The area of the region enclosed by the parabola (y - 2)^2 = x - 1$(y - 2)^2 = x - 1$ , the line x - 2y + 4 = 0$x - 2y + 4 = 0$ and the positive coordinate axes is
Numerical Answer Type:
Enter a numerical valueAnswer: 5 to 5+4 marks
Solution & Explanation
### Related Formula
textArea enclosed integrating w.r.t y-axis: int (x_textright - x_textleft) dy$\text{Area enclosed integrating w.r.t y-axis: } \int (x_{\text{right}} - x_{\text{left}}) dy$
### Core Logic
Find intersection points between the parabola x = (y-2)^2 + 1$x = (y-2)^2 + 1$ and the line x = 2y - 4$x = 2y - 4$.
Set them equal:
(y-2)^2 + 1 = 2y - 4$(y-2)^2 + 1 = 2y - 4$y^2 - 4y + 4 + 1 = 2y - 4$y^2 - 4y + 4 + 1 = 2y - 4$y^2 - 6y + 9 = 0 Rightarrow (y-3)^2 = 0$y^2 - 6y + 9 = 0 \Rightarrow (y-3)^2 = 0$
The line is tangent to the parabola at y = 3$y = 3$. At y = 3$y = 3$, x = 2(3) - 4 = 2$x = 2(3) - 4 = 2$.
We need the area enclosed by the parabola, the line, and the *positive coordinate axes*.
Let's trace the boundary intercepts.
The line x = 2y - 4$x = 2y - 4$ cuts the y-axis (x=0$x=0$) at y=2$y=2$.
Parabola vertex is at (1, 2)$(1, 2)$. Parabola cuts y-axis (x=0$x=0$) at (y-2)^2 = -1$(y-2)^2 = -1$ (No real y-intercept).
So the curve (y-2)^2 = x-1$(y-2)^2 = x-1$ only exists for x ge 1$x \ge 1$.
### Step 1: Setting up the Integral w.r.t y
Area Under Curve diagram for Q29 - JEE Main 2024 Evening
Because the area is bound by the positive coordinate axes, we track integration strictly in the first quadrant.
The required area is the area between the y-axis, the x-axis, the line, and the curve. Wait, the problem says "area enclosed by the parabola... line... and positive coordinate axes".
Let's integrate with respect to y$y$.
The boundary curves are x = (y-2)^2 + 1$x = (y-2)^2 + 1$ (which forms the rightmost boundary) and the axes.
The area bounded by the curve with y-axis is from y=0$y=0$ to y=3$y=3$ minus the triangular region cut by the line below y=2$y=2$.
The line intersects the x-axis (y=0$y=0$) at x=-4$x=-4$ and y-axis (x=0$x=0$) at y=2$y=2$.
The positive coordinate axes enforce boundaries at x ge 0, y ge 0$x \ge 0, y \ge 0$.
So the exact area is the integral of the parabola's x$x$-value from y=0$y=0$ to y=3$y=3$, minus the area of the small triangle outside the region bounded by the line x=2y-4$x=2y-4$ and the axes in the positive quadrant.
The area under the parabola (to the left towards the y-axis) from y=0$y=0$ to y=3$y=3$ is int_0^3 x_textparabola dy$\int_0^3 x_{\text{parabola}} dy$.
The line bounds the region on the left from y=2$y=2$ to y=3$y=3$. Between y=0$y=0$ and y=2$y=2$, the area goes fully to the y-axis.
So, area = int_0^3 x_textparabola dy - textArea of Delta text bounded by line and y-axis$\int_0^3 x_{\text{parabola}} dy - \text{Area of } \Delta \text{ bounded by line and y-axis}$.
The triangle bounded by x=2y-4, x=0, y=2, y=3$x=2y-4, x=0, y=2, y=3$ is: base is x=2(3)-4 = 2$x=2(3)-4 = 2$ at y=3$y=3$, height is 1$1$ (from y=2$y=2$ to y=3$y=3$). Triangle area = frac12 times 2 times 1 = 1$\frac{1}{2} \times 2 \times 1 = 1$.
Hence, Enclosed Area = int_0^3 ((y-2)^2 + 1) dy - 1$\int_{0}^{3} ((y-2)^2 + 1) dy - 1$.
