The area of the region, inside the ellipse x^2 + 4y^2 = 4$x^{2} + 4y^{2} = 4$ and outside the region bounded by the curves y = |x| - 1$y = |x| - 1$ and y = 1 - |x|$y = 1 - |x|$ , is:
A.2(pi - 1)$2(\pi - 1)$
B.2pi - frac12$2\pi - \frac{1}{2}$
C.3(pi - 1)$3(\pi - 1)$
D.2pi - 1$2\pi - 1$
Solution & Explanation
### Related Formula
Area of an ellipse fracx^2a^2 + fracy^2b^2 = 1$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by:
textArea = pi a b$$\text{Area} = \pi a b$$
Area of a rhombus bounded by |x| + |y| = a$|x| + |y| = a$ is 2a^2$2a^2$.
### Core Logic
The given curves form a bounded geometric area.
Ellipse: x^2 + 4y^2 = 4 Rightarrow fracx^24 + fracy^21 = 1$x^2 + 4y^2 = 4 \Rightarrow \frac{x^2}{4} + \frac{y^2}{1} = 1$.
Here, a = 2$a = 2$, b = 1$b = 1$.
The region to be excluded is bounded by y = |x| - 1$y = |x| - 1$ and y = 1 - |x|$y = 1 - |x|$, which rearranges to |x| + |y| = 1$|x| + |y| = 1$. This forms a square/rhombus centered at the origin with vertices at (1, 0), (0, 1), (-1, 0), (0, -1)$(1, 0), (0, 1), (-1, 0), (0, -1)$.
### Step 1: Calculate Total and Excluded Areas
Total Area of the Ellipse:
textArea = pi (2)(1) = 2pi$$\text{Area} = \pi (2)(1) = 2\pi$$
Excluded Area (Rhombus |x| + |y| = 1$|x| + |y| = 1$):
The rhombus consists of 4 identical right-angled triangles in each quadrant.
Area of one triangle = frac12 times textbase times textheight = frac12 times 1 times 1 = frac12$\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$.
Total excluded area = 4 times frac12 = 2$4 \times \frac{1}{2} = 2$.
### Step 2: Calculate Required Area
Ellipse and Modulus area diagram for Q2 - JEE Main 2026 Morning
Required Area = Area of ellipse -$-$ Shaded Area
= 2pi - 2$= 2\pi - 2$= 2(pi - 1)$= 2(\pi - 1)$
### Pattern Recognition
Transform absolute value equations y = pm(|x| - a)$y = \pm(|x| - a)$ into |x| + |y| = a$|x| + |y| = a$ to instantly recognize a standard rhombus, allowing direct geometry formulas instead of integration.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Area Under Curves
Class 11 Maths: Conic Sections
Keywords:#area of the region#JEE Main 2026 Morning Q2#Area Under Curves JEE Main 2026#Area Bounded by Ellipse and Modulus Functions JEE Main 2026
More Area Under Curves Previous-Year Questions — Page 2
Q70jee_main_2025_28_jan_morningArea Bounded by Curves and Absolute Value Functions
The area (in sq. units) of the region
\(x, y): 0 le y le 2|x| + 1, 0 le y le x^2 + 1, |x| le 3\$$\{(x, y): 0 \le y \le 2|x| + 1, 0 \le y \le x^2 + 1, |x| \le 3\}$$
is
(1) frac803$\frac{80}{3}$
(2) frac643$\frac{64}{3}$
(3) frac173$\frac{17}{3}$
(4) frac323$\frac{32}{3}$
A.frac803$\frac{80}{3}$
B.frac643$\frac{64}{3}$
C.frac173$\frac{17}{3}$
D.frac323$\frac{32}{3}$
Solution
### Related Formula
Area under a curve using definite integration bounded by curves:
textArea = 2 int_a^b f(x) dx$$\text{Area} = 2 \int_a^b f(x) dx$$
### Core Logic
Since both bounding graphs are symmetric about the y-axis, we can integrate over the positive domain (x ge 0$x \ge 0$) and double the result: Area Bounded by Curves and Absolute Value Functions diagram for Q70 - JEE Main 2025 Morning
Intersecting points: x^2 + 1 = 2x + 1 implies x = 2$x^2 + 1 = 2x + 1 \implies x = 2$.
### Step 1: Setting Up the Bound Segments
The region splits into two integral segments based on which curve sits lower:
Segment 1: From 0$0$ to 2$2$, bounded by the parabola y = x^2 + 1$y = x^2 + 1$.
Segment 2: From 2$2$ to 3$3$, bounded by the straight line y = 2x + 1$y = 2x + 1$.
