The area of the region \(x,y): |x-y| le y le 4sqrtx\$\{(x,y): |x-y| \le y \le 4\sqrt{x}\}$ is
A.512$512$
B.frac10243$\frac{1024}{3}$
C.frac5123$\frac{512}{3}$
D.frac20483$\frac{2048}{3}$
Solution & Explanation
### Related Formula
The area between two continuous curves y_1(x)$y_1(x)$ and y_2(x)$y_2(x)$ from x=a$x=a$ to x=b$x=b$ is:
textArea = int_a^b (y_upper - y_lower) \, dx$$\text{Area} = \int_a^b (y_{upper} - y_{lower}) \, dx$$
### Core Logic
Let's crack the inequalities:
1. |x - y| le y implies -y le x - y le y$|x - y| \le y \implies -y \le x - y \le y$
- From x - y le y implies y ge fracx2$x - y \le y \implies y \ge \frac{x}{2}$
- From -y le x - y
implies x ge 0$-y \le x - y
\implies x \ge 0$
2. y le 4sqrtx implies y^2 le 16x$y \le 4\sqrt{x} \implies y^2 \le 16x$
Thus, the region is bounded between y = 4sqrtx$y = 4\sqrt{x}$ and y = fracx2$y = \frac{x}{2}$.
### Step 1: Finding points of intersection
Equating the two curves:
4sqrtx = fracx2 implies 16x = fracx^24 implies x^2 - 64x = 0 implies x = 0 quad textand quad x = 64$$4\sqrt{x} = \frac{x}{2} \implies 16x = \frac{x^2}{4} \implies x^2 - 64x = 0 \implies x = 0 \quad \text{and} \quad x = 64$$
Corresponding y$y$-values: y = 0$y = 0$ to y = 32$y = 32$.
Area Under Curve diagram for Q61 - JEE Main 2025 Evening Shift
### Step 2: Area Integration
Integrating from x = 0$x = 0$ to x = 64$x = 64$:
textArea = int_0^64 left( 4sqrtx - fracx2 right) \, dx$$\text{Area} = \int_0^{64} \left( 4\sqrt{x} - \frac{x}{2} \right) \, dx$$textArea = left[ frac83 x^3/2 - fracx^24 right]_0^64$$\text{Area} = \left[ \frac{8}{3} x^{3/2} - \frac{x^2}{4} \right]_0^{64}$$textArea = frac83 (64)^3/2 - frac64^24$$\text{Area} = \frac{8}{3} (64)^{3/2} - \frac{64^2}{4}$$
Since 64^3/2 = 512$64^{3/2} = 512$:
textArea = frac8 times 5123 - 1024 = frac40963 - 1024 = frac4096 - 30723 = frac10243$$\text{Area} = \frac{8 \times 512}{3} - 1024 = \frac{4096}{3} - 1024 = \frac{4096 - 3072}{3} = \frac{1024}{3}$$
### Pattern Recognition
To verify area of parabola and line intersecting at origin: textArea = frac8a^23m^3$\text{Area} = \frac{8a^2}{3m^3}$ or similar formulas. Simply computing the standard integral int (ksqrtx - mx)dx$\int (k\sqrt{x} - mx)dx$ is extremely clean when coordinates are powers of 2.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Applications of Integrals
Class 11 Mathematics: Conic Sections
Keywords:#area bounded by parabola and line#JEE Main 2025 Evening Q61#Applications of Integrals JEE Main 2025#definite integral area under curve
More Applications of Integrals Previous-Year Questions — Page 4
Q28jee_main_2024_29_jan_morningArea Under Curves
The area (in sq. units) of the part of circle x^2+y^2=169$x^2+y^2=169$ which is below the line 5x-y=13$5x-y=13$ is fracpialpha2beta-frac652+fracalphabetasin^-1(frac1213)$\frac{\pi\alpha}{2\beta}-\frac{65}{2}+\frac{\alpha}{\beta}\sin^{-1}(\frac{12}{13})$ where alpha,beta$\alpha,\beta$ are coprime numbers. Then alpha+beta$\alpha+\beta$ is equal to
Numerical Answer.Answer: 171 to 171
Solution
### Related Formula
textStandard Integral: int sqrta^2-y^2 dy = fracy2sqrta^2-y^2 + fraca^22sin^-1left(fracyaright) + C$$\text{Standard Integral: } \int \sqrt{a^2-y^2} dy = \frac{y}{2}\sqrt{a^2-y^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{y}{a}\right) + C$$textArea of right triangle = frac12 times textbase times textheight$$\text{Area of right triangle } = \frac{1}{2} \times \text{base} \times \text{height}$$
### Core Logic
First, find the points of intersection between the circle x^2+y^2=169$x^2+y^2=169$ and the line 5x-y=13 Rightarrow y = 5x-13$5x-y=13 \Rightarrow y = 5x-13$.
