📘 Mathematics in Physics — JEE Mains PYQs
Step-by-Step Solutions with Detailed Explanations
Master Mathematics in Physics for JEE Main with previous year questions on Error Analysis, Significant Figures, Vectors, Calculus, and Graphical Analysis. Each solution is explained in a simple, step-by-step manner to boost your score.
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Mathematics in Physics — JEE Mains PYQs
JEE Main 2018–2025Q1. JEE Main 2025 (3 April Shift 2)
A physical quantity \(C\) is related to four other quantities \(p, q, r\) and \(s\) as follows:
\(C = \frac{pq^2}{r^3\sqrt{s}}\).
The percentage errors in the measurement of \(p, q, r\) and \(s\) are 1%, 2%, 3% and 2% respectively. The percentage error in the measurement of \(C\) will be _____ %.
The percentage errors in the measurement of \(p, q, r\) and \(s\) are 1%, 2%, 3% and 2% respectively. The percentage error in the measurement of \(C\) will be _____ %.
✅ Answer & Explanation:
Correct Answer: 15
Using the error propagation formula:
\(\frac{\Delta C}{C} = \frac{\Delta p}{p} + 2\frac{\Delta q}{q} + 3\frac{\Delta r}{r} + \frac{1}{2}\frac{\Delta s}{s}\)
\(\text{% error in } C = 1\% + 2(2\%) + 3(3\%) + \frac{1}{2}(2\%)\)
\(= 1 + 4 + 9 + 1 = 15\%\).
Using the error propagation formula:
\(\frac{\Delta C}{C} = \frac{\Delta p}{p} + 2\frac{\Delta q}{q} + 3\frac{\Delta r}{r} + \frac{1}{2}\frac{\Delta s}{s}\)
\(\text{% error in } C = 1\% + 2(2\%) + 3(3\%) + \frac{1}{2}(2\%)\)
\(= 1 + 4 + 9 + 1 = 15\%\).
Q2. JEE Main 2025 (29 Jan Shift 2)
A physical quantity \(Q\) is related to four observables \(a, b, c, d\) as follows :
\(Q = \frac{ab^4}{cd}\)
where, \(a = (60 \pm 3)\text{Pa}\); \(b = (20 \pm 0.1)\text{m}\); \(c = (40 \pm 0.2)\text{Nsm}^{-2}\) and \(d = (50 \pm 0.1)\text{m}\), then the percentage error in \(Q\) is \(\frac{x}{1000}\), where \(x =\) ________.
\(Q = \frac{ab^4}{cd}\)
where, \(a = (60 \pm 3)\text{Pa}\); \(b = (20 \pm 0.1)\text{m}\); \(c = (40 \pm 0.2)\text{Nsm}^{-2}\) and \(d = (50 \pm 0.1)\text{m}\), then the percentage error in \(Q\) is \(\frac{x}{1000}\), where \(x =\) ________.
✅ Answer & Explanation:
Correct Answer: 7700
Percentage error calculation:
\(\frac{\Delta Q}{Q} = \frac{\Delta a}{a} + 4\frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d}\)
\(= \frac{3}{60} + 4(\frac{0.1}{20}) + \frac{0.2}{40} + \frac{0.1}{50}\)
\(= 0.05 + 0.02 + 0.005 + 0.002 = 0.077\)
Percentage error = \(0.077 \times 100 = 7.7\%\).
Given \(\text{error} = \frac{x}{1000} \implies 7.7 = \frac{x}{1000} \implies x = 7700\).
Percentage error calculation:
\(\frac{\Delta Q}{Q} = \frac{\Delta a}{a} + 4\frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d}\)
\(= \frac{3}{60} + 4(\frac{0.1}{20}) + \frac{0.2}{40} + \frac{0.1}{50}\)
\(= 0.05 + 0.02 + 0.005 + 0.002 = 0.077\)
Percentage error = \(0.077 \times 100 = 7.7\%\).
Given \(\text{error} = \frac{x}{1000} \implies 7.7 = \frac{x}{1000} \implies x = 7700\).
