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Area Under the Curves question

2024 · 27 Jan · Shift 1 · Q56
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Area Under the Curves question

2024 · 27 Jan · Shift 1 · Q56

JEE MainMathematicsArea Under the CurvesNumerical+4 / −1
Let the area of the region {(x,y):x−2y+4⩾0,x+2y2⩾0,x+4y2≤8,y⩾0}\left\{(x, y): x-2 y+4 \geqslant 0, x+2 y^2 \geqslant 0, x+4 y^2 \leq 8, y \geqslant 0\right\}{(x,y):x−2y+4⩾0,x+2y2⩾0,x+4y2≤8,y⩾0} be mn\frac{\mathrm{m}}{\mathrm{n}}nm​, where m\mathrm{m}m and n\mathrm{n}n are coprime numbers. Then m+n\mathrm{m}+\mathrm{n}m+n is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 119

We need the area of the region

{(x,y):x−2y+4≥0,  x+2y2≥0,  x+4y2≤8,  y≥0}.\{(x,y): x-2y+4\ge 0,\; x+2y^2\ge 0,\; x+4y^2\le 8,\; y\ge 0\}.{(x,y):x−2y+4≥0,x+2y2≥0,x+4y2≤8,y≥0}.

1. Rewrite each inequality

Given:

  1. x−2y+4≥0  ⟹  x≥2y−4x-2y+4\ge 0 \implies x\ge 2y-4x−2y+4≥0⟹x≥2y−4
  2. x+2y2≥0  ⟹  x≥−2y2x+2y^2\ge 0 \implies x\ge -2y^2x+2y2≥0⟹x≥−2y2
  3. x+4y2≤8  ⟹  x≤8−4y2x+4y^2\le 8 \implies x\le 8-4y^2x+4y2≤8⟹x≤8−4y2
  4. y≥0y\ge 0y≥0

So for a fixed y≥0y\ge 0y≥0, the allowed values of xxx are

max⁡(2y−4, −2y2)≤x≤8−4y2.\max(2y-4,\,-2y^2) \le x \le 8-4y^2.max(2y−4,−2y2)≤x≤8−4y2.

Hence the horizontal width is

(8−4y2)−max⁡(2y−4,−2y2).(8-4y^2)-\max(2y-4,-2y^2).(8−4y2)−max(2y−4,−2y2).

We must determine which of 2y−42y-42y−4 and −2y2-2y^2−2y2 is larger.

2. Compare 2y−42y-42y−4 and −2y2-2y^2−2y2

Solve

2y−4=−2y2.2y-4 = -2y^2.2y−4=−2y2.

This gives

2y2+2y−4=0  ⟹  y2+y−2=0  ⟹  (y+2)(y−1)=0.2y^2+2y-4=0 \implies y^2+y-2=0 \implies (y+2)(y-1)=0.2y2+2y−4=0⟹y2+y−2=0⟹(y+2)(y−1)=0.

So intersections occur at y=−2,1y=-2,1y=−2,1. Since y≥0y\ge 0y≥0, relevant point is y=1y=1y=1.

Now test intervals:

  • For 0≤y<10\le y<10≤y<1, take y=0y=0y=0:
    2y−4=−4,−2y2=0,2y-4=-4,\quad -2y^2=0,2y−4=−4,−2y2=0, so −2y2>2y−4-2y^2>2y-4−2y2>2y−4.
  • For y>1y>1y>1, take y=2y=2y=2:
    2y−4=0,−2y2=−8,2y-4=0,\quad -2y^2=-8,2y−4=0,−2y2=−8, so 2y−4>−2y22y-4>-2y^22y−4>−2y2.

Therefore,

max⁡(2y−4,−2y2)={−2y2,0≤y≤1,2y−4,y≥1.\max(2y-4,-2y^2)= \begin{cases} -2y^2, & 0\le y\le 1,\\[4pt] 2y-4, & y\ge 1. \end{cases}max(2y−4,−2y2)={−2y2,2y−4,​0≤y≤1,y≥1.​

3. Find the valid range of yyy

We also need left endpoint ≤\le≤ right endpoint.

