CUET UG Mathematics · Paper 1

20 MCQs · Section A: Common 6, Section B1: Mathematics 7, Section B2: Applied Mathematics 7

As in the real paper: everyone answers Section A, then Section B1 (Mathematics) or B2 (Applied Mathematics).

Tap an option to answer

Section 1 · Section A: Common

Q1Modern Math › Matrices & DeterminantsEasy
Asked in CBSE Class 12 board 2026
If $A = \begin{bmatrix} 1 & a & b \\ -1 & 2 & c \\ 0 & 5 & 3 \end{bmatrix}$ is a symmetric matrix, then the value of $3a + b + c$ is
Correct option A2
A symmetric matrix satisfies $A' = A$ , that is $a_{ij} = a_{ji}$ for every entry.
Compare entries: $a = a_{21} = -1$ , $b = a_{31} = 0$ , $c = a_{32} = 5$ .
So $3a + b + c = 3(-1) + 0 + 5 = 2$ .
The Fast Way
Mirror each unknown across the main diagonal.
Read $a = -1$ , $b = 0$ , $c = 5$ straight from the lower triangle.
$-3 + 0 + 5 = 2$ .
Q2Calculus › IntegralsEasy
Asked in CBSE Class 12 board 2026
If $\int \frac{3ax}{b^2 + c^2x^2}\,dx = A\log\left|b^2 + c^2x^2\right| + K$ , then the value of $A$ is
Correct option D $\frac{3a}{2c^2}$
Put $t = b^2 + c^2x^2$ , so $dt = 2c^2x\,dx$ and $x\,dx = \frac{dt}{2c^2}$ .
The integral becomes $\frac{3a}{2c^2}\int \frac{dt}{t} = \frac{3a}{2c^2}\log|t| + K$ .
So the integral is $\frac{3a}{2c^2}\log\left|b^2 + c^2x^2\right| + K$ , and $A = \frac{3a}{2c^2}$ .
The Fast Way
Spot the pattern $\int \frac{f'(x)}{f(x)}\,dx = \log|f(x)|$ .
The denominator's derivative is $2c^2x$ , so $3ax = \frac{3a}{2c^2} \times 2c^2x$ .
So $A = \frac{3a}{2c^2}$ .
Q3Calculus › Derivatives & Their ApplicationsMedium
Two positive numbers $x$ and $y$ are such that $x + y = 16$ . The maximum value of $xy^3$ is
Correct option B6912
Put $x = 16 - y$ , so $P = (16 - y)y^3 = 16y^3 - y^4$ for $0 < y < 16$ .
$\frac{dP}{dy} = 48y^2 - 4y^3 = 4y^2(12 - y)$ , which is zero at $y = 12$ .
$\frac{d^2P}{dy^2} = 96y - 12y^2 = -576 < 0$ at $y = 12$ , so this is a maximum.
Then $x = 4$ and $P = 4 \times 12^3 = 6912$ .
Trap: splitting equally ( $x = y = 8$ ) gives only 4096.
The Fast Way
For a fixed sum, $x^m y^n$ is largest when the sum is split in the ratio $m : n$ .
Here $m : n = 1 : 3$ , so $x = 4$ and $y = 12$ .
$4 \times 1728 = 6912$ .
Q4Calculus › Differential EquationsEasy
The solution curve of the differential equation $\frac{dy}{dx} = \frac{2x}{y}$ that passes through the point $(1, 2)$ is
Correct option C $y^2 = 2x^2 + 2$
Separate the variables: $y\,dy = 2x\,dx$ .
Integrate: $\frac{y^2}{2} = x^2 + C$ .
At $(1, 2)$ : $2 = 1 + C$ , so $C = 1$ .
The curve is $\frac{y^2}{2} = x^2 + 1$ , that is $y^2 = 2x^2 + 2$ .
The Fast Way
Every option passes through $(1, 2)$ , so compare slopes there instead.
The equation needs $\frac{dy}{dx} = \frac{2 \times 1}{2} = 1$ at $(1, 2)$ .
$y^2 = 2x^2 + 2$ gives $2yy' = 4x$ , so $y' = 1$ ; the other three give $\frac{1}{2}$ , 2 and 2.
So the curve is $y^2 = 2x^2 + 2$ .
Q5Modern Math › ProbabilityEasy
Two cards are drawn one after the other, without replacement, from a well-shuffled pack of 52 cards. The probability that both cards are kings is
Correct option A $\frac{1}{221}$
There are 4 kings in 52 cards, so P(first card is a king) $= \frac{4}{52}$ .
After that, 3 kings remain among 51 cards, so P(second is a king) $= \frac{3}{51}$ .
P(both kings) $= \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}$ .
The Fast Way
Multiply the two draws and cancel at once.
$\frac{4}{52} \times \frac{3}{51} = \frac{1}{13} \times \frac{1}{17}$ .
So the probability is $\frac{1}{221}$ ; $\frac{1}{169}$ would be the with-replacement value.
Q6Modern Math › Matrices & DeterminantsMedium
If $A$ is a square matrix of order 3 with $|A| = 4$ , then $|\text{adj}(2A)|$ is equal to
Correct option B1024
For a square matrix $B$ of order $n$ , $|\text{adj}\,B| = |B|^{n-1}$ .
Take $B = 2A$ with $n = 3$ : $|2A| = 2^3|A| = 8 \times 4 = 32$ .
So $|\text{adj}(2A)| = |2A|^2 = 32^2 = 1024$ .
Trap: $|2A|$ is $2^3|A|$ , not $2|A|$ .
The Fast Way
Apply $|\text{adj}\,B| = |B|^{n-1}$ to $B = 2A$ directly.
$|2A| = 2^3 \times 4 = 32$ .
$32^2 = 1024$ .

