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Let f:[-1,2]vec[0,oo) be a continuous fu...

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)`

Text Solution

Verified by Experts

The correct Answer is:
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) `
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