Substitute y$y$ into the circle equation:
x^2 + (5x-13)^2 = 169$$x^2 + (5x-13)^2 = 169$$x^2 + 25x^2 - 130x + 169 = 169$$x^2 + 25x^2 - 130x + 169 = 169$$26x^2 - 130x = 0 Rightarrow 26x(x - 5) = 0$$26x^2 - 130x = 0 \Rightarrow 26x(x - 5) = 0$$
The solutions are x=0$x=0$ and x=5$x=5$.
When x=0, y=-13$x=0, y=-13$. Point is (0, -13)$(0, -13)$.
When x=5, y=12$x=5, y=12$. Point is (5, 12)$(5, 12)$.
The required area is bounded below the line x = fracy+135$x = \frac{y+13}{5}$ and above the right-hand boundary of the circle x = sqrt169-y^2$x = \sqrt{169-y^2}$ across the y-axis boundaries [-13, 12]$[-13, 12]$.
Area Under Curves
### Step 1: Setup Area Integral
Integrate with respect to y$y$ (from left to right curves, bounded horizontally):
Area = int_-13^12 left( sqrt169-y^2 - fracy+135 right) dy$$Area = \int_{-13}^{12} \left( \sqrt{169-y^2} - \frac{y+13}{5} \right) dy$$
Split the integral into two parts:
Part A (Circle): int_-13^12 sqrt169-y^2 dy$\int_{-13}^{12} \sqrt{169-y^2} dy$
Part B (Line): int_-13^12 fracy+135 dy$\int_{-13}^{12} \frac{y+13}{5} dy$
### Step 2: Evaluate Integrals
Part A (Circle Integral):
= left[ fracy2sqrt169-y^2 + frac1692sin^-1left(fracy13right) right]_-13^12$$= \left[ \frac{y}{2}\sqrt{169-y^2} + \frac{169}{2}\sin^{-1}\left(\frac{y}{13}\right) \right]_{-13}^{12}$$
Evaluate at upper limit 12:
= frac122sqrt169-144 + frac1692sin^-1left(frac1213right) = 6(5) + frac1692sin^-1left(frac1213right) = 30 + frac1692sin^-1left(frac1213right)$$= \frac{12}{2}\sqrt{169-144} + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}\right) = 6(5) + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}\right) = 30 + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}\right)$$
Evaluate at lower limit -13:
= 0 + frac1692sin^-1(-1) = -frac169pi4$$= 0 + \frac{169}{2}\sin^{-1}(-1) = -\frac{169\pi}{4}$$
Value of Part A = 30 + frac169pi4 + frac1692sin^-1left(frac1213right)$30 + \frac{169\pi}{4} + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}\right)$
Part B (Line Integral - matches the area of the bounded triangle geometric region):
= frac110 left[ (y+13)^2 right]_-13^12$$= \frac{1}{10} \left[ (y+13)^2 \right]_{-13}^{12}$$= frac110(12+13)^2 - 0 = frac25^210 = frac62510 = frac1252 = 62.5$$= \frac{1}{10}(12+13)^2 - 0 = \frac{25^2}{10} = \frac{625}{10} = \frac{125}{2} = 62.5$$
### Step 3: Map to Requested Format
Subtract Part B from Part A:
Area = frac169pi4 + 30 - frac1252 + frac1692sin^-1left(frac1213right)$$Area = \frac{169\pi}{4} + 30 - \frac{125}{2} + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}\right)$$Area = frac169pi4 - frac652 + frac1692sin^-1left(frac1213right)$$Area = \frac{169\pi}{4} - \frac{65}{2} + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}\right)$$
Comparing this exactly with the given format fracpialpha2beta - frac652 + fracalphabetasin^-1(frac1213)$\frac{\pi\alpha}{2\beta} - \frac{65}{2} + \frac{\alpha}{\beta}\sin^{-1}(\frac{12}{13})$:
We see that fracalphabeta = frac1692$\frac{\alpha}{\beta} = \frac{169}{2}$.