### Step 2: Performing Integration
$textArea = 2 left[ int_0^2 (x^2 + 1) dx + int_2^3 (2x + 1) dx right]$$$\text{Area} = 2 \left[ \int_0^2 (x^2 + 1) dx + \int_2^3 (2x + 1) dx \right]$$$
$= 2 left[ left( frac83 + 2 right) + (9 + 3 - 4 - 2) right] = 2 left[ frac143 + 6 right] = frac643$= 2 \left[ \left( \frac{8}{3} + 2 \right) + (9 + 3 - 4 - 2) \right] = 2 \left[ \frac{14}{3} + 6 \right] = \frac{64}{3}$$
### Pattern Recognition
Exploiting structural symmetry drops the integration limits, avoiding messy sign evaluations with absolute terms.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Area Under Curves
Q59jee_main_2025_04_april_eveningArea Bounded by Parabola and Tangent
A line passing through the point mathrmA(-2, 0)$\mathrm{A}(-2, 0)$, touches the parabola mathrmP: mathrmy^2 = mathrmx - 2$\mathrm{P}: \mathrm{y}^2 = \mathrm{x} - 2$ at the point mathrmB$\mathrm{B}$ in the first quadrant. The area, of the region bounded by the line mathrmAB$\mathrm{AB}$, parabola mathrmP$\mathrm{P}$ and the mathbfx$\mathbf{x}$-axis, is :-
A.frac73$\frac{7}{3}$
B.2$2$
C.frac83$\frac{8}{3}$
D.3$3$
Solution
### Core Logic
Let the equation of the tangent line passing through A(-2,0)$A(-2,0)$ be:
y = m(x + 2) implies x = fracym - 2$$y = m(x + 2) \implies x = \frac{y}{m} - 2$$
The equation of the parabola is y^2 = x - 2 implies x = y^2 + 2$y^2 = x - 2 \implies x = y^2 + 2$.
Substituting x$x$ from the line into the parabola:
y^2 + 2 = fracym - 2 implies y^2 - fracym + 4 = 0$$y^2 + 2 = \frac{y}{m} - 2 \implies y^2 - \frac{y}{m} + 4 = 0$$
For the line to be a tangent, the discriminant of this quadratic equation must be zero (D = 0$D = 0$):
left(-frac1mright)^2 - 4(1)(4) = 0 implies frac1m^2 = 16 implies m = pm frac14$$\left(-\frac{1}{m}\right)^2 - 4(1)(4) = 0 \implies \frac{1}{m^2} = 16 \implies m = \pm \frac{1}{4}$$
Since point B$B$ is in the first quadrant, the slope must be positive, so m = frac14$m = \frac{1}{4}$.
The line equation is y = frac14(x + 2) implies x = 4y - 2$y = \frac{1}{4}(x + 2) \implies x = 4y - 2$.
The point of tangency B$B$ is found at y = frac12m = 2$y = \frac{1}{2m} = 2$, which gives x = 6$x = 6$, so B = (6,2)$B = (6,2)$.
### Step 1: Setting up the Area Integral
Integrating with respect to y$y$ avoids splitting the region into two parts along the x-axis:
textArea = int_0^2 left(x_textparabola - x_textlineright) dy$$\text{Area} = \int_{0}^{2} \left(x_{\text{parabola}} - x_{\text{line}}\right) dy$$textArea = int_0^2 left((y^2 + 2) - (4y - 2)right) dy = int_0^2 left(y^2 - 4y + 4right) dy$$\text{Area} = \int_{0}^{2} \left((y^2 + 2) - (4y - 2)\right) dy = \int_{0}^{2} \left(y^2 - 4y + 4\right) dy$$Area under curves diagram for Q59 - JEE Main 2025 Evening
### Step 2: Evaluating the Integral
Integrating term by term:
textArea = left[ fracy^33 - 2y^2 + 4y right]_0^2$$\text{Area} = \left[ \frac{y^3}{3} - 2y^2 + 4y \right]_{0}^{2}$$textArea = left( frac83 - 2(4) + 4(2) right) - 0 = frac83 - 8 + 8 = frac83$$\text{Area} = \left( \frac{8}{3} - 2(4) + 4(2) \right) - 0 = \frac{8}{3} - 8 + 8 = \frac{8}{3}$$
### Pattern Recognition
Integrating with respect to y$y$ (horizontal strips) when dealing with horizontal parabolas or lines crossing the x-axis eliminates the need to break your area computation into multiple piecewise integrals.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Area Under Curves
Class 11 Mathematics: Conic Sections
Q71jee_main_2025_04_april_morningArea Bounded by Multiple Curves
If the area of the region \(x,y) : |x - 5| le y le 4sqrtx\$\{(x,y) : |x - 5| \le y \le 4\sqrt{x}\}$ is A, then 3A is equal to
Numerical Answer.Answer: 368 to 368
Solution
### Related Formula
Area tracking equation via horizontal slices or split verticals:
textArea = int_x_1^x_2 (y_textupper - y_textlower)\,mathrmdx$$\text{Area} = \int_{x_1}^{x_2} (y_{\text{upper}} - y_{\text{lower}})\,\mathrm{d}x$$
### Core Logic
Find the intersections of y = |x - 5|$y = |x - 5|$ and y = 4sqrtx$y = 4\sqrt{x}$:
Branch 1: 5 - x = 4sqrtx implies x + 4sqrtx - 5 = 0 implies (sqrtx + 5)(sqrtx - 1) = 0 implies x = 1, y = 4$5 - x = 4\sqrt{x} \implies x + 4\sqrt{x} - 5 = 0 \implies (\sqrt{x} + 5)(\sqrt{x} - 1) = 0 \implies x = 1, y = 4$.
