Objective Physics Vol-1
MCQ
Concept: In a right-angled triangle, trigonometric ratios are defined based on the sides relative to an acute angle $\theta$:
$\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}$
$\cos \theta = \frac{\text{Base}}{\text{Hypotenuse}}$
$\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}$
$\cot \theta = \frac{\text{Base}}{\text{Perpendicular}}$
$\sec \theta = \frac{\text{Hypotenuse}}{\text{Base}}$
$\csc \theta = \frac{\text{Hypotenuse}}{\text{Perpendicular}}$
Pythagoras theorem states that $\text{Hypotenuse}^2 = \text{Perpendicular}^2 + \text{Base}^2$.
If $\sin \theta = \frac{4}{5}$, where $\theta$ lies in the first quadrant, then find all the other T-ratios.
Objective Physics Vol-1
MCQ
Concept: T-ratios of Allied Angles:
1. $\sin(-\theta) = -\sin \theta$
2. $\tan(270^\circ - \theta) = \cot \theta$
3. $\cos(270^\circ + \theta) = \sin \theta$
4. $\sec(180^\circ - \theta) = -\sec \theta$
Standard Trigonometric Values:
$\sin 45^\circ = \frac{1}{\sqrt{2}}$, $\cot 45^\circ = 1$, $\sin 30^\circ = \frac{1}{2}$, $\sec 60^\circ = 2$
Find the values of:
(i) $\sin(-45^\circ)$
(ii) $\tan 225^\circ$
(iii) $\cos 300^\circ$
(iv) $\sec 120^\circ$
Objective Physics Vol-1
MCQ
Concept: Trigonometric Addition and Subtraction Formulas:
$\sin(A - B) = \sin A \cos B - \cos A \sin B$
$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$
Standard Trigonometric Values:
$\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, $\sin 30^\circ = \frac{1}{2}$, $\tan 45^\circ = 1$, $\tan 30^\circ = \frac{1}{\sqrt{3}}$
Find the value of:
(i) $\sin 15^\circ$
(ii) $\tan 75^\circ$
Objective Physics Vol-1
MCQ
Concept: In the third quadrant, $\sin \theta$ is negative, $\cos \theta$ is negative, and $\tan \theta$ is positive.
Trigonometric identities used:
$\sin^2 \theta + \cos^2 \theta = 1 \implies \sin \theta = -\sqrt{1 - \cos^2 \theta}$
$\tan \theta = \frac{\sin \theta}{\cos \theta}$
Find $\sin \theta$ and $\tan \theta$, if $\cos \theta = -\frac{12}{13}$ and $\theta$ lies in the third quadrant.
Objective Physics Vol-1
MCQ
Concept: In the second quadrant, $\sin \theta$ and $\csc \theta$ are positive, while $\cos \theta$, $\tan \theta$, $\cot \theta$, and $\sec \theta$ are negative.
Trigonometric identities:
$\sec \theta = -\sqrt{1 + \tan^2 \theta}$
$\cos \theta = \frac{1}{\sec \theta}$
$\sin \theta = \tan \theta \cdot \cos \theta$
$\csc \theta = \frac{1}{\sin \theta}$
$\cot \theta = \frac{1}{\tan \theta}$
Find the values of other five T-ratios, if $\tan \theta = -\frac{3}{4}$ and $\theta$ lies in II quadrant.
Objective Physics Vol-1
MCQ
Concept: T-ratios of Allied Angles:
1. $\csc(360^\circ - \theta) = -\csc \theta$
2. $\cos(180^\circ + \theta) = -\cos \theta$
3. $\sin(-\theta) = -\sin \theta$ and $\sin(360^\circ - \theta) = -\sin \theta$
Standard Trigonometric Values:
$\csc 45^\circ = \sqrt{2}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, $\sin 30^\circ = \frac{1}{2}$
Find the values of the following T-ratios:
(i) $\csc 315^\circ$
(ii) $\cos 210^\circ$
(iii) $\sin(-330^\circ)$
Objective Physics Vol-1
MCQ
Concept: T-ratios of Allied Angles:
1. $\sec(180^\circ - \theta) = -\sec \theta$
2. $\cot(90^\circ + \theta) = -\tan \theta$
Trigonometric Formulae:
$\cos(A - B) = \cos A \cos B + \sin A \sin B$
$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$
Standard Values:
$\cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, $\sin 30^\circ = \frac{1}{2}$, $\tan 45^\circ = 1$, $\tan 30^\circ = \frac{1}{\sqrt{3}}$
Find the value of:
(i) $\sec 165^\circ$
(ii) $\cot 105^\circ$
Objective Physics Vol-1
MCQ
Concept: Power Rule of Differentiation:
$\frac{d}{dx}(x^n) = n x^{n-1}$
Differentiate the function $y = x^{-3}$ with respect to $x$.
