Since y$y$ is real, y² = 1 y = ± 1$y^2 = 1 \implies y = \pm 1$.
When y = ± 1$y = \pm 1$, x = (1)/(2)$x = \frac{1}{2}$.
Intersection points are ((1)/(2), 1)$\left(\frac{1}{2}, 1\right)$ and ((1)/(2), -1)$\left(\frac{1}{2}, -1\right)$.
Step 1: Set up and Compute Area Integral
Between y = -1$y = -1$ and y = 1$y = 1$, (1)/(1+y²) ≥ (y²)/(2)$\frac{1}{1+y^2} \ge \frac{y^2}{2}$.
Whenever curves are functions of y²$y^2$, integrating along the y$y$-axis avoids dealing with messy radical functions (square roots) and naturally accounts for symmetry across the x$x$-axis.
Chapter Mix
Class 12 Mathematics: Application of Integrals
More Area Under Curves Previous-Year Questions — Page 5
Qjee_main_2024_01_february_morningArea Between Curves
The area enclosed by the curves xy+4y=16$xy+4y=16$ and x+y=6$x+y=6$ is equal to:
A.28-30 ₑ2$28-30 \log_{e}2$
B.30-28 ₑ2$30-28 \log_{e}2$
C.30-32 ₑ2$30-32 \log_{e}2$
D.32-30 ₑ2$32-30 \log_{e}2$
Solution
Related Formula
Area enclosed between two intersecting curves y₁ = f(x)$y_1 = f(x)$ and y₂ = g(x)$y_2 = g(x)$ from boundary limits x = a$x = a$ to x = b$x = b$:
Between x = -2$x = -2$ and x = 4$x = 4$, the line y = 6 - x$y = 6 - x$ lies above the curve y = (16)/(x+4)$y = \frac{16}{x+4}$.
Therefore, the required enclosed area is:
Sees: Area bounded by a straight line and a shifting rectangular hyperbola.
Shortcut: The roots of the difference equation x² - 2x - 8 = 0$x^2 - 2x - 8 = 0$ directly give the limits. Always check graph orientations to place the upper linear equation before the curve equation inside the integral bracket to preserve absolute area values.
Chapter Mix
Class 12 Mathematics: Application of Integrals
Class 11 Mathematics: Conic Sections (Hyperbola)
Qjee_main_2024_29_january_eveningArea Under Curves
Let the area of the region (x,y):0≤ x≤ 3,0≤ y≤ x² +2,2x + 2$\{(x,y):0\leq x\leq 3,0\leq y\leq \min \{x^2 +2,2x + 2\} \}$ be A$A$. Then 12A$12A$ is equal to
Numerical Answer.Answer: 164 to 164
Solution
Related Formula
Area A = ∫ f(x), g(x) dx$$\text{Area } A = \int \min\{f(x), g(x)\}\, dx$$
Core Logic
Let us find the point of intersection of the curves y = x² + 2$y = x^2 + 2$ and y = 2x + 2$y = 2x + 2$:
x² + 2 = 2x + 2 x² - 2x = 0 x = 0 or x = 2$$x^2 + 2 = 2x + 2 \implies x^2 - 2x = 0 \implies x = 0 \text{ or } x = 2$$
For min/max boundary sets, always compute intersections first to accurately split integration domains into separate regions.
Chapter Mix
Class 12 Mathematics: Application of Integrals
Qjee_main_2024_27_jan_morningArea Under Curve
Let the area of the region (x, y): x²y + 4 ≥ 0, x+2y²≥0, x+4y²≤8, y≥0$\{(x, y): x^2y + 4 \ge 0, x+2y^{2}\ge0, x+4y^{2}\le8, y\ge0\}$ be (m)/(n)$\frac{m}{n}$ where m$m$ and n$n$ are coprime numbers. Then m+n$m+n$ is equal to:
We need to find the area bounded by the curves in the first quadrant (since y ≥ 0$y \ge 0$). Note that the first inequality x²y + 4 ≥ 0$x^2y + 4 \ge 0$ is trivially satisfied for all x, y ≥ 0$x, y \ge 0$.
So we focus on bounding x$x$ using:
Right curve: x = 8 - 4y²$x = 8 - 4y^2$
Left curve: x = -2y²$x = -2y^2$
However, there might be constraints where these cross or hit the axes. We must inspect intersections.
Step 1: Finding Intersection Points
Where do the left and right parabolas intersect?
8 - 4y² = -2y² ⇒ 2y² = 8 ⇒ y² = 4 ⇒ y = 2$8 - 4y^2 = -2y^2 \Rightarrow 2y^2 = 8 \Rightarrow y^2 = 4 \Rightarrow y = 2$ (Since y ≥ 0$y \ge 0$).
Wait, does x$x$ have boundaries? Since x$x$ can't drop arbitrarily into negative territory if it's restricted by other axes. Let's check x²y + 4 ≥ 0$x^2y + 4 \ge 0$. If x = -2y²$x = -2y^2$, then y$y$ must be bounded, but wait - the question limits are actually split into two regions depending on x²y+4 ≥ 0$x^2y+4 \ge 0$? Wait, the first inequality x²y+4 ≥ 0$x^2y+4 \ge 0$ might not be trivial if x$x$ is negative.
