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Area Under the Curves question

2019 · 12 Apr · Shift 1 · Q40
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Area Under the Curves question

2019 · 12 Apr · Shift 1 · Q40

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
If the area (in sq. units) of the region {(x, y) : y2 ≤\le≤ 4x, x + y ≤\le≤ 1, x ≥\ge≥ 0, y ≥\ge≥ 0} is a 2\sqrt 22​ + b, then a – b is equal to :
  1. A
    83{8 \over 3}38​
  2. B
    −23- {2 \over 3}−32​
  3. C
    6
  4. D
    103{{10} \over 3}310​
View written solutionFree

Correct answer: C

  1. Interpret the region

The given region is

{(x,y):y2≤4x, x+y≤1, x≥0, y≥0}.\{(x,y): y^2\le 4x,\ x+y\le 1,\ x\ge 0,\ y\ge 0\}.{(x,y):y2≤4x, x+y≤1, x≥0, y≥0}.

Since x≥0x\ge 0x≥0 and y≥0y\ge 0y≥0, we are in the first quadrant.

The inequalities represent:

  • y2≤4x  ⟺  x≥y24y^2\le 4x \iff x\ge \dfrac{y^2}{4}y2≤4x⟺x≥4y2​ : region to the right of the parabola y2=4xy^2=4xy2=4x.
  • x+y≤1  ⟺  x≤1−yx+y\le 1 \iff x\le 1-yx+y≤1⟺x≤1−y : region to the left of the line x=1−yx=1-yx=1−y.

So for a fixed yyy, the admissible xxx values are

y24≤x≤1−y.\frac{y^2}{4}\le x\le 1-y.4y2​≤x≤1−y.

This is possible only when

y24≤1−y.\frac{y^2}{4}\le 1-y.4y2​≤1−y.
  1. Find the intersection of the curves

Solve

y24=1−y.\frac{y^2}{4}=1-y.4y2​=1−y.

Multiplying by 444,

y2=4−4yy^2=4-4yy2=4−4y y2+4y−4=0.y^2+4y-4=0.y2+4y−4=0.

Thus,

y=−4±16+162=−4±422=−2±22.y=\frac{-4\pm\sqrt{16+16}}{2}=\frac{-4\pm 4\sqrt2}{2}=-2\pm 2\sqrt2.y=2−4±16+16​​=2−4±42​​=−2±22​.

Since y≥0y\ge 0y≥0, the relevant value is

y0=−2+22=2(2−1).y_0=-2+2\sqrt2=2(\sqrt2-1).y0​=−2+22​=2(2​−1).
  1. Set up the area integral

Hence area is

A=∫02(2−1)[(1−y)−y24]dy.A=\int_0^{2(\sqrt2-1)}\left[(1-y)-\frac{y^2}{4}\right]dy.A=∫02(2​−1)​[(1−y)−4y2​]dy.
  1. Evaluate the integral
A=[y−y22−y312]02(2−1).A=\left[y-\frac{y^2}{2}-\frac{y^3}{12}\right]_0^{2(\sqrt2-1)}.A=[y−2y2​−12y3​]02(2​−1)​.

Let

t=2(2−1).t=2(\sqrt2-1).t=2(2​−1).

Then

A=t−t22−t312.A=t-\frac{t^2}{2}-\frac{t^3}{12}.A=t−2t2​−12t3​.

Now compute:

t=22−2.t=2\sqrt2-2.t=22​−2. t2=(22−2)2=8+4−82=12−82.t^2=(2\sqrt2-2)^2=8+4-8\sqrt2=12-8\sqrt2.t2=(22​−2)2=8+4−82​=12−82​. t3=t⋅t2=(22−2)(12−82)=402−56.t^3=t\cdot t^2=(2\sqrt2-2)(12-8\sqrt2)=40\sqrt2-56.t3=t⋅t2=(22​−2)(12−82​)=402​−56.

Substitute:

A=(22−2)−12−822−402−5612.A=(2\sqrt2-2)-\frac{12-8\sqrt2}{2}-\frac{40\sqrt2-56}{12}.A=(22​−2)−212−82​​−12402​−56​.

Simplify each term:

A=(22−2)−(6−42)−(1023−143).A=(2\sqrt2-2)-(6-4\sqrt2)-\left(\frac{10\sqrt2}{3}-\frac{14}{3}\right).A=(22​−2)−(6−42​)−(3102​​−314​). A=22−2−6+42−1023+143.A=2\sqrt2-2-6+4\sqrt2-\frac{10\sqrt2}{3}+\frac{14}{3}.A=22​−2−6+42​−3102​​+314​.

Combine rational parts:

−8+143=−103.-8+\frac{14}{3}=-\frac{10}{3}.−8+314​=−310​.

Combine 2\sqrt22​ parts:

22+42−1023=62−1023=823.2\sqrt2+4\sqrt2-\frac{10\sqrt2}{3}=6\sqrt2-\frac{10\sqrt2}{3}=\frac{8\sqrt2}{3}.22​+42​−3102​​=62​−3102​​=382​​.

Therefore,

A=823−103.A=\frac{8\sqrt2}{3}-\frac{10}{3}.A=382​​−310​.
  1. Match with the form a2+ba\sqrt2+ba2​+b

So,

A=a2+bA=a\sqrt2+bA=a2​+b

with

a=83,b=−103.a=\frac{8}{3},\qquad b=-\frac{10}{3}.a=38​,b=−310​.

Hence,

a−b=83−(−103)=183=6.a-b=\frac{8}{3}-\left(-\frac{10}{3}\right)=\frac{18}{3}=6.a−b=38​−(−310​)=318​=6.
  1. Check options
  • A: 83\frac{8}{3}38​ ❌
  • B: −23-\frac{2}{3}−32​ ❌
  • C: 666 ✅
  • D: 103\frac{10}{3}310​ ❌

Therefore the correct option is C.

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