For 0≤y≤10\le y\le 10≤y≤1

Left boundary is x=−2y2x=-2y^2x=−2y2, right boundary is x=8−4y2x=8-4y^2x=8−4y2. Then width is

(8−4y2)−(−2y2)=8−2y2>0(8-4y^2)-(-2y^2)=8-2y^2>0(8−4y2)−(−2y2)=8−2y2>0

for all 0≤y≤10\le y\le 10≤y≤1.

For y≥1y\ge 1y≥1

Left boundary is x=2y−4x=2y-4x=2y−4, right boundary is x=8−4y2x=8-4y^2x=8−4y2. Require

2y−4≤8−4y2.2y-4 \le 8-4y^2.2y−4≤8−4y2.

So

4y2+2y−12≤0  ⟹  2y2+y−6≤0  ⟹  (2y−3)(y+2)≤0.4y^2+2y-12\le 0 \implies 2y^2+y-6\le 0 \implies (2y-3)(y+2)\le 0.4y2+2y−12≤0⟹2y2+y−6≤0⟹(2y−3)(y+2)≤0.

Since y≥1y\ge 1y≥1, this gives

1≤y≤32.1\le y\le \frac32.1≤y≤23​.

Hence total range is

0≤y≤32.0\le y\le \frac32.0≤y≤23​.

4. Set up the area integral

Thus area

A=∫01[(8−4y2)−(−2y2)]dy+∫13/2[(8−4y2)−(2y−4)]dy.A=\int_0^1\Big[(8-4y^2)-(-2y^2)\Big]dy+\int_1^{3/2}\Big[(8-4y^2)-(2y-4)\Big]dy.A=∫01​[(8−4y2)−(−2y2)]dy+∫13/2​[(8−4y2)−(2y−4)]dy.

So

A=∫01(8−2y2) dy+∫13/2(12−2y−4y2) dy.A=\int_0^1 (8-2y^2)\,dy+\int_1^{3/2}(12-2y-4y^2)\,dy.A=∫01​(8−2y2)dy+∫13/2​(12−2y−4y2)dy.

5. Evaluate the first integral

∫01(8−2y2)dy=[8y−23y3]01=8−23=223.\int_0^1 (8-2y^2)dy =\left[8y-\frac{2}{3}y^3\right]_0^1 =8-\frac23 =\frac{22}{3}.∫01​(8−2y2)dy=[8y−32​y3]01​=8−32​=322​.

6. Evaluate the second integral

∫(12−2y−4y2)dy=12y−y2−43y3.\int (12-2y-4y^2)dy=12y-y^2-\frac{4}{3}y^3.∫(12−2y−4y2)dy=12y−y2−34​y3.

Therefore,

∫13/2(12−2y−4y2)dy=[12y−y2−43y3]13/2.\int_1^{3/2}(12-2y-4y^2)dy =\left[12y-y^2-\frac{4}{3}y^3\right]_1^{3/2}.∫13/2​(12−2y−4y2)dy=[12y−y2−34​y3]13/2​.

At y=32y=\frac32y=23​:

12⋅32−(32)2−43(32)3=18−94−43⋅278=18−94−92=454.12\cdot \frac32-\left(\frac32\right)^2-\frac{4}{3}\left(\frac32\right)^3 =18-\frac94-\frac{4}{3}\cdot \frac{27}{8} =18-\frac94-\frac92 =\frac{45}{4}.12⋅23​−(23​)2−34​(23​)3=18−49​−34​⋅827​=18−49​−29​=445​.

At y=1y=1y=1:

12−1−43=11−43=293.12-1-\frac43=11-\frac43=\frac{29}{3}.12−1−34​=11−34​=329​.

So the integral is

454−293=135−11612=1912.\frac{45}{4}-\frac{29}{3} =\frac{135-116}{12} =\frac{19}{12}.445​−329​=12135−116​=1219​.

7. Total area

A=223+1912=88+1912=10712.A=\frac{22}{3}+\frac{19}{12} =\frac{88+19}{12} =\frac{107}{12}.A=322​+1219​=1288+19​=12107​.

Thus

mn=10712\frac{m}{n}=\frac{107}{12}nm​=12107​

with m=107m=107m=107, n=12n=12n=12 (coprime). So

m+n=107+12=119.m+n=107+12=119.m+n=107+12=119.

8. Comparison with stored answer

Stored correct answer = 119119119.

Our derived answer also equals 119119119, so they agree.

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