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58 Questions Start Objective Physics Vol-1 Test
All Exams Objective Physics Vol-1 DPT
Q1 Objective Physics Vol-1 Trigonometry MCQ
21 Jul 2026
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.
A.
$\cos \theta = \frac{3}{5}$, $\tan \theta = \frac{4}{3}$, $\cot \theta = \frac{3}{4}$, $\sec \theta = \frac{5}{3}$, $\csc \theta = \frac{5}{4}$
B.
$\cos \theta = \frac{4}{5}$, $\tan \theta = \frac{3}{4}$, $\cot \theta = \frac{4}{3}$, $\sec \theta = \frac{5}{4}$, $\csc \theta = \frac{5}{3}$
C.
$\cos \theta = \frac{3}{4}$, $\tan \theta = \frac{5}{3}$, $\cot \theta = \frac{3}{5}$, $\sec \theta = \frac{4}{3}$, $\csc \theta = \frac{5}{4}$
D.
$\cos \theta = \frac{3}{5}$, $\tan \theta = \frac{3}{4}$, $\cot \theta = \frac{4}{3}$, $\sec \theta = \frac{5}{4}$, $\csc \theta = \frac{5}{3}$
Q2 Objective Physics Vol-1 Trigonometry MCQ
24 Jul 2026
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) $\sec 120^\circ$
A.
(i) $\frac{1}{\sqrt{2}}$, (ii) $-1$, (iii) $\frac{\sqrt{3}}{2}$, (iv) $2$
B.
(i) $-\frac{1}{\sqrt{2}}$, (ii) $1$, (iii) $\frac{1}{2}$, (iv) $-2$
C.
(i) $-\frac{1}{2}$, (ii) $1$, (iii) $\frac{1}{\sqrt{2}}$, (iv) $-2$
D.
(i) $-\frac{1}{\sqrt{2}}$, (ii) $-1$, (iii) $\frac{1}{2}$, (iv) $2$
Q3 Objective Physics Vol-1 Trigonometry MCQ
24 Jul 2026
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) $\tan 75^\circ$
A.
(i) $\frac{\sqrt{3}+1}{2\sqrt{2}}$, (ii) $\frac{\sqrt{3}-1}{\sqrt{3}+1}$
B.
(i) $\frac{\sqrt{3}-1}{2\sqrt{2}}$, (ii) $\frac{\sqrt{3}+1}{\sqrt{3}-1}$
C.
(i) $\frac{\sqrt{3}-1}{2}$, (ii) $\frac{\sqrt{3}+1}{\sqrt{3}}$
D.
(i) $\frac{1}{2\sqrt{2}}$, (ii) $\frac{\sqrt{3}-1}{\sqrt{3}+1}$
Q4 Objective Physics Vol-1 Trigonometry MCQ
24 Jul 2026
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.
A.
$\sin \theta = \frac{5}{13}$, $\tan \theta = -\frac{5}{12}$
B.
$\sin \theta = -\frac{5}{13}$, $\tan \theta = -\frac{5}{12}$
C.
$\sin \theta = -\frac{5}{13}$, $\tan \theta = \frac{5}{12}$
D.
$\sin \theta = \frac{5}{13}$, $\tan \theta = \frac{5}{12}$
Q5 Objective Physics Vol-1 Trigonometry MCQ
24 Jul 2026
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.
A.
$\sin \theta = \frac{3}{5}$, $\cos \theta = -\frac{4}{5}$, $\csc \theta = \frac{5}{3}$, $\sec \theta = -\frac{5}{4}$, $\cot \theta = -\frac{4}{3}$
B.
$\sin \theta = -\frac{3}{5}$, $\cos \theta = \frac{4}{5}$, $\csc \theta = -\frac{5}{3}$, $\sec \theta = \frac{5}{4}$, $\cot \theta = -\frac{4}{3}$
C.
$\sin \theta = \frac{4}{5}$, $\cos \theta = -\frac{3}{5}$, $\csc \theta = \frac{5}{4}$, $\sec \theta = -\frac{5}{3}$, $\cot \theta = -\frac{3}{4}$
D.
$\sin \theta = \frac{3}{5}$, $\cos \theta = \frac{4}{5}$, $\csc \theta = \frac{5}{3}$, $\sec \theta = \frac{5}{4}$, $\cot \theta = \frac{4}{3}$
Q6 Objective Physics Vol-1 Trigonometry MCQ
24 Jul 2026
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) $\sin(-330^\circ)$
A.