Section 2 · Section B1: Mathematics

Q7Algebra › Relations & FunctionsEasy
Asked in CBSE Class 12 board 2024
A function $f : \mathbb{R}_+ \to \mathbb{R}$ (where $\mathbb{R}_+$ is the set of all non-negative real numbers) defined by $f(x) = 4x + 3$ is
Correct option Aone-one but not onto
One-one: $4x_1 + 3 = 4x_2 + 3$ gives $x_1 = x_2$ , so $f$ is one-one.
Onto: for $x \ge 0$ , $f(x) = 4x + 3 \ge 3$ , so the range is $[3, \infty)$ .
The codomain is $\mathbb{R}$ , and values such as 0 are never taken, so $f$ is not onto.
Hence $f$ is one-one but not onto.
The Fast Way
Compare the range with the codomain.
A line with slope 4 is one-one, and on $x \ge 0$ it starts at $f(0) = 3$ .
Range $[3, \infty) \ne \mathbb{R}$ , so it is one-one but not onto.
Q8Vectors & 3D Geometry › VectorsEasy
Asked in CBSE Class 12 board 2026
The value of $p$ for which the vectors $\hat{i} + 2\hat{j} + 3\hat{k}$ and $2\hat{i} - p\hat{j} + \hat{k}$ are perpendicular to each other is
Correct option C $\frac{5}{2}$
Perpendicular vectors have zero dot product.
$(1)(2) + (2)(-p) + (3)(1) = 0$
$5 - 2p = 0$ , so $p = \frac{5}{2}$ .
The Fast Way
Set the dot product to zero in one line.
$2 - 2p + 3 = 0$ gives $2p = 5$ .
So $p = \frac{5}{2}$ ; a sign slip gives $-\frac{5}{2}$ .
Q9Calculus › Continuity & DifferentiabilityEasy
If a function defined by $f(x) = \begin{cases} kx + 1, & x \le \pi \\ \cos x, & x > \pi \end{cases}$ is continuous at $x = \pi$ , then the value of $k$ is
Correct option D $-\frac{2}{\pi}$
For continuity at $x = \pi$ , the left limit, the right limit and $f(\pi)$ must be equal.
Left limit $= f(\pi) = k\pi + 1$ ; right limit $= \lim_{x \to \pi^+}\cos x = \cos\pi = -1$ .
So $k\pi + 1 = -1$ , which gives $k = -\frac{2}{\pi}$ .
The Fast Way
Make the two pieces meet at the joint.
$k\pi + 1 = \cos\pi = -1$ .
So $k = -\frac{2}{\pi}$ .
Q10Calculus › IntegralsMedium
Asked in CBSE Class 12 board 2026
$\int_{-1}^{1}(1 - |x|)\,dx$ is equal to
Correct option B $2\int_{-1}^{0}(1 + x)\,dx$
$1 - |x|$ is an even function, so $\int_{-1}^{1}(1 - |x|)\,dx = 2\int_{-1}^{0}(1 - |x|)\,dx$ .
On $[-1, 0]$ , $|x| = -x$ , so $1 - |x| = 1 + x$ .
Hence the integral equals $2\int_{-1}^{0}(1 + x)\,dx$ , whose value is 1.
Check: $2\int_{0}^{1}(1 + x)\,dx = 3$ and $2\int_{-1}^{0}(1 - x)\,dx = 3$ , so neither matches.