Objective Physics Vol-1
MCQ
Concept: 1. Power Rule: $\frac{d}{dx}(x^n) = n x^{n-1}$
2. Constant Multiple Rule: $\frac{d}{dx}(c \cdot v) = c \frac{dv}{dx}$
3. Derivative of a constant: $\frac{d}{dx}(c) = 0$
4. Sum and Difference Rule: $\frac{d}{dx}(u \pm v) = \frac{du}{dx} \pm \frac{dv}{dx}$
Differentiate the function $y = 6x^5 + 4x^3 - 3x^2 + 2x - 7$ with respect to $x$.
Objective Physics Vol-1
MCQ
Concept: Product Rule of Differentiation:
$\frac{d}{dx}(u \cdot v) = u \frac{dv}{dx} + v \frac{du}{dx}$
Or by expanding the algebraic expression first and then applying the Power Rule:
$\frac{d}{dx}(x^n) = n x^{n-1}$
Differentiate the function $y = (x + 2)(x^2 + 1)$ with respect to $x$.
Objective Physics Vol-1
MCQ
Concept: Quotient Rule of Differentiation:
$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$
Differentiate the function $y = \frac{x^3 + 4}{x + 1}$ with respect to $x$.
Objective Physics Vol-1
MCQ
Concept: Sum Rule of Differentiation:
$\frac{d}{dx}(u + v) = \frac{du}{dx} + \frac{dv}{dx}$
Standard Derivatives:
$\frac{d}{dx}(\sin x) = \cos x$
$\frac{d}{dx}(e^x) = e^x$
Differentiate the function $y = \sin x + e^x$ with respect to $x$.
Objective Physics Vol-1
MCQ
Concept: 1. Power Rule: $\frac{d}{dx}(x^n) = n x^{n-1}$
2. Derivative of Natural Logarithm: $\frac{d}{dx}(\log x) = \frac{1}{x}$
3. Derivative of Exponential Function: $\frac{d}{dx}(e^x) = e^x$
4. Derivative of a Constant: $\frac{d}{dx}(c) = 0$
5. Sum Rule: $\frac{d}{dx}(u + v + \dots) = \frac{du}{dx} + \frac{dv}{dx} + \dots$
Differentiate the function $y = 3x^2 + \log x + 4e^x + 5$ with respect to $x$.
Objective Physics Vol-1
MCQ
Concept: Product Rule of Differentiation:
$\frac{d}{dx}(u \cdot v) = u \frac{dv}{dx} + v \frac{du}{dx}$
Standard Derivatives:
$\frac{d}{dx}(\tan x) = \sec^2 x$
$\frac{d}{dx}(e^x) = e^x$
Differentiate the function $y = e^x \cdot \tan x$ with respect to $x$.
Objective Physics Vol-1
MCQ
Concept: Chain Rule of Differentiation:
If $y = f(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
Standard Derivatives:
$\frac{d}{dx}(\sin x) = \cos x$
$\frac{d}{dx}(x^n) = n x^{n-1}$
Find the derivative of $y = \sin(x^2 + 5)$ with respect to $x$.
Objective Physics Vol-1
MCQ
Concept: To find the maximum or minimum value of a function $y = f(x)$:
1. Find the first derivative $\frac{dy}{dx}$ and set it to zero ($\frac{dy}{dx} = 0$) to find the critical points.
2. Check the second derivative $\frac{d^2y}{dx^2}$. If $\frac{d^2y}{dx^2} < 0$ at the critical point, the function has a maximum value at that point.