Since 169 and 2 are coprime, alpha = 169$\alpha = 169$ and beta = 2$\beta = 2$.
Calculate alpha + beta$\alpha + \beta$:
169 + 2 = 171$169 + 2 = 171$
### Pattern Recognition
When evaluating line integrals forming a triangle with horizontal bounds, bypass algebraic integration and visually calculate frac12 cdot b cdot h$\frac{1}{2} \cdot b \cdot h$. Here, base=25 along y-axis, height=5 along x-axis, area = 125/2$125/2$. Instantly saves integration time.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Application of Integrals
Class 11 Mathematics: Straight Lines
Q8jee_main_2024_30_jan_morningArea under Curves
The area (in square units) of the region bounded by the parabolay^2 = 4(x - 2)$y^2 = 4(x - 2)$ and the line y = 2x - 8$y = 2x - 8$
A.8$8$
B.9$9$
C.6$6$
D.7$7$
Solution
### Related Formula
textArea = int_y_1^y_2 (x_R - x_L) dy$$\text{Area} = \int_{y_1}^{y_2} (x_R - x_L) dy$$
### Core Logic
Area under Curves diagram for Q8 - JEE Main 2024 Morning
To simplify calculations, shift the origin. Let X = x - 2$X = x - 2$.
The equations become:
Parabola: y^2 = 4X Rightarrow X = fracy^24$y^2 = 4X \Rightarrow X = \frac{y^2}{4}$
Line: y = 2(X + 2) - 8 Rightarrow y = 2X - 4 Rightarrow X = fracy + 42$y = 2(X + 2) - 8 \Rightarrow y = 2X - 4 \Rightarrow X = \frac{y + 4}{2}$
### Step 1: Finding points of intersection
Set the X$X$ values equal to find intersection points in terms of y$y$:
fracy^24 = fracy + 42$$\frac{y^2}{4} = \frac{y + 4}{2}$$y^2 = 2y + 8$y^2 = 2y + 8$y^2 - 2y - 8 = 0$$y^2 - 2y - 8 = 0$$(y - 4)(y + 2) = 0$$(y - 4)(y + 2) = 0$$
The intersection points are at y = -2$y = -2$ and y = 4$y = 4$.
### Step 2: Area Integration
Integrate with respect to y$y$ from -2$-2$ to 4$4$:
A = int_-2^4 left( x_R - x_L right) dy$$A = \int_{-2}^{4} \left( x_R - x_L \right) dy$$A = int_-2^4 left( fracy + 42 - fracy^24 right) dy$$A = \int_{-2}^{4} \left( \frac{y + 4}{2} - \frac{y^2}{4} \right) dy$$A = left[ fracy^24 + 2y - fracy^312 right]_-2^4$$A = \left[ \frac{y^2}{4} + 2y - \frac{y^3}{12} \right]_{-2}^{4}$$
Upper limit (y=4)$(y=4)$:
frac164 + 8 - frac6412 = 4 + 8 - frac163 = 12 - frac163 = frac203$\frac{16}{4} + 8 - \frac{64}{12} = 4 + 8 - \frac{16}{3} = 12 - \frac{16}{3} = \frac{20}{3}$
Lower limit (y=-2)$(y=-2)$:
frac44 - 4 - frac-812 = 1 - 4 + frac23 = -3 + frac23 = -frac73$\frac{4}{4} - 4 - \frac{-8}{12} = 1 - 4 + \frac{2}{3} = -3 + \frac{2}{3} = -\frac{7}{3}$A = frac203 - left(-frac73right) = frac273 = 9$$A = \frac{20}{3} - \left(-\frac{7}{3}\right) = \frac{27}{3} = 9$$
The solution simplifies it directly to 9$9$ square units.
