${{\rm{R}}_2} = \int\limits_{ - 2}^3 {xf(x)dx} $
$ = \int\limits_{ - 2}^3 {(1 - x)f(1 - x)dx} $
$\left[ {{\rm{Using}}\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)dx} } \right]$
$ \Rightarrow {{\rm{R}}_2} = \int\limits_{ - 2}^3 {(1 - x)f\left( x \right)dx} $
$(\because f(x)=f(1-x) \text { on }[-2,3])$
$\therefore {{\rm{R}}_2} + {R^2} = \int\limits_{ - 2}^3 {xf\left( x \right)dx + } \int\limits_{ - 2}^3 {(1 - x)f\left( x \right)dx} $
$ = \int\limits_{ - 2}^3 {f\left( x \right)dx = {R_1}} $
$\Rightarrow 2 \mathrm{R}_{2}=\mathrm{R}_{1}$
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$x+3 y+k^{2} z=0$
$3 x+y+3 z=0$
has a non-zero solution $(x, y, z)$ for some $k \in R ,$ then $x +\left(\frac{ y }{ z }\right)$ is equal to