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Area Under the Curves question

2025 · 7 Apr · Shift 2 · Q37
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  5. /2025 · 7 Apr · Shift 2 · Q37

Area Under the Curves question

2025 · 7 Apr · Shift 2 · Q37

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
If the area of the region {(x,y):1+x2≤y≤min⁡{x+7,11−3x}}\{(x, y) : 1 + x^2 \leq y \leq \min \{x+7, 11-3x\}\}{(x,y):1+x2≤y≤min{x+7,11−3x}} is AAA, then 3A3A3A is equal to :
  1. A
    50
  2. B
    46
  3. C
    49
  4. D
    47
View written solutionFree

Correct answer: A

  1. Understand the region

    We need the area of {(x,y):1+x2≤y≤min⁡(x+7, 11−3x)}.\{(x,y): 1+x^2 \le y \le \min(x+7,\,11-3x)\}.{(x,y):1+x2≤y≤min(x+7,11−3x)}.

    So the upper boundary is the smaller of the two lines: y=x+7,y=11−3x.y=x+7, \qquad y=11-3x.y=x+7,y=11−3x.

    First, find where these two lines intersect: x+7=11−3xx+7=11-3xx+7=11−3x 4x=44x=44x=4 x=1.x=1.x=1.

    Hence,

    • for x≤1x\le 1x≤1, we have x+7≤11−3xx+7 \le 11-3xx+7≤11−3x, so upper curve is y=x+7y=x+7y=x+7;
    • for x≥1x\ge 1x≥1, upper curve is y=11−3xy=11-3xy=11−3x.

  2. Find where the parabola lies below these lines

    The lower curve is y=1+x2.y=1+x^2.y=1+x2.

    For the region to exist, we need 1+x2≤min⁡(x+7,11−3x).1+x^2 \le \min(x+7,11-3x).1+x2≤min(x+7,11−3x).

    This must be checked piecewise.

    For x≤1x\le 1x≤1:

    Need 1+x2≤x+71+x^2 \le x+71+x2≤x+7 x2−x−6≤0x^2-x-6 \le 0x2−x−6≤0 (x−3)(x+2)≤0.(x-3)(x+2)\le 0.(x−3)(x+2)≤0. So, −2≤x≤3.-2\le x\le 3.−2≤x≤3. Together with x≤1x\le 1x≤1, this gives −2≤x≤1.-2\le x\le 1.−2≤x≤1.

    For x≥1x\ge 1x≥1:

    Need 1+x2≤11−3x1+x^2 \le 11-3x1+x2≤11−3x x2+3x−10≤0x^2+3x-10\le 0x2+3x−10≤0 (x+5)(x−2)≤0.(x+5)(x-2)\le 0.(x+5)(x−2)≤0. So, −5≤x≤2.-5\le x\le 2.−5≤x≤2. Together with x≥1x\ge 1x≥1, this gives 1≤x≤2.1\le x\le 2.1≤x≤2.

    Therefore the total region extends over [−2,1]∪[1,2].[-2,1] \cup [1,2].[−2,1]∪[1,2].

  3. Set up the area integral

    Thus, A=∫−21[(x+7)−(1+x2)]dx+∫12[(11−3x)−(1+x2)]dx.A=\int_{-2}^{1}\big[(x+7)-(1+x^2)\big]dx+\int_{1}^{2}\big[(11-3x)-(1+x^2)\big]dx.A=∫−21​[(x+7)−(1+x2)]dx+∫12​[(11−3x)−(1+x2)]dx.

    Simplify: A=∫−21(−x2+x+6) dx+∫12(10−3x−x2) dx.A=\int_{-2}^{1}(-x^2+x+6)\,dx+\int_{1}^{2}(10-3x-x^2)\,dx.A=∫−21​(−x2+x+6)dx+∫12​(10−3x−x2)dx.

  4. Evaluate the first integral

    I1=∫−21(−x2+x+6)dx.I_1=\int_{-2}^{1}(-x^2+x+6)dx.I1​=∫−21​(−x2+x+6)dx.

    Antiderivative: ∫(−x2+x+6)dx=−x33+x22+6x.\int(-x^2+x+6)dx=-\frac{x^3}{3}+\frac{x^2}{2}+6x.∫(−x2+x+6)dx=−3x3​+2x2​+6x.

    So, I1=[−x33+x22+6x]−21.I_1=\left[-\frac{x^3}{3}+\frac{x^2}{2}+6x\right]_{-2}^{1}.I1​=[−3x3​+2x2​+6x]−21​.

    At x=1x=1x=1: −13+12+6=376.-\frac13+\frac12+6=\frac{37}{6}.−31​+21​+6=637​.

    At x=−2x=-2x=−2: −(−8)3+42−12=83+2−12=83−10=−223.-\frac{(-8)}{3}+\frac{4}{2}-12=\frac{8}{3}+2-12=\frac{8}{3}-10=-\frac{22}{3}.−3(−8)​+24​−12=38​+2−12=38​−10=−322​.

    Therefore, I1=376−(−223)=376+446=816=272.I_1=\frac{37}{6}-\left(-\frac{22}{3}\right)=\frac{37}{6}+\frac{44}{6}=\frac{81}{6}=\frac{27}{2}.I1​=637​−(−322​)=637​+644​=681​=227​.

  5. Evaluate the second integral

    I2=∫12(10−3x−x2)dx.I_2=\int_{1}^{2}(10-3x-x^2)dx.I2​=∫12​(10−3x−x2)dx.

    Antiderivative: ∫(10−3x−x2)dx=10x−3x22−x33.\int(10-3x-x^2)dx=10x-\frac{3x^2}{2}-\frac{x^3}{3}.∫(10−3x−x2)dx=10x−23x2​−3x3​.

    So, I2=[10x−3x22−x33]12.I_2=\left[10x-\frac{3x^2}{2}-\frac{x^3}{3}\right]_{1}^{2}.I2​=[10x−23x2​−3x3​]12​.

    At x=2x=2x=2: 20−3⋅42−83=20−6−83=14−83=343.20-\frac{3\cdot4}{2}-\frac{8}{3}=20-6-\frac83=14-\frac83=\frac{34}{3}.20−23⋅4​−38​=20−6−38​=14−38​=334​.

    At x=1x=1x=1: 10−32−13=10−116=496.10-\frac32-\frac13=10-\frac{11}{6}=\frac{49}{6}.10−23​−31​=10−611​=649​.

    Therefore, I2=343−496=686−496=196.I_2=\frac{34}{3}-\frac{49}{6}=\frac{68}{6}-\frac{49}{6}=\frac{19}{6}.I2​=334​−649​=668​−649​=619​.

  6. Compute total area

    A=I1+I2=272+196=816+196=1006=503.A=I_1+I_2=\frac{27}{2}+\frac{19}{6}=\frac{81}{6}+\frac{19}{6}=\frac{100}{6}=\frac{50}{3}.A=I1​+I2​=227​+619​=681​+619​=6100​=350​.

    Hence, 3A=3⋅503=50.3A=3\cdot \frac{50}{3}=50.3A=3⋅350​=50.

  7. Check options

    3A=50,3A=50,3A=50, which matches Option A.

  8. Compare with stored correct answer

    Stored correct answer: A

    Our derived answer: A

    So they agree.

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