Q3. JEE Main 2025 (24 Jan Shift 1)
For an experimental expression \(y = \frac{32.3 \times 1125}{27.4}\), where all the digits are significant.
Then to report the value of \(y\) we should write:
(1) 1326.19 (2) 1330
(3) 1326.186 (4) 1326.2
(1) 1326.19 (2) 1330
(3) 1326.186 (4) 1326.2
✅ Answer & Explanation:
Correct Option: (2)
In multiplication and division, the final result should be rounded off to the same number of significant figures as are possessed by the least precise term.
- \(32.3\) has 3 significant figures.
- \(27.4\) has 3 significant figures.
- \(1125\) has 4 significant figures.
The result must have 3 significant figures. Rounding the calculated value (\(\approx 1326.186\)) to 3 significant figures gives 1330.
In multiplication and division, the final result should be rounded off to the same number of significant figures as are possessed by the least precise term.
- \(32.3\) has 3 significant figures.
- \(27.4\) has 3 significant figures.
- \(1125\) has 4 significant figures.
The result must have 3 significant figures. Rounding the calculated value (\(\approx 1326.186\)) to 3 significant figures gives 1330.
Q4. JEE Main 2025 (22 Jan Shift 2)
The maximum percentage error in the measurement of density of a wire is:
[Given, mass of wire = \((0.60 \pm 0.003)\text{g}\), radius of wire = \((0.50 \pm 0.01)\text{cm}\), length of wire = \((10.00 \pm 0.05)\text{cm}\)]
(1) 8 (2) 5 (3) 4 (4) 7
[Given, mass of wire = \((0.60 \pm 0.003)\text{g}\), radius of wire = \((0.50 \pm 0.01)\text{cm}\), length of wire = \((10.00 \pm 0.05)\text{cm}\)]
(1) 8 (2) 5 (3) 4 (4) 7
✅ Answer & Explanation:
Correct Option: (2)
Density \(\rho = \frac{m}{V} = \frac{m}{\pi r^2 L}\).
Relative error \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta L}{L}\).
\(\text{% error} = \left[ \frac{0.003}{0.60} + 2\left(\frac{0.01}{0.50}\right) + \frac{0.05}{10.00} \right] \times 100\)
\(= [0.005 + 0.04 + 0.005] \times 100 = 0.05 \times 100 = 5\%\).
Density \(\rho = \frac{m}{V} = \frac{m}{\pi r^2 L}\).
Relative error \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta L}{L}\).
\(\text{% error} = \left[ \frac{0.003}{0.60} + 2\left(\frac{0.01}{0.50}\right) + \frac{0.05}{10.00} \right] \times 100\)
\(= [0.005 + 0.04 + 0.005] \times 100 = 0.05 \times 100 = 5\%\).
Q5. JEE Main 2024 (01 Feb Shift 2)
Match List - I with List - II:
Choose the correct answer from the options given below:
(1) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
(2) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(3) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
(4) (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
| List - I (Number) | List - II (Significant figure) |
|---|---|
| (A) 1001 | (I) 3 |
| (B) 010.1 | (II) 4 |
| (C) 100.100 | (III) 5 |
| (D) 0.0010010 | (IV) 6 |
(1) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
(2) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(3) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
(4) (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
✅ Answer & Explanation:
Correct Option: (3)
Rules for Significant Figures:
- (A) 1001: 4 sig figs (all non-zero digits and zeros between them are significant). [II]
- (B) 010.1: 3 sig figs (leading zeros are not significant). [I]
- (C) 100.100: 6 sig figs (trailing zeros after a decimal point are significant). [IV]
- (D) 0.0010010: 5 sig figs (leading zeros are placeholders; non-zero digits and trailing zeros after them are significant). [III]
Rules for Significant Figures:
- (A) 1001: 4 sig figs (all non-zero digits and zeros between them are significant). [II]
- (B) 010.1: 3 sig figs (leading zeros are not significant). [I]
- (C) 100.100: 6 sig figs (trailing zeros after a decimal point are significant). [IV]
- (D) 0.0010010: 5 sig figs (leading zeros are placeholders; non-zero digits and trailing zeros after them are significant). [III]
Q6. JEE Main 2022 (25 Jul Shift 2)
The maximum error in the measurement of resistance, current and time for which current flows in an electrical circuit are 1%, 2% and 3% respectively.