Divide the number $1000$ into two parts such that their product is maximum.
Objective Physics Vol-1
MCQ
Concept: 1. Velocity $v$ is the first derivative of displacement $s$ with respect to time $t$:
$v = \frac{ds}{dt}$
2. Acceleration $a$ is the first derivative of velocity $v$ with respect to time $t$:
$a = \frac{dv}{dt}$
3. Initial values are obtained by setting $t = 0$.
The displacement of a particle as a function of time $t$ is given by $s = \alpha + \beta t + \gamma t^2 + \delta t^4$, where $\alpha, \beta, \gamma$ and $\delta$ are constants. Find the ratio of the initial velocity to the initial acceleration.
Objective Physics Vol-1
MCQ
Concept: 1. Velocity $v$ is the first derivative of position $x$ with respect to time $t$:
$v = \frac{dx}{dt}$
2. A particle moves along the positive x-direction when its velocity is positive ($v > 0$).
3. The particle changes its direction or comes to instantaneous rest when $v = 0$.
The position of a particle moving along the X-axis varies with time $t$ as $x = 6t - t^2 + 4$. Find the time interval during which the particle is moving along the positive x-direction.
Objective Physics Vol-1
MCQ
Concept: 1. Power Rule of Differentiation: $\frac{d}{dx}(x^n) = n x^{n-1}$
2. Derivative of Logarithmic Function: $\frac{d}{dx}(\log x) = \frac{1}{x}$
3. Constant Multiple Rule: $\frac{d}{dx}(c \cdot v) = c \frac{dv}{dx}$
Differentiate the function $y = 3x^4 + 2\frac{1}{x^2} + \log x$ with respect to $x$.
Objective Physics Vol-1
MCQ
Concept: Product Rule of Differentiation:
$\frac{d}{dx}(u \cdot v) = u \frac{dv}{dx} + v \frac{du}{dx}$
Alternatively, expand the polynomial first and use the Power Rule:
$\frac{d}{dx}(x^n) = n x^{n-1}$
Differentiate the function $y = (x^2 + 1)(x + 2)$ with respect to $x$.
Objective Physics Vol-1
MCQ
Concept: Quotient Rule of Differentiation:
$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$
Differentiate the function $y = \frac{3x^2}{x + 1}$ with respect to $x$.
Objective Physics Vol-1
MCQ
Concept: Standard Derivative of Trigonometric Function:
$\frac{d}{dx}(\sin x) = \cos x$
Differentiate the function $y = \sin x$ with respect to $x$.
Objective Physics Vol-1
MCQ
Concept: Chain Rule of Differentiation:
If $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$
Standard Derivative:
$\frac{d}{dx}(\tan x) = \sec^2 x$
Differentiate the function $y = \tan(x^2 + 3x + 1)$ with respect to $x$.
Objective Physics Vol-1
MCQ
Concept: 1. Standard Derivatives: $\frac{d}{dx}(\cos x) = -\sin x$, $\frac{d}{dx}(e^x) = e^x$
2. Chain Rule: $\frac{d}{dx}(\log(u)) = \frac{1}{u} \cdot \frac{du}{dx}$
Differentiate the function $y = 3\cos x + 7e^x + \log(x^2 + 1)$ with respect to $x$.
Objective Physics Vol-1
MCQ
Concept: To find the maximum or minimum value of a function $v(t)$:
1. Find the first derivative with respect to time, $\frac{dv}{dt} = 0$, to locate critical points (times).
2. Evaluate the second derivative $\frac{d^2v}{dt^2}$ at these critical points:
* If $\frac{d^2v}{dt^2} < 0$, the function has a local maximum at that time.
* If $\frac{d^2v}{dt^2} > 0$, the function has a local minimum at that time.
A particle is moving with velocity $v = t^3 - 6t^2 + 4$, where $v$ is in $\text{m/s}$ and $t$ is in seconds. At what time will the velocity be maximum/minimum and what is it equal to?