Branch 2: x - 5 = 4sqrtx implies x - 4sqrtx - 5 = 0 implies (sqrtx - 5)(sqrtx + 1) = 0 implies x = 25, y = 20$x - 5 = 4\sqrt{x} \implies x - 4\sqrt{x} - 5 = 0 \implies (\sqrt{x} - 5)(\sqrt{x} + 1) = 0 \implies x = 25, y = 20$.
Area Bounded by Multiple Curves diagram for Q71 - JEE Main 2025 Morning
### Step 1: Set Up Area Definite Integral
Split integration intervals around vertex x = 5$x = 5$ or compute via simple boundary differences:
A = int_1^25 4sqrtx\,mathrmdx - textArea of Left Triangle - textArea of Right Triangle$$A = \int_{1}^{25} 4\sqrt{x}\,\mathrm{d}x - \text{Area of Left Triangle} - \text{Area of Right Triangle}$$textArea of Left Triangle = frac12 times (5 - 1) times 4 = 8$$\text{Area of Left Triangle} = \frac{1}{2} \times (5 - 1) \times 4 = 8$$textArea of Right Triangle = frac12 times (25 - 5) times 20 = 200$$\text{Area of Right Triangle} = \frac{1}{2} \times (25 - 5) \times 20 = 200$$
### Step 2: Complete Computations
int_1^25 4sqrtx\,mathrmdx = left[ frac83x^3/2 right]_1^25 = frac83(125 - 1) = frac8 times 1243 = frac9923$$\int_{1}^{25} 4\sqrt{x}\,\mathrm{d}x = \left[ \frac{8}{3}x^{3/2} \right]_{1}^{25} = \frac{8}{3}(125 - 1) = \frac{8 \times 124}{3} = \frac{992}{3}$$A = frac9923 - 208 = frac992 - 6243 = frac3683$$A = \frac{992}{3} - 208 = \frac{992 - 624}{3} = \frac{368}{3}$$3A = 368$3A = 368$
### Pattern Recognition
Subtracting standard geometric triangles beneath linear configurations from total absolute root curves saves significant time compared to managing multiple separate analytical integral pieces.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Area Under Curves
Q67jee_main_2025_24_jan_eveningArea Under Curves
The area of the region enclosed by the curves y=e^x,$y=e^{x},$y=|e^x-1|$y=|e^{x}-1|$ and y-axis is: [cite: 3389, 3390]
A.1+log_e2$1+\log_{e}2$
B.log_e2$\log_{e}2$
C.2log_e2-1$2\log_{e}2-1$
D.1-log_e2$1-\log_{e}2$
Solution
### Related Formula
The area between two 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 given by:
textArea = int_a^b |y_1(x) - y_2(x)| dx$$\text{Area} = \int_{a}^{b} |y_1(x) - y_2(x)| dx$$
### Core Logic
Analyze the curves to locate intersection points :
- Curve 1: y = e^x$y = e^x$
- Curve 2: y = |e^x - 1|$y = |e^x - 1|$
- Boundary: y-axis (x = 0$x = 0$)
For x < 0$x < 0$, e^x < 1 Rightarrow |e^x - 1| = 1 - e^x$e^x < 1 \Rightarrow |e^x - 1| = 1 - e^x$ .
Find intersection: e^x = 1 - e^x Rightarrow 2e^x = 1 Rightarrow e^x = frac12 Rightarrow x = -ln 2$e^x = 1 - e^x \Rightarrow 2e^x = 1 \Rightarrow e^x = \frac{1}{2} \Rightarrow x = -\ln 2$.