The maximum percentage error in the detection of the dissipated heat will be:
(1) 2 (2) 4
(3) 6 (4) 8
(1) 2 (2) 4
(3) 6 (4) 8
✅ Answer & Explanation:
Correct Option: (4)
Dissipated heat is given by Joule's law: \(H = I^2Rt\).
The relative error in heat calculation is:
\(\frac{\Delta H}{H} = 2\frac{\Delta I}{I} + \frac{\Delta R}{R} + \frac{\Delta t}{t}\)
Substituting the given percentage errors:
\(\text{% error in } H = 2(2\%) + 1\% + 3\%\)
\(= 4\% + 1\% + 3\% = 8\%\).
Dissipated heat is given by Joule's law: \(H = I^2Rt\).
The relative error in heat calculation is:
\(\frac{\Delta H}{H} = 2\frac{\Delta I}{I} + \frac{\Delta R}{R} + \frac{\Delta t}{t}\)
Substituting the given percentage errors:
\(\text{% error in } H = 2(2\%) + 1\% + 3\%\)
\(= 4\% + 1\% + 3\% = 8\%\).
Q7. JEE Main 2021 (31 Aug Shift 2)
Statement-I: Two forces \((\vec{P}+\vec{Q})\) and \((\vec{P}-\vec{Q})\) where \(\vec{P} \perp \vec{Q}\),
when act at an angle \(\theta_1\) each other, the magnitude of their resultant is \(\sqrt{3(P^2+Q^2)}\),
when they act at an angle \(\theta_2\), the magnitude of their resultant becomes \(\sqrt{2(P^2+Q^2)}\).
This is possible only when \(\theta_1 < \theta_2\).
Statement-II: In the situation given above, \(\theta_1 = 60^\circ\) and \(\theta_2 = 90^\circ\).
In the light of the above statement, choose the most appropriate answer from the options given below :
(1) Statement I is false but Statement II is true.
(2) Both Statement I and Statement II are true.
(3) Both Statement I and Statement II are false.
(4) Statement I is true but Statement II is false.
Statement-II: In the situation given above, \(\theta_1 = 60^\circ\) and \(\theta_2 = 90^\circ\).
In the light of the above statement, choose the most appropriate answer from the options given below :
(1) Statement I is false but Statement II is true.
(2) Both Statement I and Statement II are true.
(3) Both Statement I and Statement II are false.
(4) Statement I is true but Statement II is false.
✅ Answer & Explanation:
Correct Option: (2)
Since \(\vec{P} \perp \vec{Q}\), the square of the magnitude of both forces is \(A^2 = B^2 = P^2 + Q^2\).
Resultant \(R = \sqrt{A^2 + B^2 + 2AB\cos\theta} = \sqrt{2(P^2+Q^2)(1+\cos\theta)}\).
- For \(R = \sqrt{3(P^2+Q^2)}\): \(3 = 2(1+\cos\theta_1) \implies \cos\theta_1 = 1/2 \implies \theta_1 = 60^\circ\).
- For \(R = \sqrt{2(P^2+Q^2)}\): \(2 = 2(1+\cos\theta_2) \implies \cos\theta_2 = 0 \implies \theta_2 = 90^\circ\).
Since \(60^\circ < 90^\circ\), both statements are correct.
Since \(\vec{P} \perp \vec{Q}\), the square of the magnitude of both forces is \(A^2 = B^2 = P^2 + Q^2\).
Resultant \(R = \sqrt{A^2 + B^2 + 2AB\cos\theta} = \sqrt{2(P^2+Q^2)(1+\cos\theta)}\).
- For \(R = \sqrt{3(P^2+Q^2)}\): \(3 = 2(1+\cos\theta_1) \implies \cos\theta_1 = 1/2 \implies \theta_1 = 60^\circ\).
