JEE MAIN - Mathematics (2019 - 12th January Morning Slot - No. 3)
Let f and g be continuous functions on [0, a] such that f(x) = f(a – x) and g(x) + g(a – x) = 4, then $$\int\limits_0^a \, $$f(x) g(x) dx is equal to :
4$$\int\limits_0^a \, $$f(x)dx
$$-$$ 3$$\int\limits_0^a \, $$f(x)dx
$$\int\limits_0^a \, $$f(x)dx
2$$\int\limits_0^a \, $$f(x)dx
Explanation
$${\rm I} = \int_0^a {f\left( x \right)g\left( x \right)dx} $$
$${\rm I} = \int_0^a {f\left( {a - x} \right)g\left( {a - x} \right)dx} $$
$${\rm I} = \int_0^a {f\left( x \right)\left( {4 - g\left( x \right.} \right)dx} $$
$${\rm I} = 4\int_0^a {f\left( x \right)dx - {\rm I}} $$
$$ \Rightarrow {\rm I} = 2\int_0^a {f\left( x \right)dx} $$
$${\rm I} = \int_0^a {f\left( {a - x} \right)g\left( {a - x} \right)dx} $$
$${\rm I} = \int_0^a {f\left( x \right)\left( {4 - g\left( x \right.} \right)dx} $$
$${\rm I} = 4\int_0^a {f\left( x \right)dx - {\rm I}} $$
$$ \Rightarrow {\rm I} = 2\int_0^a {f\left( x \right)dx} $$
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