### Step 2: Evaluating the Integral
textArea = int_0^3 (y^2 - 4y + 5) dy - 1$\text{Area} = \int_{0}^{3} (y^2 - 4y + 5) dy - 1$= left[ fracy^33 - 2y^2 + 5y right]_0^3 - 1$= \left[ \frac{y^3}{3} - 2y^2 + 5y \right]_0^3 - 1$= left( frac273 - 2(9) + 5(3) right) - 0 - 1$= \left( \frac{27}{3} - 2(9) + 5(3) \right) - 0 - 1$= (9 - 18 + 15) - 1 = 6 - 1 = 5$= (9 - 18 + 15) - 1 = 6 - 1 = 5$
### Pattern Recognition
Instead of evaluating horizontal strips with a broken left-bound, evaluate the total area mapped to the y-axis int_0^3 x dy$\int_0^3 x dy$ and simply subtract the basic geometric triangle removed by the straight line.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Integral Calculus
For xin(-fracpi2,fracpi2)$x\in(-\frac{\pi}{2},\frac{\pi}{2})$, if y(x)=intfracoperatornamecosec x+sin xoperatornamecosec x sec x+tan x sin^2 xdx$y(x)=\int\frac{\operatorname{cosec} x+\sin x}{\operatorname{cosec} x \sec x+\tan x \sin^2 x}dx$ and lim_xrightarrow(fracpi2)^-y(x)=0$\lim_{x\rightarrow(\frac{\pi}{2})^-}y(x)=0$ then y(fracpi4)$y(\frac{\pi}{4})$ is equal to
### Related Formula
int frac1u^2 + a^2 du = frac1a tan^-1left(fracuaright) + C$\int \frac{1}{u^2 + a^2} du = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$
### Core Logic
Simplify the given integrand by converting all trigonometric ratios into sin x$\sin x$ and cos x$\cos x$:
I = fracoperatornamecosec x + sin xoperatornamecosec x sec x + tan x sin^2 x$I = \frac{\operatorname{cosec} x + \sin x}{\operatorname{cosec} x \sec x + \tan x \sin^2 x}$
Numerator: frac1sin x + sin x = frac1 + sin^2 xsin x$\frac{1}{\sin x} + \sin x = \frac{1 + \sin^2 x}{\sin x}$
Denominator: frac1sin x cos x + fracsin xcos x cdot sin^2 x = frac1 + sin^4 xsin x cos x$\frac{1}{\sin x \cos x} + \frac{\sin x}{\cos x} \cdot \sin^2 x = \frac{1 + \sin^4 x}{\sin x \cos x}$
Divide Numerator by Denominator:
I = fracfrac1 + sin^2 xsin xfrac1 + sin^4 xsin x cos x = frac(1 + sin^2 x)cos x1 + sin^4 x$I = \frac{\frac{1 + \sin^2 x}{\sin x}}{\frac{1 + \sin^4 x}{\sin x \cos x}} = \frac{(1 + \sin^2 x)\cos x}{1 + \sin^4 x}$
### Step 1: Apply Substitution
Now rewrite the integral:
y(x) = int frac(1 + sin^2 x)cos x1 + sin^4 x dx$y(x) = \int \frac{(1 + \sin^2 x)\cos x}{1 + \sin^4 x} dx$
Let sin x = t$\sin x = t$, then cos x \, dx = dt$\cos x \, dx = dt$. The integral becomes:
y(x) = int frac1 + t^21 + t^4 dt$y(x) = \int \frac{1 + t^2}{1 + t^4} dt$
Divide numerator and denominator by t^2$t^2$:
y(x) = int frac1 + frac1t^2t^2 + frac1t^2 dt$y(x) = \int \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} dt$
We know that t^2 + frac1t^2 = left(t - frac1tright)^2 + 2$t^2 + \frac{1}{t^2} = \left(t - \frac{1}{t}\right)^2 + 2$. Let u = t - frac1t$u = t - \frac{1}{t}$, then du = left(1 + frac1t^2right) dt$du = \left(1 + \frac{1}{t^2}\right) dt$.
y(x) = int fracduu^2 + (sqrt2)^2 = frac1sqrt2tan^-1left(fracusqrt2right) + C$y(x) = \int \frac{du}{u^2 + (\sqrt{2})^2} = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{u}{\sqrt{2}}\right) + C$y(x) = frac1sqrt2tan^-1left(fract - 1/tsqrt2right) + C = frac1sqrt2tan^-1left(fracsin x - operatornamecosec xsqrt2right) + C$y(x) = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t - 1/t}{\sqrt{2}}\right) + C = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sin x - \operatorname{cosec} x}{\sqrt{2}}\right) + C$
### Step 2: Solve for Constant C
We are given that lim_xrightarrow(pi/2)^- y(x) = 0$\lim_{x\rightarrow(\pi/2)^-} y(x) = 0$.