- For \(R = \sqrt{2(P^2+Q^2)}\): \(2 = 2(1+\cos\theta_2) \implies \cos\theta_2 = 0 \implies \theta_2 = 90^\circ\).
Since \(60^\circ < 90^\circ\), both statements are correct.
Q8. JEE Main 2021 (22 Jul Shift 1)
Three students \(S_1, S_2\) and \(S_3\) perform an experiment for determining the acceleration due to gravity \((g)\) using a simple pendulum.
They use different lengths of pendulum and record time for different number of oscillations. The observations are:
(Least count of length = 0.1 cm, least count for time = 0.1 s)
If \(E_1, E_2\) and \(E_3\) are the percentage errors in \(g\) for students 1, 2 and 3, respectively, then the minimum percentage error is obtained by student no _____.
| Student | Length (cm) | Oscillations (n) | Total time (s) |
|---|---|---|---|
| 1 | 64.0 | 8 | 128.0 |
| 2 | 64.0 | 4 | 64.0 |
| 3 | 20.0 | 4 | 36.0 |
If \(E_1, E_2\) and \(E_3\) are the percentage errors in \(g\) for students 1, 2 and 3, respectively, then the minimum percentage error is obtained by student no _____.
✅ Answer & Explanation:
Correct Answer: 1
Using \(g = 4\pi^2 \frac{L}{T^2}\), where \(T = \frac{t_{total}}{n}\).
Percentage error \(\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta t}{t}\).
- Student 1: \(\frac{0.1}{64} + 2(\frac{0.1}{128}) = 0.00156 + 0.00156 = 0.00312\)
- Student 2: \(\frac{0.1}{64} + 2(\frac{0.1}{64}) = 0.00156 + 0.00312 = 0.00468\)
- Student 3: \(\frac{0.1}{20} + 2(\frac{0.1}{36}) = 0.005 + 0.0055 = 0.0105\)
Student 1 has the minimum error.
Using \(g = 4\pi^2 \frac{L}{T^2}\), where \(T = \frac{t_{total}}{n}\).
Percentage error \(\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta t}{t}\).
- Student 1: \(\frac{0.1}{64} + 2(\frac{0.1}{128}) = 0.00156 + 0.00156 = 0.00312\)
- Student 2: \(\frac{0.1}{64} + 2(\frac{0.1}{64}) = 0.00156 + 0.00312 = 0.00468\)
- Student 3: \(\frac{0.1}{20} + 2(\frac{0.1}{36}) = 0.005 + 0.0055 = 0.0105\)
Student 1 has the minimum error.
Q9. JEE Main 2019 (Jan Shift 1)
Two vectors \(\vec{A}\) and \(\vec{B}\) have magnitudes 3 and 4 respectively. What should be the angle between them for the magnitude of their resultant to be 5?
✅ Answer & Explanation:
Correct Answer: \(90^\circ\)
Resultant magnitude: \(R = \sqrt{A^2 + B^2 + 2AB\cos\theta}\).
Given \(R = 5, A = 3, B = 4\):
\(25 = 9 + 16 + 2(3)(4)\cos\theta \implies 25 = 25 + 24\cos\theta\)
\(\cos\theta = 0 \implies \theta = 90^\circ\).
This is the classic 3-4-5 triangle.
Resultant magnitude: \(R = \sqrt{A^2 + B^2 + 2AB\cos\theta}\).
Given \(R = 5, A = 3, B = 4\):
\(25 = 9 + 16 + 2(3)(4)\cos\theta \implies 25 = 25 + 24\cos\theta\)
\(\cos\theta = 0 \implies \theta = 90^\circ\).
This is the classic 3-4-5 triangle.
Q10. JEE Main 2020 (Sep Shift 2)
If \(\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k}\) and \(\vec{B} = -\hat{i} + 2\hat{j} + 4\hat{k}\), find the value of \((\vec{A} \cdot \vec{B})\).