Area bounded by exponential curves diagram for Q67 - JEE Main 2025 Evening
### Step 1: Set up the Definite Integral
The integration spans from x = -ln 2$x = -\ln 2$ to x = 0$x = 0$. In this interval, e^x ge 1 - e^x$e^x \ge 1 - e^x$:
textArea = int_-ln 2^0 left[ e^x - (1 - e^x) right] dx$$\text{Area} = \int_{-\ln 2}^{0} \left[ e^x - (1 - e^x) \right] dx$$textArea = int_-ln 2^0 (2e^x - 1) dx$$\text{Area} = \int_{-\ln 2}^{0} (2e^x - 1) dx$$
### Step 2: Integration Evaluation
Integrate term-by-term :
textArea = left[ 2e^x - x right]_-ln 2^0$$\text{Area} = \left[ 2e^x - x \right]_{-\ln 2}^{0}$$= left(2e^0 - 0right) - left(2e^-ln 2 - (-ln 2)right)$$= \left(2e^0 - 0\right) - \left(2e^{-\ln 2} - (-\ln 2)\right)$$= 2 - left(2left(frac12right) + ln 2right) = 2 - (1 + ln 2) = 1 - ln 2$$= 2 - \left(2\left(\frac{1}{2}\right) + \ln 2\right) = 2 - (1 + \ln 2) = 1 - \ln 2$$
### Pattern Recognition
Modulus graphs split at their critical point (e^x = 1 Rightarrow x = 0$e^x = 1 \Rightarrow x = 0$). Recognizing that the required segment lies strictly in the negative quadrant simplifies the definition of the upper and lower functions immediately.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Application of Integrals
Q63jee_main_2025_24_jan_morningArea Bounded by Multiple Curves
The area of the region \(x,y)colon x^2 + 4x + 2 leq y leq |x + 2|\$\{(x,y)\colon x^2 + 4x + 2 \leq y \leq |x + 2|\}$ is equal to :
A.7$7$
B.24/5$24/5$
C.20/3$20/3$
D.5$5$
Solution
### Related Formula
The net area between two intersecting functions y_2(x)$y_2(x)$ and y_1(x)$y_1(x)$ from lower limit a$a$ to upper limit b$b$ is evaluated using the definite integral:
textArea = int_a^b [y_2(x) - y_1(x)] dx$$\text{Area} = \int_{a}^{b} [y_2(x) - y_1(x)] dx$$
### Core Logic
Let's perform a horizontal shift transformation by defining X = x + 2$X = x + 2$ to simplify the expressions:
- Parabola: y = x^2 + 4x + 2 = (x+2)^2 - 2 implies y = X^2 - 2$y = x^2 + 4x + 2 = (x+2)^2 - 2 \implies y = X^2 - 2$
- Absolute curve: y = |x + 2| implies y = |X|$y = |x + 2| \implies y = |X|$
This coordinate translation preserves area completely while shifting the axis to the origin.
### Step 1: Compute Points of Intersection
Find where the transformed curves intersect by equating the functions:
|X|^2 - 2 = |X|$|X|^2 - 2 = |X|$|X|^2 - |X| - 2 = 0$$|X|^2 - |X| - 2 = 0$$(|X| - 2)(|X| + 1) = 0$$(|X| - 2)(|X| + 1) = 0$$
Since |X| geq 0$|X| \geq 0$, we discard |X| = -1$|X| = -1$. This gives |X| = 2 implies X = pm 2$|X| = 2 \implies X = \pm 2$.
### Step 2: Set up and Evaluate the Transformed Integral
Since both curves are symmetric about the vertical line X = 0$X = 0$, we can compute the area for the positive half and double it:
textArea = 2 int_0^2 left[ X - (X^2 - 2) right] dX = 2 int_0^2 left( 2 + X - X^2 right) dX$$\text{Area} = 2 \int_{0}^{2} \left[ X - (X^2 - 2) \right] dX = 2 \int_{0}^{2} \left( 2 + X - X^2 \right) dX$$
Integrate the polynomial row-by-row:
= 2 left[ 2X + fracX^22 - fracX^33 right]_0^2$$= 2 \left[ 2X + \frac{X^2}{2} - \frac{X^3}{3} \right]_{0}^{2}$$= 2 left[ 2(2) + frac42 - frac83 right] = 2 left[ 4 + 2 - frac83 right]$$= 2 \left[ 2(2) + \frac{4}{2} - \frac{8}{3} \right] = 2 \left[ 4 + 2 - \frac{8}{3} \right]$$= 2 left[ 6 - frac83 right] = 2 cdot frac103 = frac203$$= 2 \left[ 6 - \frac{8}{3} \right] = 2 \cdot \frac{10}{3} = \frac{20}{3}$$
### Pattern Recognition
Applying a horizontal variable substitution matching X = x + ± c$X = x + ± c$ shifts complex quadratic forms to centered formats, which avoids tedious arithmetic across asymmetrical integration limits.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Application of Integrals
More Area Under Curves Questions — jee_main_2026_21_jan_morning
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