(i) $\sqrt{2}$, (ii) $\frac{\sqrt{3}}{2}$, (iii) $-\frac{1}{2}$
B.
(i) $-\sqrt{2}$, (ii) $-\frac{\sqrt{3}}{2}$, (iii) $\frac{1}{2}$
C.
(i) $-\frac{1}{\sqrt{2}}$, (ii) $-\frac{1}{2}$, (iii) $\frac{\sqrt{3}}{2}$
D.
(i) $-\sqrt{2}$, (ii) $\frac{1}{2}$, (iii) $-\frac{\sqrt{3}}{2}$
Q7 Objective Physics Vol-1 Trigonometry MCQ
24 Jul 2026
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) $\cot 105^\circ$
A.
(i) $\sqrt{2} + \sqrt{6}$, (ii) $\frac{1+\sqrt{3}}{1-\sqrt{3}}$
B.
(i) $\sqrt{2} - \sqrt{6}$, (ii) $\frac{1-\sqrt{3}}{1+\sqrt{3}}$
C.
(i) $\sqrt{6} - \sqrt{2}$, (ii) $\frac{\sqrt{3}-1}{1+\sqrt{3}}$
D.
(i) $-\sqrt{2} - \sqrt{6}$, (ii) $\frac{\sqrt{3}+1}{1-\sqrt{3}}$
Q8 Objective Physics Vol-1 Differentiation MCQ
24 Jul 2026
Concept: Power Rule of Differentiation:
$\frac{d}{dx}(x^n) = n x^{n-1}$
Differentiate the function $y = x^{-3}$ with respect to $x$.
A.
$-3x^{-4}$
B.
$-3x^{-2}$
C.
$3x^{-4}$
D.
$-\frac{3}{x^3}$
Q9 Objective Physics Vol-1 Differentiation MCQ
24 Jul 2026
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 $x$.
A.
$30x^5 + 12x^3 - 6x^2 + 2x$
B.
$30x^4 + 12x^2 - 6x + 2$
C.
$30x^4 + 12x^2 - 6x$
D.
$24x^4 + 12x^2 - 6x + 2$
Q10 Objective Physics Vol-1 Differentiation MCQ
24 Jul 2026
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 $x$.
A.
$3x^2 + 2x + 1$
B.
$2x^2 + 4x + 1$
C.
$3x^2 + 4x + 1$
D.
$3x^2 + 4x$
Q11 Objective Physics Vol-1 Differentiation MCQ
24 Jul 2026
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 $x$.
A.
$\frac{3x^3 + 3x^2 - 4}{(x + 1)^2}$
B.
$\frac{2x^3 + 3x^2 + 4}{(x + 1)^2}$
C.
$\frac{x^3 + 3x^2 - 4}{(x + 1)^2}$
D.
$\frac{2x^3 + 3x^2 - 4}{(x + 1)^2}$
Q12 Objective Physics Vol-1 Differentiation MCQ
24 Jul 2026
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 $x$.
A.
$\cos x + e^x$
B.
$-\cos x + e^x$
C.
$\cos x - e^x$
D.
$\sin x + e^x$
Q13 Objective Physics Vol-1 Differentiation MCQ
24 Jul 2026
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 $x$.
A.
$6x + \frac{1}{x} + e^x$
B.
$6x + \frac{1}{x} + 4e^x + 5$
C.
$6x + \frac{1}{x} + 4e^x$
D.
$3x + \frac{1}{x} + 4e^x$
Q14 Objective Physics Vol-1 Differentiation MCQ
24 Jul 2026
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 $x$.
A.
$e^x (\sec^2 x - \tan x)$
B.
$e^x \sec^2 x$
C.
$e^x (\tan x - \sec^2 x)$
D.
$e^x (\sec^2 x + \tan x)$
Q15 Objective Physics Vol-1 Differentiation MCQ
24 Jul 2026
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 $x$.
A.
$2x \sin(x^2 + 5)$
B.
$2x \cos(x^2 + 5)$
C.
$\cos(x^2 + 5)$
D.
$-2x \cos(x^2 + 5)$
Q16 Objective Physics Vol-1 Differentiation MCQ
24 Jul 2026
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.
A.
$400$ and $600$
B.
$500$ and $500$
C.
$300$ and $700$
D.