As x to fracpi2$x \to \frac{\pi}{2}$, sin x to 1$\sin x \to 1$ and operatornamecosec x to 1$\operatorname{cosec} x \to 1$.
Thus, sin x - operatornamecosec x to 0$\sin x - \operatorname{cosec} x \to 0$.
0 = frac1sqrt2tan^-1left(frac1 - 1sqrt2right) + C$0 = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{1 - 1}{\sqrt{2}}\right) + C$0 = frac1sqrt2tan^-1(0) + C Rightarrow C = 0$0 = \frac{1}{\sqrt{2}}\tan^{-1}(0) + C \Rightarrow C = 0$
### Step 3: Find y(pi/4)
Substitute x = fracpi4$x = \frac{\pi}{4}$ into the exact solution y(x) = frac1sqrt2tan^-1left(fracsin x - operatornamecosec xsqrt2right)$y(x) = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sin x - \operatorname{cosec} x}{\sqrt{2}}\right)$:
At x = fracpi4$x = \frac{\pi}{4}$, sinleft(fracpi4right) = frac1sqrt2$\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and operatornamecosecleft(fracpi4right) = sqrt2$\operatorname{cosec}\left(\frac{\pi}{4}\right) = \sqrt{2}$.
yleft(fracpi4right) = frac1sqrt2tan^-1left(fracfrac1sqrt2 - sqrt2sqrt2right)$y\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\frac{1}{\sqrt{2}} - \sqrt{2}}{\sqrt{2}}\right)$= frac1sqrt2tan^-1left(fracfrac1 - 2sqrt2sqrt2right) = frac1sqrt2tan^-1left(frac-1/sqrt2sqrt2right)$= \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\frac{1 - 2}{\sqrt{2}}}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{-1/\sqrt{2}}{\sqrt{2}}\right)$= frac1sqrt2tan^-1left(-frac12right)$= \frac{1}{\sqrt{2}}\tan^{-1}\left(-\frac{1}{2}\right)$
### Pattern Recognition
A high-degree polynomial of sin x$\sin x$ or cos x$\cos x$ in the denominator matched with their derivative elements on top is the hallmark of the t+1/t$t+1/t$ or t-1/t$t-1/t$ structural substitution form. Dividing out the middle power (t^2$t^2$) aligns the differential instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Integral Calculus
Q18jee_main_2024_29_jan_morningProperties of Definite Integrals
If the value of the integralint_-fracpi2^fracpi2left(fracx^2cos x1+pi^x+frac1+sin^2x1+e^sin x^2023right)dx=fracpi4(pi+a)-2$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{x^2\cos x}{1+\pi^x}+\frac{1+\sin^2x}{1+e^{\sin x^{2023}}}\right)dx=\frac{\pi}{4}(\pi+a)-2$, then the value of a$a$ is
A.3$3$
B.-frac32$-\frac{3}{2}$
C.2$2$
D.frac32$\frac{3}{2}$
Solution
### Related Formula
textKing's Rule Variant: int_-a^a f(x) dx = int_0^a (f(x) + f(-x)) dx$\text{King's Rule Variant: } \int_{-a}^{a} f(x) dx = \int_{0}^{a} (f(x) + f(-x)) dx$
### Core Logic
Let the integral be I$I$.
We apply the property int_-a^a f(x) dx = int_0^a (f(x)+f(-x)) dx$\int_{-a}^a f(x) dx = \int_0^a (f(x)+f(-x)) dx$ to each fractional term independently.
For the first term f_1(x) = fracx^2cos x1+pi^x$f_1(x) = \frac{x^2\cos x}{1+\pi^x}$:
f_1(x) + f_1(-x) = fracx^2cos x1+pi^x + frac(-x)^2cos(-x)1+pi^-x = x^2cos x left( frac11+pi^x + fracpi^xpi^x+1 right) = x^2cos x$f_1(x) + f_1(-x) = \frac{x^2\cos x}{1+\pi^x} + \frac{(-x)^2\cos(-x)}{1+\pi^{-x}} = x^2\cos x \left( \frac{1}{1+\pi^x} + \frac{\pi^x}{\pi^x+1} \right) = x^2\cos x$
For the second term f_2(x) = frac1+sin^2x1+e^sin x^2023$f_2(x) = \frac{1+\sin^2x}{1+e^{\sin x^{2023}}}$:
Observe the exponent: sin(-x)^2023 = sin(-x^2023) = -sin x^2023$\sin(-x)^{2023} = \sin(-x^{2023}) = -\sin x^{2023}$.
