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JEE Main & Advanced JEE Main Paper (Held On 25 April 2013)

  • question_answer
                    Let f: \[\left[ -2,3 \right]\to \left[ 8,\infty  \right]\]be a continuous function such that \[f(1-x)=f(x)\] for all \[x\in [-1,3].\]                 If \[{{\operatorname{R}}_{1}}\]is the numerical value of the area of the region bounded by \[y=f(x),x=-2\], \[x=3\]and          the axis of \[x\] and\[{{R}_{2}}=\int\limits_{-2}^{3}{xf(x)\operatorname{d}x,\operatorname{then}:}\]     JEE Main Online Paper ( Held On 25  April 2013 )

    A)                 \[3{{R}_{1}}=2{{R}_{2}}\]                             

    B)                                        \[2{{R}_{1}}=3{{R}_{2}}\]

    C)                                        \[{{R}_{1}}={{R}_{2}}\]                  

    D)                                        \[{{R}_{1}}=2{{R}_{2}}\]

    Correct Answer: D

    Solution :

                    We have \[{{R}_{2}}=\int\limits_{-2}^{3}{x}f(x)dx=\int\limits_{-2}^{3}{(1-x)f(1-x)dx}\] \[\left[ \text{Using}\,\int\limits_{a}^{b}{{}}f(x)dx=\int\limits_{a}^{b}{f(a+b-x)dx} \right]\] \[\Rightarrow \]\[{{R}_{2}}=\int\limits_{-2}^{3}{(1-x)f(x)dx}\] \[(\because f(x)=f(1-x)on[-2,3])\] \[\therefore \]\[{{R}_{2}}+{{R}_{2}}=\int\limits_{-2}^{3}{xf(x)dx}+\int\limits_{-2}^{3}{(1-x)}f(x)dx\] \[=\int\limits_{-2}^{3}{f(x)dx}={{R}_{1}}\]\[\Rightarrow \]\[2{{R}_{2}}={{R}_{1}}\]


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