✅ Answer & Explanation:
Correct Answer: 0
Dot product: \(\vec{A} \cdot \vec{B} = A_xB_x + A_yB_y + A_zB_z\)
\(= (2)(-1) + (3)(2) + (-1)(4) = -2 + 6 - 4 = 0\).
Since the dot product is zero, the vectors are perpendicular.
Dot product: \(\vec{A} \cdot \vec{B} = A_xB_x + A_yB_y + A_zB_z\)
\(= (2)(-1) + (3)(2) + (-1)(4) = -2 + 6 - 4 = 0\).
Since the dot product is zero, the vectors are perpendicular.
Q11. JEE Main 2018 (April Shift 2)
Find a unit vector perpendicular to both \(\vec{A} = \hat{i} + \hat{j} + \hat{k}\) and \(\vec{B} = \hat{i} - \hat{j} + \hat{k}\).
✅ Answer & Explanation:
Correct Answer: \(\frac{\hat{i} - \hat{k}}{\sqrt{2}}\)
A vector perpendicular to both is \(\vec{A} \times \vec{B}\):
\(\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(1\cdot1 - 1\cdot(-1)) - \hat{j}(1\cdot1 - 1\cdot1) + \hat{k}(1\cdot(-1) - 1\cdot1)\)
\(= \hat{i}(1+1) - \hat{j}(0) + \hat{k}(-2) = 2\hat{i} - 2\hat{k}\).
Magnitude: \(|2\hat{i} - 2\hat{k}| = \sqrt{4+4} = 2\sqrt{2}\).
Unit vector = \(\frac{2\hat{i} - 2\hat{k}}{2\sqrt{2}} = \frac{\hat{i} - \hat{k}}{\sqrt{2}}\).
A vector perpendicular to both is \(\vec{A} \times \vec{B}\):
\(\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(1\cdot1 - 1\cdot(-1)) - \hat{j}(1\cdot1 - 1\cdot1) + \hat{k}(1\cdot(-1) - 1\cdot1)\)
\(= \hat{i}(1+1) - \hat{j}(0) + \hat{k}(-2) = 2\hat{i} - 2\hat{k}\).
Magnitude: \(|2\hat{i} - 2\hat{k}| = \sqrt{4+4} = 2\sqrt{2}\).
Unit vector = \(\frac{2\hat{i} - 2\hat{k}}{2\sqrt{2}} = \frac{\hat{i} - \hat{k}}{\sqrt{2}}\).
Q12. JEE Main 2021 (March Shift 1)
The position of a particle is given by \(x(t) = 2t^3 - 9t^2 + 12t + 5\). Find the time when the particle's velocity is zero.
✅ Answer & Explanation:
Correct Answer: \(t = 1\) s and \(t = 2\) s
Velocity \(v(t) = \frac{dx}{dt} = 6t^2 - 18t + 12\).
Set \(v(t) = 0\): \(6t^2 - 18t + 12 = 0 \implies t^2 - 3t + 2 = 0\)
\(\implies (t-1)(t-2) = 0 \implies t = 1\) s and \(t = 2\) s.
Velocity \(v(t) = \frac{dx}{dt} = 6t^2 - 18t + 12\).
Set \(v(t) = 0\): \(6t^2 - 18t + 12 = 0 \implies t^2 - 3t + 2 = 0\)
\(\implies (t-1)(t-2) = 0 \implies t = 1\) s and \(t = 2\) s.
Q13. JEE Main 2022 (July Shift 1)
A force \(F(x) = 4x + 3\) N acts on a particle. Calculate the work done by the force in moving the particle from \(x=1\) m to \(x=3\) m.
✅ Answer & Explanation:
Correct Answer: 22 J
Work done \(W = \int_{x=1}^{3} F(x) \, dx = \int_{1}^{3} (4x + 3) \, dx\)
\(= [2x^2 + 3x]_{1}^{3} = (2(9) + 9) - (2(1) + 3) = (18+9) - (2+3) = 27 - 5 = 22\) J.
