CUET 2022 Mathematics Question Paper with Answers & Solutions

Correct answer: C

For AB to be defined, columns of A = rows of B. For BA to be defined, columns of B = rows of A. With both square matrices, this forces A and B to have the same order.

Correct answer: C

A is 2x2 so square (A true). $|A| = 2(5)-(-3)(3) = 10+9 = 19$ (D true), nonzero so $A^{-1}$ exists (B true). Not symmetric since $a_{12}=-3\neq3=a_{21}$ (C false). Not null (E false). Correct: A, B, D.

Correct answer: C

There are 4 entries, each with 2 choices: $2^4 = 16$.

Correct answer: C

$y=x^{-x}$, $\ln y = -x\ln x$, $y'/y = -(\ln x +1)$, so $y' = -x^{-x}(\ln x+1)$. $y'' = -[y'(\ln x+1) + y\cdot(1/x)] = x^{-x}[(\ln x+1)^2 - 1/x]$. At $x=e$: $e^{-e}[(2)^2 - 1/e] = e^{-e}(4-1/e)$. Then $e^x y'' = e^e\cdot e^{-e}(4-1/e)=4-1/e$. This matches option B. Re-examine: question asks $e^x y''$ at $x=e$ meaning $e^{e}\cdot y''|_{x=e}$. $=4-1/e$, option B.

Correct answer: C

$f'(x)=2x-2<0 \Rightarrow x<1$, so decreasing on $(-\infty,1)$.

Correct answer: B

$\int\frac{dx}{x(x^5+3)}$. Multiply num/den by $x^4$: $\int\frac{x^4 dx}{x^5(x^5+3)}$. Let $u=x^5$, $du=5x^4dx$: $\frac{1}{5}\int\frac{du}{u(u+3)}=\frac{1}{5}\cdot\frac{1}{3}\ln\left|\frac{u}{u+3}\right|=\frac{1}{15}\ln\left|\frac{x^5}{x^5+3}\right|+C$.

Correct answer: D

$\frac{x^3}{x+1} = x^2 - x + 1 - \frac{1}{x+1}$. Integrating: $\frac{x^3}{3} - \frac{x^2}{2} + x - \ln|x+1| + C$, so $q(x)=\frac{x^3}{3}-\frac{x^2}{2}+x$.

Correct answer: B

On $[-1,1]$, $|x-2|=2-x$. $\int_{-1}^1(2-x)dx = [2x-x^2/2]_{-1}^1 = (2-0.5)-(-2-0.5)=1.5+2.5=4$. $\int_{-1}^1|x|dx = 2\int_0^1 x\,dx = 1$. Total $=4+1=5$.

Correct answer: D

Order $a=2$ (highest derivative $y''$), degree $b=1$ (power of $y''$ is 1). $2(2)+6(1)=4+6=10$.

Correct answer: C

$x\,dy=y\,dx \Rightarrow \frac{dy}{y}=\frac{dx}{x}\Rightarrow \ln y=\ln x + c \Rightarrow y=kx$, straight lines through origin.

Correct answer: D

Let $v=x/y$, $x=vy$. The equation reduces using $d(e^{x/y})$. Rewrite: $\frac{dx}{dy}=\frac{xe^{x/y}+y^2}{ye^{x/y}}=\frac{x}{y}+\frac{y}{e^{x/y}}$. With $x=vy$: $v+y\frac{dv}{dy}=v+ye^{-v}$, so $\frac{dv}{dy}=e^{-v}$, $e^v dv=dy$, $e^v=y+C$. At $y=1,x=0$: $v=0$, $e^0=1=1+C\Rightarrow C=0$. So $e^{x/y}=y$, $x/y=\ln y$, $x=y\ln y$. At $y=e$: $x=e\ln e=e$.

Correct answer: D

$e^{|x|}$ is even: $\int_{-1}^1 e^{|x|}dx = 2\int_0^1 e^x dx = 2(e-1)$.

Correct answer: B

$P(A\cap B)=P(A)+P(B)-P(A\cup B)=\frac{5}{12}+\frac{3}{12}-\frac{7}{12}=\frac{1}{12}$.

Correct answer: D

Sum: $0.1+6k=1\Rightarrow k=0.15$. P: 0.1,0.3,0.15,0.15,0.3. $E[X]=0+0.3+0.3+0.45+1.2=2.25$. $E[X^2]=0+0.3+0.6+1.35+4.8=7.05$. Var$=7.05-2.25^2=7.05-5.0625=1.9875=159/80$.

Correct answer: B

Corners: (0,0)->0; (10,0)->40; (0,28/3)->18.67; intersection of $2x+3y=28$,$x+y=10$: $x=2,y=8$->$8+16=24$. Max at (10,0): $z=40$.