The Fast Way
Picture it: $y = 1 - |x|$ on $[-1, 1]$ is a triangle of base 2 and height 1, so the value is 1.
$2\int_{-1}^{0}(1 + x)\,dx = 2 \times \frac{1}{2} = 1$ ; the other two integrals give 3.
So it equals $2\int_{-1}^{0}(1 + x)\,dx$ .
Q11Vectors & 3D Geometry › Three-Dimensional GeometryMedium
The angle between the lines $\frac{x - 1}{2} = \frac{y + 3}{-1} = \frac{z - 2}{2}$ and $\frac{1 - x}{1} = \frac{y - 2}{2} = \frac{z + 1}{2}$ is
Correct option D $\frac{\pi}{2}$
Write the second line in standard form: $\frac{1 - x}{1} = \frac{x - 1}{-1}$ , so its direction ratios are $-1, 2, 2$ .
The first line has direction ratios $2, -1, 2$ .
$\cos\theta = \frac{|(2)(-1) + (-1)(2) + (2)(2)|}{3 \times 3} = \frac{0}{9} = 0$ .
So $\theta = \frac{\pi}{2}$ .
Trap: reading the second set as $1, 2, 2$ gives $\cos^{-1}\frac{4}{9}$ .
The Fast Way
Fix the $1 - x$ form first: it flips the sign of the $x$ direction ratio.
$(2, -1, 2) \cdot (-1, 2, 2) = -2 - 2 + 4 = 0$ .
A zero dot product means the angle is $\frac{\pi}{2}$ .
Q12Modern Math › ProbabilityMedium
A factory has two machines, X and Y. Machine X makes 60% of the items and machine Y makes 40%. Of the items made by X, 2% are defective, and of the items made by Y, 5% are defective. An item picked at random is found to be defective. The probability that it was made by machine Y is
Correct option A $\frac{5}{8}$
Let $E_1$ : made by X, $E_2$ : made by Y, and $F$ : the item is defective.
$P(E_1) = 0.6$ , $P(E_2) = 0.4$ , $P(F|E_1) = 0.02$ , $P(F|E_2) = 0.05$ .
Total probability: $P(F) = 0.6 \times 0.02 + 0.4 \times 0.05 = 0.012 + 0.020 = 0.032$ .
Bayes' theorem: $P(E_2|F) = \frac{0.020}{0.032} = \frac{5}{8}$ .
The Fast Way
Run Bayes on 1000 imaginary items.
X gives $600 \times 2\% = 12$ defectives and Y gives $400 \times 5\% = 20$ .
Share from Y $= \frac{20}{12 + 20} = \frac{5}{8}$ .

Q13Geometry › Inverse Trigonometric FunctionsMedium

Match List I with List II.

List I	List II
P. $\sin^{-1}\left(\sin\frac{3\pi}{5}\right)$	1. $-\frac{\pi}{4}$
Q. $\cos^{-1}\left(\cos\frac{7\pi}{6}\right)$	2. $\frac{5\pi}{6}$
R. $\tan^{-1}\left(\tan\frac{3\pi}{4}\right)$	3. $\frac{\pi}{3}$
S. $\cos^{-1}\left(\cos\left(-\frac{\pi}{3}\right)\right)$	4. $\frac{2\pi}{5}$

Choose the correct answer.