### Pattern Recognition
For a horizontal parabola interacting with a line, integrating along the y-axis is always cleaner than splitting it into multiple integrals along the x-axis.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Application of Integrals
Q11jee_main_2024_31_jan_eveningArea Under Curves
The area of the region enclosed by the parabola y = 4x - x^2$y = 4x - x^2$ and 3y = (x - 4)^2$3y = (x - 4)^2$ is equal to
Q5jee_main_2024_31_jan_morningArea bounded by Parabolas and Inequalities
The area of the region left\(x,y): y^2 le 4x, x < 4, fracxy(x - 1)(x - 2)(x - 3)(x - 4) > 0, x neq 3right\$\left\{(x,y): y^2 \le 4x, x < 4, \frac{xy(x - 1)(x - 2)}{(x - 3)(x - 4)} > 0, x \neq 3\right\}$ is
A.frac163$$\frac{16}{3}$$
B.frac643$$\frac{64}{3}$$
C.frac83$$\frac{8}{3}$$
D.frac323$$\frac{32}{3}$$
Solution
### Core Logic
Given y^2 le 4x$y^2 \le 4x$ and x < 4$x < 4$.
Analyze the inequality fracxy(x-1)(x-2)(x-3)(x-4) > 0$\frac{xy(x-1)(x-2)}{(x-3)(x-4)} > 0$ considering y > 0$y > 0$ and y < 0$y < 0$ separately.
Area bounded by Parabolas and Inequalities diagram for Q5 - JEE Main 2024 Morning
### Step 1: Case I (y > 0)
If y > 0$y > 0$, the inequality reduces to fracx(x-1)(x-2)(x-3)(x-4) > 0$\frac{x(x-1)(x-2)}{(x-3)(x-4)} > 0$.
Using wavy curve method and given x in (0, 4)$x \in (0, 4)$:
x in (0, 1) cup (2, 3)$x \in (0, 1) \cup (2, 3)$.
### Step 2: Case II (y < 0)
If y < 0$y < 0$, the inequality reduces to fracx(x-1)(x-2)(x-3)(x-4) < 0$\frac{x(x-1)(x-2)}{(x-3)(x-4)} < 0$.
Using wavy curve method and given x in (0, 4)$x \in (0, 4)$:
x in (1, 2) cup (3, 4)$x \in (1, 2) \cup (3, 4)$.
### Step 3: Area Computation
Because the regions map perfectly without overlap in opposite quadrants relative to the x-axis, they form complete parabolic strips when combined:
Area = 2 int_0^4 sqrtx dx = 2 cdot frac23[x^3/2]_0^4 = frac43 cdot 8 = frac323$= 2 \int_{0}^{4} \sqrt{x} dx = 2 \cdot \frac{2}{3}[x^{3/2}]_{0}^{4} = \frac{4}{3} \cdot 8 = \frac{32}{3}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Area Under Curves
Class 11 Maths: Linear Inequalities
We Map Every Repeating Question in Competitive Exams.
Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.
Select Your Target Exam
Choose an exam track below to find formulas per chapter and patterns.
Syncing Exam Intelligence
Mapping formulas and patterns across all tracks…
PATH A — FULL LENGTH PRACTICE
Full Mock Test Hub
Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