If x = -2y²$x = -2y^2$, then (-2y²)² y + 4 ≥ 0 ⇒ 4y⁵ + 4 ≥ 0$(-2y^2)^2 y + 4 \ge 0 \Rightarrow 4y^5 + 4 \ge 0$, which is true for all y ≥ 0$y \ge 0$.
Wait, there seems to be a misinterpretation of the first inequality. Let's look closer at the PDF solution boundaries.
The integration is broken at y=1$y=1$.
Why? x+2y² ≥ 0 ⇒ x ≥ -2y²$x+2y^2 \ge 0 \Rightarrow x \ge -2y^2$. Wait, the solution states another left curve: 2y-4$2y-4$. Is x²y+4 ≥ 0$x^2y+4 \ge 0$ actually x + 2y - 4 ≥ 0$x + 2y - 4 \ge 0$?
Yes, OCR shows `x^2y+4` but the solution integrates `(2y-4)`. Thus the original condition is likely x - 2y + 4 ≥ 0 ⇒ x ≥ 2y - 4$x - 2y + 4 \ge 0 \Rightarrow x \ge 2y - 4$!
Let's assume the left boundary splits between x = -2y²$x = -2y^2$ and x = 2y - 4$x = 2y - 4$.
Step 2: Region Bounds
Intersection of x = -2y²$x = -2y^2$ and x = 2y - 4$x = 2y - 4$:
-2y² = 2y - 4 ⇒ y² + y - 2 = 0 ⇒ (y+2)(y-1) = 0 ⇒ y = 1$-2y^2 = 2y - 4 \Rightarrow y^2 + y - 2 = 0 \Rightarrow (y+2)(y-1) = 0 \Rightarrow y = 1$.
Intersection of x = 2y - 4$x = 2y - 4$ and x = 8 - 4y²$x = 8 - 4y^2$:
2y - 4 = 8 - 4y² ⇒ 4y² + 2y - 12 = 0 ⇒ 2y² + y - 6 = 0 ⇒ (2y-3)(y+2) = 0 ⇒ y = 3/2$2y - 4 = 8 - 4y^2 \Rightarrow 4y^2 + 2y - 12 = 0 \Rightarrow 2y^2 + y - 6 = 0 \Rightarrow (2y-3)(y+2) = 0 \Rightarrow y = 3/2$.
So the region shifts left boundary at y=1$y=1$ and closes entirely at y=3/2$y=3/2$.
This implies m = 107$m = 107$ and n = 12$n = 12$.
Since 107 and 12 are coprime, m + n = 107 + 12 = 119$m + n = 107 + 12 = 119$.
Pattern Recognition
When dealing with multiple inequalities bounded by y ≥ 0$y \ge 0$, always project horizontally (integrate wrt y$y$) as the bounds natively trace left-to-right distances. Find intersection nodes to partition the integral correctly.
Chapter Mix
Class 12 Maths: Application of Integrals
Q28jee_main_2024_29_jan_morningArea Under Curves
The area (in sq. units) of the part of circle x²+y²=169$x^2+y^2=169$ which is below the line 5x-y=13$5x-y=13$ is (πα)/(2β)-(65)/(2)+(α)/(β) ⁻¹((12)/(13))$\frac{\pi\alpha}{2\beta}-\frac{65}{2}+\frac{\alpha}{\beta}\sin^{-1}(\frac{12}{13})$ where α,β$\alpha,\beta$ are coprime numbers. Then α+β$\alpha+\beta$ is equal to
Numerical Answer.Answer: 171 to 171
Solution
Related Formula
Standard Integral: ∫ √(a²-y²) dy = (y)/(2)√(a²-y²) + (a²)/(2) ⁻¹((y)/(a)) + C$$\text{Standard Integral: } \int \sqrt{a^2-y^2} dy = \frac{y}{2}\sqrt{a^2-y^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{y}{a}\right) + C$$Area of right triangle = (1)/(2) × base × height$$\text{Area of right triangle } = \frac{1}{2} \times \text{base} \times \text{height}$$
Core Logic
First, find the points of intersection between the circle x²+y²=169$x^2+y^2=169$ and the line 5x-y=13 ⇒ y = 5x-13$5x-y=13 \Rightarrow y = 5x-13$.
Substitute y$y$ into the circle equation:
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 = (y+13)/(5)$x = \frac{y+13}{5}$ and above the right-hand boundary of the circle x = √(169-y²)$x = \sqrt{169-y^2}$ 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):
Split the integral into two parts:
Part A (Circle): ∫₋₁₃¹² √(169-y²) dy$\int_{-13}^{12} \sqrt{169-y^2} dy$
Part B (Line): ∫₋₁₃¹² (y+13)/(5) dy$\int_{-13}^{12} \frac{y+13}{5} dy$
Comparing this exactly with the given format (πα)/(2β) - (65)/(2) + (α)/(β) ⁻¹((12)/(13))$\frac{\pi\alpha}{2\beta} - \frac{65}{2} + \frac{\alpha}{\beta}\sin^{-1}(\frac{12}{13})$:
We see that (α)/(β) = (169)/(2)$\frac{\alpha}{\beta} = \frac{169}{2}$.
Since 169 and 2 are coprime, α = 169$\alpha = 169$ and β = 2$\beta = 2$.
When evaluating line integrals forming a triangle with horizontal bounds, bypass algebraic integration and visually calculate (1)/(2) · b · 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.
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 parabolay² = 4(x - 2)$y^2 = 4(x - 2)$ and the line y = 2x - 8$y = 2x - 8$
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.
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.
