JEE Advance - Mathematics (2011 - Paper 2 Offline - No. 8)
Let f $$:$$$$\left[ { - 1,2} \right] \to \left[ {0,\infty } \right]$$ be a continuous function such that
$$f\left( x \right) = f\left( {1 - x} \right)$$ for all $$x \in \left[ { - 1,2} \right]$$
$$f\left( x \right) = f\left( {1 - x} \right)$$ for all $$x \in \left[ { - 1,2} \right]$$
Let $${R_1} = \int\limits_{ - 1}^2 {xf\left( x \right)dx,} $$ and $${R_2}$$ be the area of the region bounded by $$y=f(x),$$ $$x=-1,$$ $$x=2,$$ and the $$x$$-axis. Then
$${R_1} = 2{R_2}$$
$${R_1} = 3{R_2}$$
$${2R_1} = {R_2}$$
$${3R_1} = {R_2}$$
Explanation
$${R_1} = \int\limits_{ - 1}^2 {xf(x)dx = \int\limits_{ - 1}^2 {(2 - 1 - x)f(2 - 1 - x)dx} } $$
$$ = \int\limits_{ - 1}^2 {(1 - x)f(1 - x)dx = \int\limits_{ - 1}^2 {(1 - x)f(x)dx} } $$
Hence, $$2{R_1} = \int\limits_{ - 1}^2 {f(x)dx = {R_2}} $$.
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