Let `f:[-1,2]vec[0,oo)`be a continuous function such that `f(x)=f(1-x)fora l lx in [-1,2]dot`Let `R_1=int_(-1)^2xf(x)dx ,`and `R_2`be the area of the region bounded by `y=f(x),x=-1,x=2,`and the `x-a xi s`. Then`R_1=2R_2`(b) `R_1=3R_2``2R_1`(d) `3R_1=R_2`A. `R_(1)=2R_(2)`B. `R_(1)=3R_(2)`C. `2R_(1)=R_(2)`D. `3R_(1)=R_(2)`

Let `f:[-1,2]vec[0,oo)`be a continuous function such that `f(x)=f(1-x)

1.	Let `f:[-1,2]vec[0,oo)`be a continuous function such that `f(x)=f(1-x)fora l lx in [-1,2]dot`Let `R_1=int_(-1)^2xf(x)dx ,`and `R_2`be the area of the region bounded by `y=f(x),x=-1,x=2,`and the `x-a xi s`. Then`R_1=2R_2`(b) `R_1=3R_2``2R_1`(d) `3R_1=R_2`A. `R_(1)=2R_(2)`B. `R_(1)=3R_(2)`C. `2R_(1)=R_(2)`D. `3R_(1)=R_(2)`
Answer» Correct Answer - C `R_(1)=int_(-1)^(2)xf(x)dx " "...(i)` Using ` int_(a)^(b)f(x)dx =int_(a)^(b)f(a+b-x)dx ` ` R_(1)=int_(-1)^(2)(1-x)f(1-x)dx ` ` therefore R_(1)=int_(-1)^(2)(1-x)f(x)dx " "...(ii) ` `[f(x)=f(1-x), " given "]` Given, ` R_(2) ` is area bounded by ` f(x),x=-1 " and " x=2 .` ` therefore R_(2)=int_(-1)^(2)f(x)dx " "...(iii)` On adding Eqs.(i) and (ii), we get `2R_(1)=int_(-1)^(2)f(x)dx " "...(iv)` From Eqs.(iii) and (iv), we get ` 2R_(1)=R_(2) `