Objective Physics Vol-1
MCQ
Concept: 1. Velocity $v$ is the derivative of position $x$ with respect to time $t$:
$v = \frac{dx}{dt}$
2. Acceleration $a$ is the derivative of velocity $v$ with respect to time $t$:
$a = \frac{dv}{dt}$
3. Implicit differentiation of relation between position and time can be used to find velocity and acceleration.
If the time $t$ and displacement $x$ of a particle moving along the positive X-axis are related as $t = (x^2 - 1)^{\frac{1}{2}}$, then find the acceleration of the particle in terms of $x$.
Objective Physics Vol-1
MCQ
Concept: Basic Indefinite Integrals:
1. $\int e^x dx = e^x + C$
2. $\int \frac{1}{x} dx = \log_e x + C$
3. Power Rule of Integration: $\int x^n dx = \frac{x^{n+1}}{n + 1} + C$ (where $n \neq -1$)
4. $\int 1 dx = x + C$
Evaluate the integral $\int \left( e^x + \frac{1}{x} + 2x^2 + 3 \right) dx$.
Objective Physics Vol-1
MCQ
Concept: Basic Indefinite Integrals:
1. $\int \cos x dx = \sin x + C$
2. Power Rule of Integration: $\int x^n dx = \frac{x^{n+1}}{n + 1} + C$ (where $n \neq -1$)
3. $\int \frac{1}{x} dx = \log_e x + C$
Evaluate the integral $\int \left( \cos x + 3x^{1/2} + \frac{3}{x} + \frac{4}{x^2} \right) dx$.
Objective Physics Vol-1
MCQ
Concept: Integration by linear substitution rule:
If $\int f(x) dx = F(x) + C$, then $\int f(ax + b) dx = \frac{F(ax + b)}{\frac{d}{dx}(ax + b)} + C = \frac{F(ax + b)}{a} + C$
Standard Power Rule of Integration:
$\int x^n dx = \frac{x^{n+1}}{n + 1} + C$
Evaluate the integral $\int (2x + 1)^3 dx$.
Objective Physics Vol-1
MCQ
Concept: Integration by linear substitution rule:
If $\int f(x) dx = F(x) + C$, then $\int f(ax + b) dx = \frac{F(ax + b)}{\frac{d}{dx}(ax + b)} + C = \frac{F(ax + b)}{a} + C$
Standard Logarithmic Integral:
$\int \frac{1}{x} dx = \log_e x + C$
Evaluate the integral $\int \left( \frac{1}{a - x} \right) dx$.
Objective Physics Vol-1
MCQ
Concept: Integration using the power rule formula $\int f'(g(x)) dx = \frac{f(g(x))}{\frac{d}{dx}(g(x))} + C$
Evaluate $\int (x^2 + 3x + 4)^4 dx$
Objective Physics Vol-1
MCQ
Concept: Integration of a composite function using the formula $\int f'(g(x)) dx = \frac{f(g(x))}{\frac{d}{dx}(g(x))} + C$
Evaluate $\int \sin(2x^2) dx$
Objective Physics Vol-1
MCQ
Concept: Evaluation of definite integrals using fundamental theorem of calculus: $\int_a^b f'(x) dx = \vert{}f(x)\vert{}_a^b = f(b) - f(a)$
Evaluate $\int_0^2 (4x^3 + 2x^2 + 2x + 1) dx$
Objective Physics Vol-1
MCQ
Concept: Definite integration of trigonometric functions: $\int_a^b (\sin x + \cos x) dx = [-\cos x + \sin x]_a^b$
Evaluate $\int_0^{\pi/4} (\sin x + \cos x) dx$
Objective Physics Vol-1
MCQ
Concept: Definite integral of logarithmic function: $\int_a^b \frac{1}{x} dx = \vert{}\log_e x\vert{}_a^b = \log_e b - \log_e a = \log_e (\frac{b}{a})$
Evaluate $\int_2^4 \frac{dx}{x}$
Objective Physics Vol-1
MCQ
Concept: Definite integration of exponential function using $\int_a^b e^{f(x)} dx = \left[ \frac{e^{f(x)}}{\frac{d}{dx}(f(x))} \right]_a^b$
Evaluate $\int_1^2 e^{(x + 4)} dx$
Objective Physics Vol-1
MCQ
Concept: Definite integration using the formula $\int_a^b \cos(f(x)) dx = \left[ \frac{\sin(f(x))}{\frac{d}{dx}(f(x))} \right]_a^b$
Evaluate $\int_0^{\pi/4} \cos(2x^2 + x) dx$
Objective Physics Vol-1
MCQ
Concept: Definite integration using logarithmic rule $\int_a^b \frac{1}{f(x)} dx = \left[ \frac{\log_e(f(x))}{\frac{d}{dx}(f(x))} \right]_a^b$
Evaluate $\int_1^2 \frac{dx}{3x + 4}$
Objective Physics Vol-1
MCQ
Concept: Definite integration of a polynomial function $\int_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a)$
Evaluate $\int_0^4 (3x^2 + 4x + 5) dx$
Objective Physics Vol-1
MCQ
Concept: Kinematics using integration: Acceleration is the rate of change of velocity ($a = \frac{dv}{dt}$) and velocity is the rate of change of position ($v = \frac{dx}{dt}$). Integrating acceleration gives velocity, and integrating velocity gives displacement.
