If f and g are defined as f(x) = f (a -x) and g(x) + g (a -x) = 4, then ∫0a f(x)g(x)dx is equal to

If f and g are defined as f(x) = f (a -x) and g(x) + g (a -x) = 4, then ${\int }_{0}^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}$ is equal to

1. A

${\int }_{0}^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$

2. B

$2{\int }_{0}^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$

3. C

$2{\int }_{0}^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}$

4. D

None of these

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Solution:

Let,

$\begin{array}{l}\mathrm{I}={\int }_{0}^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}----\left(\mathrm{i}\right)\\ \mathrm{I}={\int }_{0}^{\mathrm{a}} \mathrm{f}\left(\mathrm{a}-\mathrm{x}\right)\mathrm{g}\left(\mathrm{a}-\mathrm{x}\right)\mathrm{dx}\\ \left[\because {\int }_{0}^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_{0}^{\mathrm{a}} \mathrm{f}\left(\mathrm{a}-\mathrm{x}\right)\mathrm{dx}\right]\end{array}$

On adding Eqs. (i) and (ir), we get

$2\mathrm{I}={\int }_{0}^{\mathrm{a}} 4\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}⇒\mathrm{I}=2{\int }_{0}^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$

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