Correct option CP-4, Q-2, R-1, S-3

P: $\frac{3\pi}{5}$ is outside $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , and $\sin\frac{3\pi}{5} = \sin\left(\pi - \frac{3\pi}{5}\right) = \sin\frac{2\pi}{5}$ , so P gives $\frac{2\pi}{5}$ .

Q: $\frac{7\pi}{6}$ is outside $[0, \pi]$ , and $\cos\frac{7\pi}{6} = \cos\left(2\pi - \frac{7\pi}{6}\right) = \cos\frac{5\pi}{6}$ , so Q gives $\frac{5\pi}{6}$ .

R: $\tan\frac{3\pi}{4} = \tan\left(\frac{3\pi}{4} - \pi\right) = \tan\left(-\frac{\pi}{4}\right)$ , and $-\frac{\pi}{4}$ lies in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ , so R gives $-\frac{\pi}{4}$ .

S: $\cos\left(-\frac{\pi}{3}\right) = \cos\frac{\pi}{3}$ and $\frac{\pi}{3} \in [0, \pi]$ , so S gives $\frac{\pi}{3}$ .

So the match is P-4, Q-2, R-1, S-3.

The Fast Way

Settle one row, then let the options do the rest.

$\sin^{-1}\left(\sin\frac{3\pi}{5}\right) = \pi - \frac{3\pi}{5} = \frac{2\pi}{5}$ , so P-4 and only two options survive.

$\tan^{-1}\left(\tan\frac{3\pi}{4}\right) = \frac{3\pi}{4} - \pi = -\frac{\pi}{4}$ , so R-1.

That leaves P-4, Q-2, R-1, S-3.