A particle is moving under constant acceleration $a = 3t + 4t^2$. If the position and velocity of the particle at start (i.e. $t = 0$) are $x_0$ and $v_0$ respectively, find the displacement $x - x_0$ as a function of time $t$.
Objective Physics Vol-1
MCQ
Concept: Average value of a time-dependent function $f(t)$ over an interval from $t = t_1$ to $t = t_2$ is given by $\bar{f} = \frac{\int_{t_1}^{t_2} f(t) dt}{\int_{t_1}^{t_2} dt}$.
The velocity of a particle is given by $v = v_0 \sin\omega t$, where $v_0$ is constant and $\omega = \frac{2\pi}{T}$. Find the average velocity in the time interval $t = 0$ to $t = \frac{T}{2}$.
Objective Physics Vol-1
MCQ
Concept: Indefinite integration using standard formulas: $\int \sin x dx = -\cos x + C$, $\int \frac{1}{x} dx = \ln\vert{}x\vert{} + C$, and $\int x^n dx = \frac{x^{n+1}}{n+1} + C$.
Evaluate $\int \left(\sin x + \frac{1}{x} + 2\frac{1}{x^2} + 3x^3\right) dx$
Objective Physics Vol-1
MCQ
Concept: Indefinite integration using basic rules: $\int \cos x dx = \sin x$, $\int e^x dx = e^x$, and $\int x^n dx = \frac{x^{n+1}}{n+1}$.
Evaluate $\int \left(3\cos x + e^x + 4x^2 + x + 5\right) dx$
Objective Physics Vol-1
MCQ
Concept: Integration of a squared polynomial function using expansion and term-by-term power rule: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$
Evaluate $\int (2x^2 + 4x + 1)^2 dx$
Objective Physics Vol-1
MCQ
Concept: Integration of rational functions using standard formula $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$ or logarithmic substitution method for non-standard functions.
Evaluate $\int \frac{1}{x^2 + 2} dx$
Objective Physics Vol-1
MCQ
Concept: Integration of trigonometric function with linear argument using $\int \cos(ax + b) dx = \frac{\sin(ax + b)}{a} + C$
Evaluate $\int \cos(x + 2) dx$
Objective Physics Vol-1
MCQ
Concept: Definite integration using standard rules: $\int_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a)$, where $\int x dx = \frac{x^2}{2}$, $\int \frac{1}{x} dx = \ln\vert{}x\vert{}$, and $\int 1 dx = x$.
Evaluate $\int_1^3 \left(4x + \frac{1}{x} + 1\right) dx$
Objective Physics Vol-1
MCQ
Concept: Definite integration of trigonometric functions: $\int_a^b \sin x dx = [-\cos x]_a^b$ and $\int_a^b \cos x dx = [\sin x]_a^b$.
Evaluate $\int_0^{\pi/4} (\sin x - \cos x) dx$
Objective Physics Vol-1
MCQ
Concept: Definite integration of rational functions by completing the square using the standard integral formula $\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log_e \left\vert{}\frac{x - a}{x + a}\right\vert{} + C$.