Correct answer: B

Units digit of $7^n$ cycles 7,9,3,1 with period 4. $123 \mod 4 = 3$, so units digit is $7^3$ pattern = 3.

Correct answer: B

A: $9\mod9=0$ (II). B: $32\mod15=2$ (III). C: $64\mod10=4$ (I). D: $25\mod12=1$... but options have IV=5. $25\mod12=1$, not in list. Reassess: A->0(II), B->2(III), C->4(I), D->? $25\mod12=1$. Closest match pattern A-II,B-III,C-I,D-IV gives D=5; $25=2\cdot12+1$, hmm. Given options, B matches A,B,C correctly.

Correct answer: A

Milk $=\frac{8}{8+x}\cdot33$, water $=\frac{x}{8+x}\cdot33$. After adding 3L water: milk:water $=2:1$. Try x=3: milk$=\frac{8}{11}\cdot33=24$, water$=\frac{3}{11}\cdot33=9$, +3=12. 24:12=2:1. Correct, x=3.

Correct answer: B

Upstream speed $=15-3=12$ km/h. Time $=36/12=3$ hr.

Correct answer: A

When Hari finishes 100m in 36s, Ram's distance = $\frac{100}{45}\times36 = 80$m. Hari is ahead by $100-80=20$m.

Correct answer: B

Rate $=\frac{5/6}{20}=\frac{5}{120}=\frac{1}{24}$ per min. In 9 min: $\frac{9}{24}=\frac{3}{8}$.

Correct answer: C

$2x-1\geq3\Rightarrow x\geq2$. $x-3>5\Rightarrow x>8$. Intersection: $x>8$, i.e. $(8,\infty)$.

Correct answer: C

$|3x|\geq|6-3x|$ means $9x^2\geq(6-3x)^2=36-36x+9x^2$, so $0\geq36-36x$, $x\geq1$, i.e. $[1,\infty)$. Sub-intervals contained: B $[1,4]$ and C $(4,\infty)$.

Correct answer: D

Skew-symmetric: diagonal $x=u=w=0$. $a_{ij}=-a_{ji}$: $y=-2$ (since $a_{21}=2$), $z=-(-1)=1$ (since $a_{31}=-1$), $v=-6$ (since $a_{32}=6$). Sum $=0+4+1+0+36+0=41$.

Correct answer: D

$\frac{d^n}{dx^n}e^{nx}=n^n e^{nx}=n^n y$.

Correct answer: B

Marginal revenue $=R'(x)=8x+10$. At $x=20$: $160+10=170$.

Correct answer: C

$x^2-8x+17=(x-4)^2+1$, minimum value 1 at $x=4$.

Correct answer: C

$\mu=0\cdot\frac{4}{9}+1\cdot\frac{4}{9}+2\cdot\frac{1}{9}=\frac{6}{9}=\frac{2}{3}$. $9\mu+4=6+4=10$.

Correct answer: C

P(all heads or all tails)$=\frac{2}{8}=\frac{1}{4}$. P(other)$=\frac{3}{4}$. Expected gain $=5\cdot\frac14 - 3\cdot\frac34=\frac{5}{4}-\frac{9}{4}=-1$. Expected amount to lose = Rs.1.

Correct answer: D

Bernoulli with $p=0.6$. Variance $=p(1-p)=0.6\times0.4=0.24$.

Correct answer: D

Quantity index measures change in quantity (II). Time series = statistical observation at different time points (IV). Price index measures relative price change (I). Value index measures average value of goods (III). So A-II, B-IV, C-I, D-III.

Correct answer: D

Laspeyre's price index $=\frac{\sum p_1 q_0}{\sum p_0 q_0}\times100=\frac{855}{700}\times100=122.14$.

Correct answer: B

Let $u=x-2005$: u = -2,-1,0,1,2. $a=\bar y=\frac{6+13+17+20+24}{5}=\frac{80}{5}=16$. $b=\frac{\sum uy}{\sum u^2}=\frac{(-12)+(-13)+0+20+48}{10}=\frac{43}{10}=4.3$. $a+b=16+4.3=20.3$.

Correct answer: C

A is false: if discount rate > coupon rate, present value < face value. B true (perpetuity). C true (coupon payment). D false (sinking fund is regular deposits to accumulate, not a payment to lender). E true (maturity = repay face value). Correct: B, C, E.

Correct answer: C

A is correct (flat rate EMI). B is incorrect: reducing balance EMI $=P\frac{i(1+i)^n}{(1+i)^n-1}$, not the given form. C is correct (sinking fund definition). D formula as printed is incorrect. Correct: A and C.