f_2(x) + f_2(-x) = frac1+sin^2x1+e^p + frac1+sin^2(-x)1+e^-p = (1+sin^2x) left( frac11+e^p + frace^pe^p+1 right) = 1+sin^2x$f_2(x) + f_2(-x) = \frac{1+\sin^2x}{1+e^{p}} + \frac{1+\sin^2(-x)}{1+e^{-p}} = (1+\sin^2x) \left( \frac{1}{1+e^p} + \frac{e^p}{e^p+1} \right) = 1+\sin^2x$
Thus, the integral radically simplifies to:
I = int_0^pi/2 (x^2cos x + 1 + sin^2 x) dx$I = \int_0^{\pi/2} (x^2\cos x + 1 + \sin^2 x) dx$
### Step 1: Execute Integration by Parts
Evaluate int_0^pi/2 x^2cos x \, dx$\int_0^{\pi/2} x^2\cos x \, dx$ using integration by parts:
u = x^2 Rightarrow du = 2x \, dx$u = x^2 \Rightarrow du = 2x \, dx$dv = cos x \, dx Rightarrow v = sin x$dv = \cos x \, dx \Rightarrow v = \sin x$int x^2cos x dx = x^2sin x - int 2xsin x dx$\int x^2\cos x dx = x^2\sin x - \int 2x\sin x dx$
Applying parts again to int 2xsin x dx$\int 2x\sin x dx$:
int 2xsin x dx = 2x(-cos x) - int 2(-cos x)dx = -2xcos x + 2sin x$\int 2x\sin x dx = 2x(-\cos x) - \int 2(-\cos x)dx = -2x\cos x + 2\sin x$
Substituting back:
left[ x^2sin x + 2xcos x - 2sin x right]_0^pi/2$\left[ x^2\sin x + 2x\cos x - 2\sin x \right]_0^{\pi/2}$
Evaluate at upper limit pi/2$\pi/2$:
(pi/2)^2(1) + 0 - 2(1) = fracpi^24 - 2$(\pi/2)^2(1) + 0 - 2(1) = \frac{\pi^2}{4} - 2$
Evaluate at lower limit 0:
0 + 0 - 0 = 0$0 + 0 - 0 = 0$
So, int_0^pi/2 x^2cos x \, dx = fracpi^24 - 2$\int_0^{\pi/2} x^2\cos x \, dx = \frac{\pi^2}{4} - 2$.
### Step 2: Execute Standard Integrals
Evaluate int_0^pi/2 (1 + sin^2 x) dx$\int_0^{\pi/2} (1 + \sin^2 x) dx$:
= int_0^pi/2 dx + int_0^pi/2 sin^2 x dx$= \int_0^{\pi/2} dx + \int_0^{\pi/2} \sin^2 x dx$= fracpi2 + left[ frac12 cdot fracpi2 right] = fracpi2 + fracpi4 = frac3pi4$= \frac{\pi}{2} + \left[ \frac{1}{2} \cdot \frac{\pi}{2} \right] = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$
### Step 3: Combine and Compare
Sum the components to get the total integral I$I$:
I = left(fracpi^24 - 2right) + frac3pi4 = fracpi^24 + frac3pi4 - 2$I = \left(\frac{\pi^2}{4} - 2\right) + \frac{3\pi}{4} = \frac{\pi^2}{4} + \frac{3\pi}{4} - 2$
Factor out fracpi4$\frac{\pi}{4}$:
I = fracpi4(pi + 3) - 2$I = \frac{\pi}{4}(\pi + 3) - 2$
The problem states I = fracpi4(pi + a) - 2$I = \frac{\pi}{4}(\pi + a) - 2$. Comparing the two expressions gives:
a = 3$a = 3$
### Pattern Recognition
Frightening denominators like 1+a^f(x)$1+a^{f(x)}$ in an integral from -L$-L$ to L$L$ where f(x)$f(x)$ is an odd function are almost exclusively designed to cancel out and leave 1 via the f(x)+f(-x)$f(x)+f(-x)$ expansion. Strip the ugly denominators immediately.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Integral Calculus
Let y = f(x)$y = f(x)$ be a thrice differentiable function in (-5, 5)$(-5, 5)$ . Let the tangents to the curve y = f(x)$y = f(x)$ at (1, f(1))$(1, f(1))$ and (3, f(3))$(3, f(3))$ make angles fracpi6$\frac{\pi}{6}$ and fracpi4$\frac{\pi}{4}$ , respectively with positive x-axis. If
