IIT-JEE 2011 · Paper 2 · Mathematics Q46

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MathematicsSingle correct+3 / -1Question 46 of 60

Let $f:[-1,2]\to[0,\infty)$ be a continuous function such that $f(x)=f(1-x)$ for all $x\in[-1,2]$ . Let $R_1=\displaystyle\int_{-1}^2xf(x)\,dx$ , and $R_2$ be the area of the region bounded by $y=f(x)$ , $x=-1$ , $x=2$ , and the $x$ -axis. Then

R_1=2R_2

R_1=3R_2

2R_1=R_2

3R_1=R_2

Answer and worked solution

Official answer: C

Use the symmetry $x\to1-x$ , which maps $[-1,2]$ to itself:
$R_1=\int_{-1}^2xf(x)\,dx=\int_{-1}^2(1-x)f(1-x)\,dx=\int_{-1}^2(1-x)f(x)\,dx.$

Add the two forms. $2R_1=\displaystyle\int_{-1}^2f(x)\,dx=R_2$ (since $f\ge0$ ).

(C)

Topics:Properties of definite integrals Area under a curve

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IIT-JEE 2011 · Paper 2, Mathematics question 46