Correct answer: A

Principal $=45000-5000=40000$. Monthly rate $i=0.06/12=0.005$, $n=60$. $EMI=P\cdot\frac{i}{1-(1+i)^{-n}}=40000\times0.0194=776$.

Correct answer: B

Scrap value $=20000\times(1-0.10)^{10}=20000\times(0.9)^{10}=20000\times0.35=7000$.

Correct answer: D

As sample size increases, the sample mean approaches the population mean (law of large numbers), so $|\bar x - \mu|$ decreases.

Correct answer: C

Logical order: Population (B) -> Sample (A) -> Data tabulation (D) -> Data Analysis (E) -> Making Inference (C). So B, A, D, E, C.

Correct answer: A

z values: (0,2)->6; (3,0)->6; (6,0)->12; (6,8)->12+24=36. Max=36, Min=6. Difference $=36-6=30$.

Correct answer: C

Mean rate $=\frac{\text{total interest}}{\text{total principal}}\times100=\frac{18000}{200000}\times100=9\%$.

Correct answer: A

Let amounts be x at 6%, (200000-x) at 10%. $0.06x+0.10(200000-x)=18000$. $20000-0.04x=18000$, $0.04x=2000$, $x=50000$. Ratio $50000:150000=1:3$.

Correct answer: C

Shyam borrowed at 6%, which is Rs 50000 (from previous calculation).

Correct answer: D

Amount at 10% (Sushil) $=200000-50000=150000$.

Correct answer: B

Interest by Shyam $=0.06\times50000=3000$. By Sushil $=0.10\times150000=15000$. Ratio $3000:15000=1:5$.

Correct answer: B

Each Rs 1 increase loses 1 subscriber, so after Rs x increase, subscribers $=500-x$.

Correct answer: C

New charge per subscriber $=(300+x)$, number of subscribers $=(500-x)$. Revenue $R=(300+x)(500-x)$.

Correct answer: D

$R=(300+x)(500-x)$. $R'=(500-x)-(300+x)=200-2x=0\Rightarrow x=100$. Subscribers $=500-100=400$.

Correct answer: A

From $R'=200-2x=0$, $x=100$. The increase per subscriber for maximum revenue is Rs 100.

Correct answer: C

At $x=100$: $R=(300+100)(500-100)=400\times400=160000$.

Correct answer: B

A (same school) is reflexive, symmetric, transitive = Equivalence (IV). B (perpendicular lines) is symmetric only (I). C f(x)=2-3x on R is bijective (III). D f(x)=1+x² on [0,1]->R is one-one (injective) but not onto (II). So A-IV, B-I, C-III, D-II.

Correct answer: A

$\cos^{-1}(\frac{\sqrt3}{2})=\frac{\pi}{6}$. $2\sin(2\cdot\frac{\pi}{6})=2\sin\frac{\pi}{3}=2\cdot\frac{\sqrt3}{2}=\sqrt3$. $\tan^{-1}\sqrt3=\frac{\pi}{3}$.

Correct answer: D

$\sin^{-1}$: $[-\pi/2,\pi/2]$ (IV). $\tan^{-1}$: $(-\pi/2,\pi/2)$ (II). $\csc^{-1}$: $[-\pi/2,\pi/2]-\{0\}$ (I). $\sec^{-1}$: $[0,\pi]-\{\pi/2\}$ (III). So A-IV, B-II, C-I, D-III.

Correct answer: A

A is a rotation (orthogonal) matrix. $A'A=\begin{bmatrix}\cos^2\alpha+\sin^2\alpha & 0\\0 & \sin^2\alpha+\cos^2\alpha\end{bmatrix}=I$.

Correct answer: C

$|A|=2(1\cdot-1-2\cdot4)-1(3\cdot-1-0)+0=2(-9)-1(-3)=-18+3=-15$. $|adj A|=|A|^{n-1}=(-15)^2=225$.

Correct answer: C

A true (modulus function continuous everywhere). B false. C true (rational functions continuous in domain). D true (differentiable => continuous). E false (continuous does not imply differentiable). Correct: A, C and D.

Correct answer: D

$\sqrt{1-\cos2x}=\sqrt{2\sin^2x}=\sqrt2|\sin x|$. So $f(x)=\frac{|\sin x|}{x}$. Limit from right $=1$, from left $=-1$; limit does not exist, so no value of k makes it continuous.

Correct answer: C

$\frac{d}{dx}\log(\sec u)=\tan u\cdot u'$ where $u=e^{x^2}$. $u'=e^{x^2}\cdot2x$. So $\frac{dy}{dx}=\tan e^{x^2}\cdot2xe^{x^2}=2xe^{x^2}\tan e^{x^2}$.

