Q24 / 60 · JEE Advanced 2011 (Paper 2) · Mathematics

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

  • A
    $R_1=2 R_2$
  • B
    $R_1=3 R_2$
  • Correct answer
    C
    $2 R_1=R_2$
  • D
    $3 R_1=R_2$

Solution

Here, $R_1=\int_{-1}^2 x f(x) d x$ Using, $\quad \int_a^b f(x) d x=\int_a^b f(a+b-x) d x$ $$ \begin{aligned} & R_1=\int_{-1}^2(1-x) f(1-x) d x, \\ \therefore \quad & \left.R_1=\int_{-1}^2(1-x) f(x) d x \quad \ldots(1-x)\right] \end{aligned} $$ Given, $R_2$ is area bounded by $$ \begin{aligned} & f(x), x-1 \text { and } x=2 \\ \therefore & R_2=\int_{-1}^2 f(x) d x \end{aligned} $$ Adding Eqs. (i) and (ii), we get $$ 2 R_1=\int_{-1}^2 f(x) d x $$ $\therefore$ From Eqs. (iii) and (iv), we get $$ 2 R_1=R_2 $$

Hence, the correct answer is option: 'C'.

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