g(x)[3−h(x)]dxI=32∫2a0f(x)g(x)dx=3∫a0f(x)g(x)dx

Let f(x), g(x), h(x) be continous in [0, 2a] and satisfies $f (2 a - x) = f (x), g (2 a - x) = g (x), h (x) + h (2 a - x) = 3, f (2 a - x) g (2 a - x) = f (x) g (x) t h e n \int_{0}^{2 a} f (x) g (x) h (x) d x =$

The correct option is B $3 \int_{0}^{a} f (x) g (x) d x$
$I = \int_{0}^{2 a} f (x) g (x) h (x) d x = \int_{0}^{2 a} f (2 a - x) g (2 a - x) h (2 a - x) d x = \int_{0}^{2 a} f (x) g (x) [3 - h (x)] d x I = \frac{3}{2} \int_{0}^{2 a} f (x) g (x) d x = 3 \int_{0}^{a} f (x) g (x) d x$