Section 3 · Section B2: Applied Mathematics

Q14Number System › Modulo ArithmeticEasy
The value of $-70 \bmod 13$ is
Correct option C8
$a \bmod n$ is the remainder $r$ with $0 \le r < n$ when $a = nq + r$ .
$-70 = 13 \times (-6) + 8$ , and $0 \le 8 < 13$ .
So $-70 \bmod 13 = 8$ .
Trap: $-70 = 13 \times (-5) - 5$ gives $-5$ , but the remainder must be non-negative.
The Fast Way
Reduce $70$ first, then take the complement from 13.
$70 = 5 \times 13 + 5$ , so $-70 \equiv -5 \equiv 13 - 5 \pmod{13}$ .
So the value is 8.
Q15Financial Mathematics › DepreciationEasy
Asked in CBSE Class 12 board 2026
A machine costs ₹45,000 with an estimated useful life of 5 years and a scrap value of ₹10,000. The annual depreciation of the machine is
Correct option B₹7,000
Linear method: annual depreciation $= \frac{\text{cost} - \text{scrap value}}{\text{useful life}}$ .
$= \frac{45000 - 10000}{5} = \frac{35000}{5} = 7000$ .
So the annual depreciation is ₹7,000.
The Fast Way
Spread the depreciable amount evenly over the life.
$(45 - 10)$ thousand $\div 5 = 7$ thousand.
So it is ₹7,000 a year.
Q16Arithmetic › Boats & StreamsMedium
A boat goes 24 km upstream and 36 km downstream in 6 hours. It goes 36 km upstream and 24 km downstream in 6 hours 30 minutes. The speed of the boat in still water is
Correct option D10 km/h
Let the upstream speed be $u$ km/h and the downstream speed be $d$ km/h.
$\frac{24}{u} + \frac{36}{d} = 6$ and $\frac{36}{u} + \frac{24}{d} = 6.5$ .
Adding: $60\left(\frac{1}{u} + \frac{1}{d}\right) = 12.5$ ; subtracting: $12\left(\frac{1}{u} - \frac{1}{d}\right) = 0.5$ .
So $\frac{1}{u} + \frac{1}{d} = \frac{5}{24}$ and $\frac{1}{u} - \frac{1}{d} = \frac{1}{24}$ , giving $u = 8$ and $d = 12$ .
Boat speed $= \frac{u + d}{2} = \frac{8 + 12}{2} = 10$ km/h (and the stream flows at 2 km/h).
The Fast Way
Guess round speeds that make the first trip whole hours: upstream 8 (24 km in 3 h) and downstream 12 (36 km in 3 h), total 6 h.
Check the second trip: $\frac{36}{8} + \frac{24}{12} = 4.5 + 2 = 6.5$ hours, so both fit.
Boat speed $= \frac{8 + 12}{2} = 10$ km/h.
Q17Financial Mathematics › EMIMedium
A loan of ₹1,00,000 is to be repaid in 12 equal monthly instalments by the reducing balance method, at 12% per annum compounded monthly. Taking $(1.01)^{-12} = 0.8874$ , the EMI is approximately
Correct option A₹8,881
Monthly rate $i = \frac{12\%}{12} = 0.01$ and $n = 12$ .
EMI $= \frac{P \times i}{1 - (1 + i)^{-n}} = \frac{100000 \times 0.01}{1 - 0.8874} = \frac{1000}{0.1126}$ .
EMI $\approx 8881$ , that is about ₹8,881.
The Fast Way
Use EMI $= \frac{Pi}{1 - (1 + i)^{-n}}$ with $i = 0.01$ .
$\frac{1000}{0.1126} \approx 8881$ .
It must lie between ₹8,333 (no interest) and ₹9,333 (flat interest), so ₹8,881.
Q18Statistics › Probability DistributionsEasy
The marks of students in a test are normally distributed with mean 60 and standard deviation 8. Given $P(0 < Z < 1.5) = 0.4332$ , the percentage of students who scored more than 72 marks is
Correct option C6.68%
Standardise: $Z = \frac{X - \mu}{\sigma} = \frac{72 - 60}{8} = 1.5$ .
$P(X > 72) = P(Z > 1.5) = 0.5 - P(0 < Z < 1.5) = 0.5 - 0.4332 = 0.0668$ .
So 6.68% of the students scored more than 72.
The Fast Way
Convert to $Z$ , then use tail $= 0.5 -$ table area.
$Z = \frac{12}{8} = 1.5$ , so the tail is $0.5 - 0.4332 = 0.0668$ .
That is 6.68%.
Q19Calculus › Consumer & Producer SurplusMedium
The demand function for a product is $p = 50 - x^2$ , where $p$ is the price (in ₹) and $x$ is the quantity demanded. If the market price is ₹14, the consumer surplus is
Correct option D₹144
At $p_0 = 14$ : $50 - x^2 = 14$ , so $x_0 = 6$ .
CS $= \int_0^{6}(50 - x^2)\,dx - p_0x_0$ .
$\int_0^{6}(50 - x^2)\,dx = 300 - \frac{216}{3} = 300 - 72 = 228$ .
CS $= 228 - 14 \times 6 = 228 - 84 = 144$ , that is ₹144.
The Fast Way
For $p = a - x^2$ , CS $= \frac{2}{3}x_0^3$ (the area between the curve and the price line).
$50 - x^2 = 14$ gives $x_0 = 6$ .
CS $= \frac{2}{3} \times 216 = 144$ , that is ₹144.
Q20Statistics › Time SeriesMedium
The sales (in ₹ lakh) of a firm for the years 2019 to 2023 are 12, 15, 14, 18 and 21 respectively. A straight-line trend $y = a + bx$ is fitted by the method of least squares. The trend value for 2024 is
Correct option B22.3
Take the origin at 2021, so $x = -2, -1, 0, 1, 2$ and $\sum x = 0$ .
$\sum y = 80$ , $\sum x^2 = 10$ , $\sum xy = -24 - 15 + 0 + 18 + 42 = 21$ .
$a = \frac{\sum y}{n} = 16$ and $b = \frac{\sum xy}{\sum x^2} = \frac{21}{10} = 2.1$ .
For 2024, $x = 3$ : $y = 16 + 2.1 \times 3 = 22.3$ (₹ lakh).
The Fast Way
Centre the years so $\sum x = 0$ ; then $a = \bar{y}$ and $b = \frac{\sum xy}{\sum x^2}$ .
$a = 16$ and $b = \frac{21}{10} = 2.1$ .
2024 is 3 steps from 2021: $16 + 6.3 = 22.3$ .