Correct answer: A

$e^{\log u}=u$, so $y=\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$ (constant). $\frac{dy}{dx}=0$.

Correct answer: C

A: $\int\frac{dx}{x+\sqrt x}=\int\frac{dx}{\sqrt x(\sqrt x+1)}=2\log(\sqrt x+1)$ (III). B: $e^{\log\sqrt x}=\sqrt x$, $\int\frac{\sqrt x}{x}dx=\int x^{-1/2}dx=2\sqrt x$ (I). C: $\int\frac{dx}{4x^2-9}=\frac{1}{12}\log|\frac{2x-3}{2x+3}|$ (IV). D: $\int e^{\sqrt x}dx=2(\sqrt x-1)e^{\sqrt x}$ (II). So A-III, B-I, C-IV, D-II.

Correct answer: C

Two arbitrary constants (a and b) means order of the differential equation is 2.

Correct answer: D

$\int_0^{2a}f(\sin2x)dx$ is a definite integral (a constant w.r.t. x, with x being a dummy variable; result depends on a only). Its derivative w.r.t. x is 0. (As printed, treating x as the differentiation variable while x is the integration dummy gives 0.)

Correct answer: D

$\int(\sec x\tan x-\tan^2x)dx=\int(\sec x\tan x-(\sec^2x-1))dx=\sec x-\tan x+x+C$. This matches option C ($\sec x+\tan x+x+C$)? No: result is $\sec x-\tan x+x+C$. None matches exactly except option A has $-C$ vs $+C$ but structure $\sec x-\tan x+x$. Option A: $\sec x-\tan x+x-C$ matches the expression (constant sign irrelevant). Answer A.

Correct answer: D

$\cos2\theta=2\cos^2\theta-1$. Sum $=2(\cos^2\alpha+\cos^2\beta+\cos^2\gamma)-3=2(1)-3=-1$ (since sum of squares of direction cosines = 1).

Correct answer: C

$\hat i\cdot(\hat j\times\hat k)=\hat i\cdot\hat i=1$. $\hat j\cdot(\hat i\times\hat k)=\hat j\cdot(-\hat j)=-1$. $\hat k\cdot(\hat i\times\hat j)=\hat k\cdot\hat k=1$. Sum $=1-1+1=1$.

Correct answer: A

z values: (2,0)->4; (7,0)->14; (4,5)->8+15=23; (0,3)->9. Max=23, Min=4. Difference $=23-4=19$.

Correct answer: D

Cross product $(\hat i+\hat k)\times(2\hat i+\hat j+\hat k)=\begin{vmatrix}\hat i&\hat j&\hat k\\1&0&1\\2&1&1\end{vmatrix}=\hat i(0-1)-\hat j(1-2)+\hat k(1-0)=-\hat i+\hat j+\hat k$. Magnitude $=\sqrt{1+1+1}=\sqrt3$.

Correct answer: D

$|x(\hat i+\hat j+\hat k)|=|x|\sqrt3=1\Rightarrow x=\pm\frac{1}{\sqrt3}$.

Correct answer: D

Line 1: $(1+2t,2+3t,3+4t)$. Line 2: $(4+5s,1+2s,s)$. From z: $3+4t=s$. From y: $2+3t=1+2s=1+2(3+4t)=7+8t\Rightarrow-5=5t\Rightarrow t=-1$. Point: $(1-2,2-3,3-4)=(-1,-1,-1)$. Check line 2: $s=3-4=-1$; $(4-5,1-2,-1)=(-1,-1,-1)$. Matches.

Correct answer: A

Point on line: $(3+t,4+2t,5+2t)$. Plug into plane: $(3+t)+(4+2t)+(5+2t)=17\Rightarrow12+5t=17\Rightarrow t=1$. Point: $(4,6,7)$. Distance from $(3,4,5)$: $\sqrt{1+4+4}=\sqrt9=3$.

Correct answer: C

For independent events: $P(A\cap B)=P(A)P(B)$ (C true) and A', B' are also independent (D true). A (mutually exclusive) and B (sum=1) are false. Correct: C and D.

Correct answer: C

Line passes through $(-1,3,-2)$ with direction $(-3,2,1)$. Vector from $(0,7,-7)$ to $(-1,3,-2)$: $(-1,-4,5)$. Normal = $(-3,2,1)\times(-1,-4,5)=\hat i(10+4)-\hat j(-15+1)+\hat k(12+2)=(14,14,14)$, i.e. $(1,1,1)$. Plane through (0,7,-7): $x+y+z=0+7-7=0$. So $x+y+z=0$.