$200$ and $800$
Q17 Objective Physics Vol-1 Differentiation MCQ
24 Jul 2026
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.
A.
$\frac{\beta}{2\gamma}$
B.
$\frac{2\gamma}{\beta}$
C.
$\frac{\beta}{\gamma}$
D.
$\frac{\alpha}{2\beta}$
Q18 Objective Physics Vol-1 Differentiation MCQ
24 Jul 2026
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.
A.
$t = 0$ to $t = 6\text{ s}$
B.
$t = 0$ to $t = 3\text{ s}$
C.
$t = 3\text{ s}$ to $t = 6\text{ s}$
D.
$t > 3\text{ s}$
Q19 Objective Physics Vol-1 Differentiation MCQ
24 Jul 2026
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 $x$.
A.
$12x^3 - \frac{4}{x^3} + \frac{1}{x}$
B.
$12x^3 + \frac{4}{x^3} + \frac{1}{x}$
C.
$12x^3 - \frac{2}{x^3} + \frac{1}{x}$
D.
$3x^3 - \frac{4}{x^3} + \log x$
Q20 Objective Physics Vol-1 Differentiation MCQ
24 Jul 2026
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 $x$.
A.
$3x^2 + 2x + 1$
B.
$3x^2 + 4x + 1$
C.
$2x^2 + 4x + 1$
D.
$3x^2 + 4x$
Q21 Objective Physics Vol-1 Differentiation MCQ
24 Jul 2026
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 $x$.
A.
$\frac{3x^2 + 3x}{(x + 1)^2}$
B.
$\frac{6x^2 + 6x}{(x + 1)^2}$
C.
$\frac{3x^2 + 6x}{(x + 1)^2}$
D.
$\frac{3x^2 - 6x}{(x + 1)^2}$
Q22 Objective Physics Vol-1 Differentiation MCQ
24 Jul 2026
Concept: Standard Derivative of Trigonometric Function:
$\frac{d}{dx}(\sin x) = \cos x$
Differentiate the function $y = \sin x$ with respect to $x$.
A.
$-\cos x$
B.
$\cos x$
C.
$\tan x$
D.
$-\sin x$
Q23 Objective Physics Vol-1 Differentiation MCQ
24 Jul 2026
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 $x$.
A.
$(2x + 3) \sec^2(x^2 + 3x + 1)$
B.
$\sec^2(x^2 + 3x + 1)$
C.
$(2x + 3) \tan(x^2 + 3x + 1)$
D.
$(x + 3) \sec^2(x^2 + 3x + 1)$
Q24 Objective Physics Vol-1 Differentiation MCQ
24 Jul 2026
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 $x$.
A.
$3\sin x + 7e^x + \frac{2x}{x^2 + 1}$
B.
$-3\sin x + 7e^x + \frac{1}{x^2 + 1}$
C.
$-3\sin x + 7e^x + \frac{2x}{x^2 + 1}$
D.
$-3\sin x + e^x + \frac{2x}{x^2 + 1}$
Q25 Objective Physics Vol-1 Differentiation MCQ
24 Jul 2026
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?
A.
$v_{\text{max}} = 4\text{ m/s}$ at $t = 4\text{ s}$ and $v_{\text{min}} = -28\text{ m/s}$ at $t = 0\text{ s}$
B.
$v_{\text{max}} = 4\text{ m/s}$ at $t = 0\text{ s}$ and $v_{\text{min}} = -28\text{ m/s}$ at $t = 4\text{ s}$
C.
$v_{\text{max}} = 6\text{ m/s}$ at $t = 0\text{ s}$ and $v_{\text{min}} = -20\text{ m/s}$ at $t = 4\text{ s}$
D.
$v_{\text{max}} = 4\text{ m/s}$ at $t = 2\text{ s}$ and $v_{\text{min}} = -16\text{ m/s}$ at $t = 4\text{ s}$
Q26 Objective Physics Vol-1 Differentiation MCQ
24 Jul 2026
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 $x$.
A.
$\frac{1}{x^2}$
B.
$\frac{1}{x^3}$
C.
$\frac{1}{x^4}$
D.
$-\frac{1}{x^3}$
Q27 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
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 $\int \left( e^x + \frac{1}{x} + 2x^2 + 3 \right) dx$.
A.
$e^x + \log_e x + \frac{2}{3}x^3 + 3x + C$
B.
$e^x - \log_e x + \frac{2}{3}x^3 + 3x + C$
C.
