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Application of Integration question

2011 · Shift 2 · Q28
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  5. /2011 · Shift 2 · Q28

Application of Integration question

2011 · Shift 2 · Q28

JEE AdvancedMathematicsApplication of IntegrationMCQ+4 / −1
Let f ::: [−1,2]→[0,∞]\left[ { - 1,2} \right] \to \left[ {0,\infty } \right][−1,2]→[0,∞] be a continuous function such that f(x)=f(1−x)f\left( x \right) = f\left( {1 - x} \right)f(x)=f(1−x) for all x∈[−1,2]x \in \left[ { - 1,2} \right]x∈[−1,2] Let R1=∫−12xf(x)dx,{R_1} = \int\limits_{ - 1}^2 {xf\left( x \right)dx,}R1​=−1∫2​xf(x)dx, and R2{R_2}R2​ be the area of the region bounded by y=f(x),x=−1,x=2,y=f(x),x=-1,x=2,y=f(x),x=−1,x=2, and the xxx-axis. Then
  1. A
    R1=2R2{R_1} = 2{R_2}R1​=2R2​
  2. B
    R1=3R2{R_1} = 3{R_2}R1​=3R2​
  3. C
    2R1=R2{2R_1} = {R_2}2R1​=R2​
  4. D
    3R1=R2{3R_1} = {R_2}3R1​=R2​
View written solutionFree

Correct answer: C

  1. Given symmetry of the function

We are given f(x)=f(1−x)for all x∈[−1,2].f(x)=f(1-x) \quad \text{for all } x\in[-1,2].f(x)=f(1−x)for all x∈[−1,2].

This means fff is symmetric about the vertical line x=12.x=\frac12.x=21​.

  1. Interpret R2R_2R2​

Since f:[−1,2]→[0,∞)f:[-1,2]\to[0,\infty)f:[−1,2]→[0,∞), the graph lies above the xxx-axis. Hence the area bounded by y=f(x)y=f(x)y=f(x), x=−1x=-1x=−1, x=2x=2x=2, and the xxx-axis is simply R2=∫−12f(x) dx.R_2=\int_{-1}^{2} f(x)\,dx.R2​=∫−12​f(x)dx.

  1. Use the symmetry to evaluate R1R_1R1​

We have R1=∫−12xf(x) dx.R_1=\int_{-1}^{2} x f(x)\,dx.R1​=∫−12​xf(x)dx.

Now apply the substitution x=1−t⇒dx=−dt.x=1-t \quad \Rightarrow \quad dx=-dt.x=1−t⇒dx=−dt.

When x=−1x=-1x=−1, t=2t=2t=2; when x=2x=2x=2, t=−1t=-1t=−1. So

=\int_{-1}^{2} (1-t)f(1-t)\,dt.$$ Using $f(1-t)=f(t)$, $$R_1=\int_{-1}^{2} (1-t)f(t)\,dt.$$ Rename $t$ as $x$: $$R_1=\int_{-1}^{2} (1-x)f(x)\,dx.$$ 4. **Add the two expressions for $R_1$** We now have $$R_1=\int_{-1}^{2} x f(x)\,dx$$ and $$R_1=\int_{-1}^{2} (1-x)f(x)\,dx.$$ Adding, $$2R_1=\int_{-1}^{2} \bigl(x+(1-x)\bigr)f(x)\,dx =\int_{-1}^{2} f(x)\,dx.$$ But from step 2, $$\int_{-1}^{2} f(x)\,dx=R_2.$$ Therefore, $$2R_1=R_2.$$ 5. **Check options** - A: $R_1=2R_2$ — false - B: $R_1=3R_2$ — false - C: $2R_1=R_2$ — true - D: $3R_1=R_2$ — false Hence the correct option is **C**.
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