Work done \(W = \int_{x=1}^{3} F(x) \, dx = \int_{1}^{3} (4x + 3) \, dx\)
\(= [2x^2 + 3x]_{1}^{3} = (2(9) + 9) - (2(1) + 3) = (18+9) - (2+3) = 27 - 5 = 22\) J.
Q14. JEE Main 2023 (April Shift 1)
A velocity-time graph for a particle is shown. Which of the following is the correct acceleration-time graph?
[Graph: A straight line with positive slope starting from origin]
(1) [Graph: Horizontal line] (2) [Graph: Line with positive slope]
(3) [Graph: Line with negative slope] (4) [Graph: Parabola]
[Graph: A straight line with positive slope starting from origin]
(1) [Graph: Horizontal line] (2) [Graph: Line with positive slope]
(3) [Graph: Line with negative slope] (4) [Graph: Parabola]
✅ Answer & Explanation:
Correct Option: (1)
The velocity-time graph is a straight line with a positive slope. The slope of this graph gives acceleration. Since the slope is constant, the acceleration is constant and positive.
Therefore, the acceleration-time graph will be a horizontal line (constant value) above the time axis.
The velocity-time graph is a straight line with a positive slope. The slope of this graph gives acceleration. Since the slope is constant, the acceleration is constant and positive.
Therefore, the acceleration-time graph will be a horizontal line (constant value) above the time axis.
Q15. JEE Main 2024 (Jan Shift 1)
In an experiment, the relation between two quantities is \(y = a e^{bx}\). To plot a straight-line graph, what should be plotted on the y-axis and x-axis?
✅ Answer & Explanation:
Correct Answer: \(\ln(y)\) on y-axis, \(x\) on x-axis
To linearize the equation \(y = a e^{bx}\), take the natural logarithm of both sides:
\(\ln(y) = \ln(a) + bx\).
This is of the form \(Y = mX + c\), where \(Y = \ln(y)\), \(X = x\), the slope \(m = b\), and the intercept \(c = \ln(a)\).
Therefore, plotting \(\ln(y)\) versus \(x\) will give a straight line.
To linearize the equation \(y = a e^{bx}\), take the natural logarithm of both sides:
\(\ln(y) = \ln(a) + bx\).
This is of the form \(Y = mX + c\), where \(Y = \ln(y)\), \(X = x\), the slope \(m = b\), and the intercept \(c = \ln(a)\).
Therefore, plotting \(\ln(y)\) versus \(x\) will give a straight line.
Q16. JEE Main 2020 (Sep Shift 2)
Solve the differential equation: \(\frac{dy}{dx} + 2y = 4\).
✅ Answer & Explanation:
Correct Answer: \(y = 2 + Ce^{-2x}\)
This is a linear differential equation of the form \(\frac{dy}{dx} + Py = Q\), where \(P=2, Q=4\).
The integrating factor is \(I.F. = e^{\int P dx} = e^{2x}\).
The solution is \(y \cdot e^{2x} = \int 4 \cdot e^{2x} dx + C\)
\(= 2e^{2x} + C\).
Therefore, \(y = 2 + Ce^{-2x}\).
This is a linear differential equation of the form \(\frac{dy}{dx} + Py = Q\), where \(P=2, Q=4\).
The integrating factor is \(I.F. = e^{\int P dx} = e^{2x}\).
The solution is \(y \cdot e^{2x} = \int 4 \cdot e^{2x} dx + C\)
\(= 2e^{2x} + C\).
Therefore, \(y = 2 + Ce^{-2x}\).
📘 Exam Preparation Tip for Mathematics in Physics
- Error Analysis: Memorize propagation formulas for addition, subtraction, multiplication, and division. Practice maximum possible error problems.
- Significant Figures: Know the rules for counting sig figs and rounding off. Pay attention to zeroes.
- Vectors: Master addition, subtraction, dot product, and cross product. Practice unit vector problems.
- Calculus: Be comfortable with differentiation (polynomials, trig, exponential) and integration. Apply them to physics problems like velocity, acceleration, work.
- Graphs: Interpret slope and area under the curve. Learn to linearize equations (e.g., using logarithms).