$e^x + \log_e x + 2x^3 + 3x + C$
D.
$e^x + \frac{1}{x^2} + \frac{2}{3}x^3 + 3x + C$
Q28 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
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 $\int \left( \cos x + 3x^{1/2} + \frac{3}{x} + \frac{4}{x^2} \right) dx$.
A.
$\sin x + 2x^{3/2} + 3\log_e x + \frac{4}{x} + C$
B.
$\sin x + 3x^{3/2} + 3\log_e x - \frac{4}{x} + C$
C.
$\sin x + 2x^{3/2} + 3\log_e x - \frac{4}{x} + C$
D.
$-\sin x + 2x^{3/2} + 3\log_e x - \frac{4}{x} + C$
Q29 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
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 $\int (2x + 1)^3 dx$.
A.
$\frac{(2x + 1)^4}{4} + C$
B.
$\frac{(2x + 1)^4}{8} + C$
C.
$\frac{(2x + 1)^3}{6} + C$
D.
$\frac{(2x + 1)^4}{2} + C$
Q30 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
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 $\int \left( \frac{1}{a - x} \right) dx$.
A.
$\log_e(a - x) + C$
B.
$-\frac{1}{(a - x)^2} + C$
C.
$-\log_e(a - x) + C$
D.
$\frac{1}{a}\log_e(a - x) + C$
Q31 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
Concept: Integration using the power rule formula $\int f'(g(x)) dx = \frac{f(g(x))}{\frac{d}{dx}(g(x))} + C$
Evaluate $\int (x^2 + 3x + 4)^4 dx$
A.
$\frac{(x^2 + 3x + 4)^5}{5} + C$
B.
$\frac{(x^2 + 3x + 4)^5}{2x + 3} + C$
C.
$\frac{(x^2 + 3x + 4)^5}{5(2x + 3)} + C$
D.
$\frac{(x^2 + 3x + 4)^4}{5(2x + 3)} + C$
Q32 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
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 $\int \sin(2x^2) dx$
A.
$\frac{-\cos(2x^2)}{4x} + C$
B.
$\frac{\cos(2x^2)}{4x} + C$
C.
$\frac{-\sin(2x^2)}{4x} + C$
D.
$\frac{-\cos(2x^2)}{2x} + C$
Q33 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
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 $\int_0^2 (4x^3 + 2x^2 + 2x + 1) dx$
A.
$\frac{80}{3}$
B.
$\frac{82}{3}$
C.
$\frac{76}{3}$
D.
$\frac{85}{3}$
Q34 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
Concept: Definite integration of trigonometric functions: $\int_a^b (\sin x + \cos x) dx = [-\cos x + \sin x]_a^b$
Evaluate $\int_0^{\pi/4} (\sin x + \cos x) dx$
A.
$\sqrt{2} - 1$
B.
$\sqrt{2} + 1$
C.
$1$
D.
$0$
Q35 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
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 $\int_2^4 \frac{dx}{x}$
A.
$\log_e 4$
B.
$\log_e 3$
C.
$\log_e 2$
D.
$\log_e 8$
Q36 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
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 $\int_1^2 e^{(x + 4)} dx$
A.
$e^5 (e - 1)$
B.
$e^4 (e - 1)$
C.
$e^6 - e^4$
D.
$e^5 (e + 1)$
Q37 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
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 $\int_0^{\pi/4} \cos(2x^2 + x) dx$
A.
$\frac{1}{\pi + 1} \sin\left( \frac{\pi^2 + 8\pi}{16} \right)$
B.
$\frac{1}{\pi + 1} \sin\left( \frac{\pi^2 + 2\pi}{8} \right)$
C.
$\frac{1}{\pi + 1} \sin\left( \frac{\pi^2 + 4\pi}{8} \right)$
D.
$\frac{1}{\pi + 2} \sin\left( \frac{\pi^2 + 2\pi}{8} \right)$
Q38 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
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 $\int_1^2 \frac{dx}{3x + 4}$
A.
$\frac{1}{3} \log_e\left(\frac{10}{7}\right)$
B.
$\frac{1}{3} \log_e\left(\frac{7}{10}\right)$
C.
$\log_e\left(\frac{10}{7}\right)$
D.
$\frac{1}{4} \log_e\left(\frac{10}{7}\right)$
Q39 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
Concept: Definite integration of a polynomial function $\int_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a)$
Evaluate $\int_0^4 (3x^2 + 4x + 5) dx$
A.
