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Area Under the Curves question

2015 · Shift 0 · Q29
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Area Under the Curves question

2015 · Shift 0 · Q29

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
The area (in sq. units) of the region described by {(x,y):y2≤2x\left\{ {\left( {x,y} \right):{y^2} \le 2x} \right.{(x,y):y2≤2x and y≥4x−1}\left. {y \ge 4x - 1} \right\}y≥4x−1} is :
  1. A
    1564{{15} \over {64}}6415​
  2. B
    932{{9} \over {32}}329​
  3. C
    732{{7} \over {32}}327​
  4. D
    564{{5} \over {64}}645​
View written solutionFree

Correct answer: B

  1. Interpret the region

The inequalities are y2≤2xy^2 \le 2xy2≤2x and y≥4x−1.y \ge 4x-1.y≥4x−1.

From y2≤2x,y^2 \le 2x,y2≤2x, we get x≥y22,x \ge \frac{y^2}{2},x≥2y2​, which is the region to the right of the parabola y2=2x.y^2=2x.y2=2x.

From y≥4x−1,y \ge 4x-1,y≥4x−1, we get x≤y+14,x \le \frac{y+1}{4},x≤4y+1​, which is the region to the left of the line y=4x−1.y=4x-1.y=4x−1.

So for a fixed yyy, the region lies between x=y22andx=y+14.x=\frac{y^2}{2} \quad \text{and} \quad x=\frac{y+1}{4}.x=2y2​andx=4y+1​.

Hence the horizontal width is y+14−y22.\frac{y+1}{4}-\frac{y^2}{2}.4y+1​−2y2​.

  1. Find the limits of integration

The curves intersect where y2=2xandy=4x−1.y^2=2x \quad \text{and} \quad y=4x-1.y2=2xandy=4x−1.

Using x=y+14x=\frac{y+1}{4}x=4y+1​ in y2=2xy^2=2xy2=2x: y2=2⋅y+14=y+12.y^2=2\cdot \frac{y+1}{4}=\frac{y+1}{2}.y2=2⋅4y+1​=2y+1​.

So 2y2=y+12y^2=y+12y2=y+1 2y2−y−1=02y^2-y-1=02y2−y−1=0 (2y+1)(y−1)=0.(2y+1)(y-1)=0.(2y+1)(y−1)=0.

Thus, y=1ory=−12.y=1 \quad \text{or} \quad y=-\frac12.y=1ory=−21​.

Therefore the required area is A=∫−1/21(y+14−y22)dy.A=\int_{-1/2}^{1}\left(\frac{y+1}{4}-\frac{y^2}{2}\right)dy.A=∫−1/21​(4y+1​−2y2​)dy.

  1. Evaluate the integral

A=∫−1/21(y4+14−y22)dyA=\int_{-1/2}^{1}\left(\frac y4+\frac14-\frac{y^2}{2}\right)dyA=∫−1/21​(4y​+41​−2y2​)dy

Antiderivative:

=\frac{y^2}{8}+\frac y4-\frac{y^3}{6}.$$ Now substitute the limits: At $y=1$, $$\frac{1}{8}+\frac{1}{4}-\frac{1}{6} =\frac{3+6-4}{24} =\frac{5}{24}.$$ At $y=-\frac12$, $$\frac{(1/4)}{8}+\frac{-1/2}{4}-\frac{(-1/8)}{6} =\frac{1}{32}-\frac{1}{8}+\frac{1}{48}.$$ Taking LCM $96$: $$\frac{1}{32}-\frac{1}{8}+\frac{1}{48} =\frac{3-12+2}{96}=-\frac{7}{96}.$$ Therefore, $$A=\frac{5}{24}-\left(-\frac{7}{96}\right) =\frac{20}{96}+\frac{7}{96} =\frac{27}{96} =\frac{9}{32}.$$ 4. **Match with the options** $$\frac{9}{32}$$ is **Option B**. 5. **Comparison with stored answer** Stored correct answer: **B** Our derived answer is also **B**. So they agree.
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