27int_1^3left(left(f'(t)right)^2 + 1right)f''(t)dt = alpha + beta sqrt3 quad textwhere alpha, beta text are integers, then the value of alpha +beta text equals$27\int_{1}^{3}\left(\left(f'(t)\right)^{2} + 1\right)f''(t)dt = \alpha + \beta \sqrt{3} \quad \text{where } \alpha, \beta \text{ are integers, then the value of } \alpha +\beta \text{ equals}$
A.-14$-14$
B.26$26$
C.-16$-16$
D.36$36$
Solution
### Related Formula
textSlope of tangent at x=a text is f'(a) = tan theta$\text{Slope of tangent at } x=a \text{ is } f'(a) = \tan \theta$int u^n du = fracu^n+1n+1$\int u^n du = \frac{u^{n+1}}{n+1}$
### Core Logic
Given y=f(x)$y=f(x)$, the derivatives at the given points correspond to the slope of tangents:
left. fracdydx right|_x=1 = f'(1) = tanleft(fracpi6right) = frac1sqrt3$\left. \frac{dy}{dx} \right|_{x=1} = f'(1) = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$left. fracdydx right|_x=3 = f'(3) = tanleft(fracpi4right) = 1$\left. \frac{dy}{dx} \right|_{x=3} = f'(3) = \tan\left(\frac{\pi}{4}\right) = 1$
We must evaluate the integral:
I = int_1^3 left( (f'(t))^2 + 1 right) f''(t) dt$I = \int_{1}^{3} \left( (f'(t))^2 + 1 \right) f''(t) dt$
### Step 1: Integration by Substitution
Let z = f'(t)$z = f'(t)$. Then dz = f''(t)dt$dz = f''(t)dt$.
The limits of integration change accordingly:
When t = 1$t = 1$, z = f'(1) = frac1sqrt3$z = f'(1) = \frac{1}{\sqrt{3}}$
When t = 3$t = 3$, z = f'(3) = 1$z = f'(3) = 1$
The integral becomes:
I = int_1/sqrt3^1 (z^2 + 1) dz$I = \int_{1/\sqrt{3}}^{1} (z^2 + 1) dz$I = left[ fracz^33 + z right]_1/sqrt3^1$I = \left[ \frac{z^3}{3} + z \right]_{1/\sqrt{3}}^{1}$
### Step 2: Evaluating the Definite Integral
Plug in the limits:
I = left( frac1^33 + 1 right) - left( frac13 cdot frac13sqrt3 + frac1sqrt3 right)$I = \left( \frac{1^3}{3} + 1 \right) - \left( \frac{1}{3} \cdot \frac{1}{3\sqrt{3}} + \frac{1}{\sqrt{3}} \right)$I = frac43 - left( frac19sqrt3 + frac99sqrt3 right)$I = \frac{4}{3} - \left( \frac{1}{9\sqrt{3}} + \frac{9}{9\sqrt{3}} \right)$I = frac43 - frac109sqrt3 = frac43 - frac10sqrt327$I = \frac{4}{3} - \frac{10}{9\sqrt{3}} = \frac{4}{3} - \frac{10\sqrt{3}}{27}$
### Step 3: Finding alpha and beta
We are given that 27 cdot I = alpha + betasqrt3$27 \cdot I = \alpha + \beta\sqrt{3}$.
27 left( frac43 - frac10sqrt327 right) = 36 - 10sqrt3$27 \left( \frac{4}{3} - \frac{10\sqrt{3}}{27} \right) = 36 - 10\sqrt{3}$
Comparing with alpha + betasqrt3$\alpha + \beta\sqrt{3}$, we get:
alpha = 36$\alpha = 36$beta = -10$\beta = -10$
Therefore, alpha + beta = 36 - 10 = 26$\alpha + \beta = 36 - 10 = 26$.
### Pattern Recognition
Integrals containing a function derivative alongside its second derivative are classic substitution traps. Set u = f'(x)$u = f'(x)$ directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Integral Calculus
Class 12 Maths: Application of Derivatives
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