$112$
B.
$120$
C.
$116$
D.
$124$
Q40 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
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 $t$.
A.
$v_0 t + t^3 + \frac{1}{3} t^4$
B.
$v_0 t + \frac{3}{2} t^2 + \frac{1}{3} t^4$
C.
$v_0 t + \frac{3}{2} t^2 + \frac{4}{3} t^3$
D.
$v_0 t + \frac{1}{2} t^3 + \frac{1}{3} t^4$
Q41 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
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 $t = \frac{T}{2}$.
A.
$\frac{v_0}{\pi}$
B.
$\frac{2v_0}{\pi}$
C.
$\frac{v_0}{2\pi}$
D.
$\frac{3v_0}{\pi}$
Q42 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
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 $\int \left(\sin x + \frac{1}{x} + 2\frac{1}{x^2} + 3x^3\right) dx$
A.
$-\cos x + \log_e x - \frac{2}{x} + \frac{3}{4} x^4 + C$
B.
$\cos x + \log_e x - \frac{2}{x} + \frac{3}{4} x^4 + C$
C.
$-\cos x - \log_e x + \frac{2}{x} + \frac{3}{4} x^4 + C$
D.
$-\cos x + \log_e x + \frac{2}{x} + 3x^4 + C$
Q43 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
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 $\int \left(3\cos x + e^x + 4x^2 + x + 5\right) dx$
A.
$3\sin x + e^x + 4x^3 + \frac{x^2}{2} + 5x + C$
B.
$-3\sin x + e^x + \frac{4}{3} x^3 + \frac{x^2}{2} + 5x + C$
C.
$3\sin x + e^x + \frac{4}{3} x^3 + \frac{x^2}{2} + 5x + C$
D.
$3\sin x + e^x + \frac{4}{3} x^3 + x^2 + 5x + C$
Q44 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
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 $\int (2x^2 + 4x + 1)^2 dx$
A.
$\frac{1}{12} \frac{(2x^2 + 4x + 1)^3}{(x + 1)} + C$
B.
$\frac{4}{5} x^5 + 4x^4 + \frac{20}{3} x^3 + 4x^2 + x + C$
C.
$\frac{1}{12} (2x^2 + 4x + 1)^3 + C$
D.
$\frac{4}{5} x^5 + 2x^4 + \frac{20}{3} x^3 + 2x^2 + x + C$
Q45 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
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 $\int \frac{1}{x^2 + 2} dx$
A.
$\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x}{\sqrt{2}}\right) + C$
B.
$\frac{1}{2x} \log_e(x^2 + 2) + C$
C.
$\frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + C$
D.
$\frac{1}{\sqrt{2}} \log_e(x^2 + 2) + C$
Q46 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
Concept: Integration of trigonometric function with linear argument using $\int \cos(ax + b) dx = \frac{\sin(ax + b)}{a} + C$
Evaluate $\int \cos(x + 2) dx$
A.
$-\sin(x + 2) + C$
B.
$\cos(x + 2) + C$
C.
$\sin(x + 2) + C$
D.
$\frac{\sin(x + 2)}{2} + C$
Q47 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
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 $\int_1^3 \left(4x + \frac{1}{x} + 1\right) dx$
A.
$16 + \log_e 3$
B.
$18 + \log_e 3$
C.
$18 - \log_e 3$
D.
$20 + \log_e 3$
Q48 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
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 $\int_0^{\pi/4} (\sin x - \cos x) dx$
A.
$\sqrt{2} - 1$
B.
$1 + \sqrt{2}$
C.
$1 - \sqrt{2}$
D.
$-\sqrt{2}$
Q49 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
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 $\int_0^2 \frac{dx}{x^2 + 4x + 1}$
A.
$\frac{1}{2\sqrt{3}} \log_e\left(\frac{13}{8}\right)$
B.
$\frac{1}{\sqrt{3}} \log_e\left(\frac{13}{8}\right)$
C.
$\frac{1}{2\sqrt{3}} \log_e 13$
D.
$\frac{1}{2} \log_e\left(\frac{13}{8}\right)$
Q50 Objective Physics Vol-1 Integrations MCQ
24 Jul 2026
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 $\int_1^3 (4x^3 + 3x^2 + 2x + 1) dx$
A.
$112$
B.
$116$
C.
$120$
D.
$124$