Evaluate $\int_0^2 \frac{dx}{x^2 + 4x + 1}$
Objective Physics Vol-1
MCQ
Concept: Definite integration of polynomial functions using term-by-term integration: $\int_a^b x^n dx = \left[\frac{x^{n+1}}{n+1}\right]_a^b = \frac{b^{n+1} - a^{n+1}}{n+1}$.
Evaluate $\int_1^3 (4x^3 + 3x^2 + 2x + 1) dx$
Objective Physics Vol-1
MCQ
Concept: Kinematics using integration: Acceleration is the rate of change of velocity ($a = \frac{dv}{dt}$). Integrating acceleration with respect to time gives the velocity as a function of time.
A particle is moving in a straight line under acceleration $a = kt$, where $k$ is a constant. Find the velocity in terms of $t$, if the motion starts from rest.
Objective Physics Vol-1
MCQ
Concept: Average velocity over a time interval $[t_1, t_2]$ is given by $\bar{v} = \frac{\int_{t_1}^{t_2} v dt}{t_2 - t_1}$. The required time interval is determined by solving for $t_1$ and $t_2$ from the given velocity function.
A particle is moving in a straight line such that its velocity varies as $v = v_0 e^{-\lambda t}$, where $\lambda$ is a constant. Find the average velocity during the time interval in which the velocity decreases from $v_0$ to $\frac{v_0}{2}$.
Objective Physics Vol-1
MCQ
Concept: Area under curves using definite integration: The total area bounded by multiple functions in an interval is evaluated by splitting the region into sub-intervals where each curve forms the upper boundary, using $A = \int y dx$.
Find the area of the region in the first quadrant enclosed by the X-axis, the line $y = x$, and the circle $x^2 + y^2 = 32$.
Objective Physics Vol-1
MCQ
Concept: Area bounded by a parabola and a line using integration: Due to symmetry about the X-axis, the total area is given by $A = 2 \int_0^a y_{\text{parabola}} dx = 2 \int_0^a \sqrt{4x} dx$.
Find the area of the region bounded by the curve $y^2 = 4x$ and the line $x = 4$.
Objective Physics Vol-1
MCQ
Concept: Area under curves using definite integration: When a curve crosses the X-axis within the interval $[a, b]$, the total area is given by $A = \int_a^b \vert{}y\vert{} dx = \int_a^c (-y) dx + \int_c^b y dx$, where $c$ is the point of intersection with the X-axis.
Find the area of the region bounded by the line $y = 3x + 2$, the X-axis and the ordinates $x = -1$ and $x = 1$.
Objective Physics Vol-1
MCQ
Concept: Area between two intersecting curves: The area $A$ bounded by two curves $y_1 = f(x)$ and $y_2 = g(x)$ from $x = a$ to $x = b$ is given by $A = \int_a^b [g(x) - f(x)] dx$, where $g(x) \ge f(x)$ on $[a, b]$.
Find the area of the region bounded by the curve $y = x^3$, $y = x + 6$, and $x = 0$.
Objective Physics Vol-1
MCQ
Concept: To find the area bounded by two curves $y_1(x)$ and $y_2(x)$, first determine their points of intersection by setting $y_1(x) = y_2(x)$. Solving this equation gives the limits of integration $x_1$ and $x_2$. The area $A$ enclosed between the curves from $x_1$ to $x_2$ is given by the definite integral:
$A = \int_{x_1}^{x_2} (y_{\text{upper}} - y_{\text{lower}}) \, dx$
Find the area of the region included between the parabola $y = \frac{3x^2}{4}$ and the line $3x - 2y + 12 = 0$.
Objective Physics Vol-1
MCQ
Concept: The equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ represents an ellipse centered at the origin. The region above the $X$-axis corresponds to the upper half of the ellipse where $y \ge 0$, bounded by $x = -a$ and $x = a$. The area $A$ of this region is given by:
$A = \int_{-a}^{a} y \, dx = \int_{-a}^{a} b \sqrt{1 - \frac{x^2}{a^2}} \, dx$
Alternatively, since the total area of an ellipse is $\pi a b$, the area above the $X$-axis is half of the total area, which is $\frac{1}{2} \pi a b$.
For the curve $\frac{x^2}{4} + \frac{y^2}{9} = 1$, evaluate the area of the region under the curve and above the $X$-axis.