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JEE Advanced 2011Paper 2
Exam
JEE Advanced
Year
2011
Paper
Paper 2
Subject
Mathematics
Chapter
Definite Integration

JEE Advanced 2011 · Paper 2 · Mathematics · Definite Integration

Let f:[-1,2] arrow[0, ∞) be a continuous function such that...

Let f:[−1,2]→[0,∞)f:[-1,2] \rightarrow[0, \infty) be a continuous function such that f(x)=f(1−x)f(x)=f(1-x) for all x∈[−1,2]x \in[-1,2]. Let R1=∫−12xf(x)dxR_1=\int_{-1}^2 x f(x) d x and R2R_2 be the area of the region bounded by y=f(x),x=−1,x=2y=f(x), x=-1, x=2 and the XX-axis. Then,

Options

  1. AR1=2R2R_1=2 R_2
  2. BR1=3R2R_1=3 R_2
  3. C2R1=R22 R_1=R_2
  4. D3R1=R23 R_1=R_2

Correct answer

C. 2R1=R22 R_1=R_2

Step-by-step solution

Here, R1=∫−12xf(x)dxR_1=\int_{-1}^2 x f(x) d x Using, ∫abf(x)dx=∫abf(a+b−x)dx\quad \int_a^b f(x) d x=\int_a^b f(a+b-x) d xR1=∫−12(1−x)f(1−x)dx,∴R1=∫−12(1−x)f(x)dx…(1−x)]\begin{aligned} & R_1=\int_{-1}^2(1-x) f(1-x) d x, \\ \therefore \quad & \left.R_1=\int_{-1}^2(1-x) f(x) d x \quad \ldots(1-x)\right] \end{aligned}Given,Given,R_2isareaboundedbyis area bounded byf(x),x−1 and  x=2∴R2=∫−12f(x)dx\begin{aligned} & f(x), x-1 \text{\text{ and } } x=2 \\ \therefore & R_2=\int_{-1}^2 f(x) d x \end{aligned}AddingEqs.(i)and(ii),weget2R1=∫−12f(x)dxAdding Eqs. (i) and (ii), we get 2 R_1=\int_{-1}^2 f(x) d x \thereforeFromEqs.(iii)and(iv),weget2R1=R2From Eqs. (iii) and (iv), we get 2 R_1=R_2
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