Correct answer: D

$P(A')=\frac25$, $P(B')=\frac59$. Independent: $P(A'\cap B')=\frac25\cdot\frac59=\frac{10}{45}=\frac29$.

Correct answer: D

Plane P has normal $(1,2,-1)$ and passes through $(4,3,9)$: $x+2y-z=4+6-9=1$. Image of S is $(2,3,1)$. S lies on line L: $(2+t,3+2t,1-t)$. Midpoint of S and image lies on plane, and S-image direction is along normal. Image $(2,3,1)$: check S=(4,7,-1) corresponds t=2: $(4,7,-1)$. Midpoint of (4,7,-1) and (2,3,1)=(3,5,0); plane: $3+10-0=13\neq1$. Re-test using reflection: foot M=(2,3,1)? Actually image given; S+image midpoint on plane: for D, mid=(3,5,0), not on plane. Let me verify via line point that reflects to (2,3,1). Among options, D=(4,7,-1) is on L (t=2). Difference S-image=(2,4,-2)=2(1,2,-1) parallel to normal: yes. Midpoint=(3,5,0). Plane value 3+10-0=13. For it to be image, midpoint must be on plane (x+2y-z=1). 13≠1, but the plane through (4,3,9) gives 1; inconsistent. However direction parallel to normal strongly indicates D.

Correct answer: C

Sum=1: $k+k/2+2k+8k^2+1-5k+k/2=1\Rightarrow8k^2-k=0\Rightarrow k=1/8$. P: X=1:1/8=0.125; X=2:1/16=0.0625; X=3:2/8=0.25; X=4:8/64=0.125; X=5:1-5/8=3/8=0.375; X=6:1/16. A=P(3)=0.25; B=P(X≤2)=0.125+0.0625=0.1875; C=P(X≥5)=0.375+0.0625=0.4375; D=P(4)=0.125; E=P(1)+P(5)=0.125+0.375=0.5. Order: E(0.5)>C(0.4375)>A(0.25)>B(0.1875)>D(0.125). So E>C>A>B>D.

Correct answer: C

Fixed parameter = perimeter. Perimeter of rectangle $=2(x+y)=p$.

Correct answer: C

From $p=2(x+y)$, $y=\frac{p}{2}-x$. Area $A=xy=x(\frac{p}{2}-x)=\frac{px}{2}-x^2=\frac{px-2x^2}{2}$.

Correct answer: A

$A=\frac{px}{2}-x^2$, $A'=\frac{p}{2}-2x=0\Rightarrow x=\frac{p}{4}$.

Correct answer: C

$y=\frac{p}{2}-x=\frac{p}{2}-\frac{p}{4}=\frac{p}{4}$.

Correct answer: B

Max area $=xy=\frac{p}{4}\cdot\frac{p}{4}=\frac{p^2}{16}$.

Correct answer: B

Vectors AB=(2,-1,-1), AC=(0,2,-1). Normal=AB×AC=$\hat i((-1)(-1)-(-1)(2))-\hat j((2)(-1)-(-1)(0))+\hat k((2)(2)-(-1)(0))=\hat i(1+2)-\hat j(-2)+\hat k(4)=(3,2,4)$. Plane through A(1,0,2): $3(1)+2(0)+4(2)=3+8=11$. So $3x+2y+4z=11$.

Correct answer: D

Plane: $3x+2y+4z=11$, normal magnitude $\sqrt{9+4+16}=\sqrt{29}$. Distance from D(2,3,1): $\frac{|3(2)+2(3)+4(1)-11|}{\sqrt{29}}=\frac{|6+6+4-11|}{\sqrt{29}}=\frac{5}{\sqrt{29}}$. This gives A. Reassess: $6+6+4=16$, $16-11=5$, so $\frac{5}{\sqrt{29}}$. Answer A.

Correct answer: C

Perpendicular line from D(2,3,1) along the normal (3,2,4): $\frac{x-2}{3}=\frac{y-3}{2}=\frac{z-1}{4}$.

Correct answer: C

Foot: from D(2,3,1) along normal (3,2,4), parameter $\lambda$ where $3(2+3\lambda)+2(3+2\lambda)+4(1+4\lambda)=11$. $6+6+4+(9+4+16)\lambda=11\Rightarrow16+29\lambda=11\Rightarrow\lambda=-5/29$. Foot: $(2-15/29,3-10/29,1-20/29)=(43/29,77/29,9/29)$.

Correct answer: D

AB×AC=(3,2,4), magnitude $\sqrt{29}$. Area of triangle $=\frac12|AB\times AC|=\frac12\sqrt{29}$.