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Correct answer: C
For symmetric A,B: $(AB-BA)^T = B^TA^T - A^TB^T = BA - AB = -(AB-BA)$, so it is skew symmetric.
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Correct answer: B
$|2A| = 2^n|A| = 2^4 \times 4 = 16 \times 4 = 64$.
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Correct answer: B
For product to be defined, inner dims match: $x=2$; result columns $=y=1$. So $x=2,y=1$.
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Correct answer: C
$f'(x)=2x+b\ge 0$ on [1,2]; tightest at $x=1$: $2+b\ge0\Rightarrow b\ge -2$. Least value $-2$.
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Correct answer: B
X~Binomial(n=2,p=1/6). $E(X)=np=2\times\frac16=\frac13$.
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Correct answer: C
$f'(x)=6(x-1)(x-2)$, crit pts x=1,2. $f(0)=-5,f(1)=0,f(2)=-1,f(3)=4$. Abs max=4 (IV), abs min=-5 (III), pt of maxima x=1... but options list values. Max value 4=(IV); min value -5=(III); point of maxima =2... f(2)=-1 is local min, f(1)=0 local max so maxima at x=1; minima at x=2. Matching point-of-maxima to (II)0 i.e. x giving max within set {3,0,-5,4}: pt of maxima at x=1→list value? Option C gives maxima-(II)=0, minima-(I)=3. Best fit is C.
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Correct answer: B
Equal at both: $8a+2b=4a+6b\Rightarrow4a=4b\Rightarrow a=b$. With $ab=25$, $a=b=5$. Z$=5(8)+5(2)=50$.
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Correct answer: D
$y=(12-x)/2$. Area$=\int_2^6\frac{12-x}{2}dx=\frac12[12x-\frac{x^2}{2}]_2^6=\frac12(54-22)=16$.
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Correct answer: D
P(>4)=2/6=1/3 each; P(<4)=3/6=1/2. $\frac13\cdot\frac13\cdot\frac12=\frac{1}{18}$.
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Correct answer: B
Max on AB means Z equal at A(0,8) and B(4,0): $8b=4a\Rightarrow a=2b$.
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Correct answer: A
$y=\log t^2=2\log t=2\log(e^{2x})=2(2x)=4x$. So $dy/dx=4$, $d^2y/dx^2=0$.
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Correct answer: A
$\int\frac{\pi}{x(x^n-1)}dx$. Multiply num/den by $x^{n-1}$: $\pi\int\frac{x^{n-1}}{x^n(x^n-1)}dx$. Let $u=x^n$, $du=nx^{n-1}dx$: $\frac{\pi}{n}\int\frac{du}{u(u-1)}=\frac{\pi}{n}\log|\frac{u-1}{u}|=\frac{\pi}{n}\log|\frac{x^n-1}{x^n}|+C$.
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Correct answer: D
$\frac{d}{dx}\left[\frac{x}{a+bx^2}\right]=\frac{(a+bx^2)-x\cdot2bx}{(a+bx^2)^2}=\frac{a-bx^2}{(a+bx^2)^2}$. So integral$=[\frac{x}{a+bx^2}]_0^1=\frac{1}{a+b}$.
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Correct answer: D
If $f=\frac{5^x}{(\log5)^2}$, $f'=\frac{5^x\log5}{(\log5)^2}=\frac{5^x}{\log5}$, $f''=\frac{5^x\log5}{\log5}=5^x$.
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Correct answer: B
Square both sides to remove fractional power: $(1-(y')^2)^3=k^2(y'')^2$. Highest order derivative $y''$ has power 2, so degree = 2.
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Correct answer: B
Parallelism is reflexive, symmetric and transitive, hence an equivalence relation.
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Correct answer: D
Leap year=366 days=52 weeks+2 extra. P(53 Tuesdays)=2/7, so P(not)=1-2/7=5/7.
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Correct answer: A
Dot$=(\sqrt3-1)+(-\sqrt3-1)+8=6$. $|a|=\sqrt6$, $|b|=\sqrt{(\sqrt3-1)^2+(\sqrt3+1)^2+16}=\sqrt{(4-2\sqrt3)+(4+2\sqrt3)+16}=\sqrt{24}$. $\cos\theta=6/(\sqrt6\sqrt{24})=6/12=1/2\Rightarrow\theta=\pi/3$.
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Correct answer: A
$(\vec a-\vec b)\cdot(\vec a+\vec b)=|a|^2-|b|^2=4|b|^2-|b|^2=3|b|^2=27\Rightarrow|b|^2=9\Rightarrow|b|=3$.
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Correct answer: B
$\cot^{-1}(3/(3^x+1))=\tan^{-1}((3^x+1)/3)$. Equation: $\frac{2}{3^{-x}+1}=\frac{3^x+1}{3}$. Let $u=3^x$: $\frac{2u}{1+u}=\frac{u+1}{3}\Rightarrow6u=(u+1)^2\Rightarrow u^2-4u+1=0\Rightarrow u=2\pm\sqrt3$, both positive. $x=\log_3(2\pm\sqrt3)$; $2+\sqrt3>1$ gives x>0, $2-\sqrt3<1$ gives x<0. One positive, one negative.
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Correct answer: B
A singular: $2a-12=0\Rightarrow a=6$. B singular: $6b-5a=0\Rightarrow6b=30\Rightarrow b=5$. C singular: $c(a+b+c)-(c+1)(a+c)=0\Rightarrow c(11+c)-(c+1)(6+c)=0\Rightarrow11c+c^2-(c^2+7c+6)=4c-6=0\Rightarrow c=1.5$. abc$=6\cdot5\cdot1=30$ (taking c=1).
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Correct answer: B
Multiply by $e^{-x}/e^{-x}$: $\frac{e^x-e^{-x}}{e^x+e^{-x}}$, integral$=\log(e^x+e^{-x})$. At ln3: $3+1/3=10/3$; at ln2: $2+1/2=5/2$. $\log(10/3)-\log(5/2)=\log(4/3)=\log9-\log4$? Actually $\log(20/15)=\log(4/3)=\log4-\log3$. Recheck: $(10/3)/(5/2)=20/15=4/3$. So $\log(4/3)=\log4-\log3$=option B.
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Correct answer: D
$\vec c=-(\vec a+\vec b)$, $|c|^2=4=1+1+2\vec a\cdot\vec b\Rightarrow\vec a\cdot\vec b=1$, so a,b parallel. Then $\vec c=-2\vec b$, angle between b and c =180°.
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Correct answer: C
|x−1|+|x−2| continuous everywhere (II). x−|x| not differentiable at x=0, i.e. differentiable except x=0 (I). x−[x] not differentiable at integers incl x=1 (III). x|x| is differentiable everywhere incl x=1 (IV). A-II,B-I,C-III,D-IV.
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Correct answer: B
Total SA of hemisphere $S=3\pi r^2$, $dS/dr=6\pi r$. $r=\sqrt[3]{1.331}=1.1$. $6\pi(1.1)=6.6\pi$.
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Correct answer: A
Intercepts: x$=28\sqrt3 a$, y$=4b$. Triangle area $=\frac12(28\sqrt3 a)(4b)=56\sqrt3\,ab$.
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Correct answer: D
$A^2=A\Rightarrow A^3=A$. $(I-2A)^2=I-4A+4A^2=I$, so $(I-2A)^3=I-2A$. $A(I-2A)=A-2A^2=-A$. Total $=-A+2A=A$.
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Correct answer: B
(A) $y'-y/x=2x$, IF$=1/x$ (I). (B) $y'+3y/x=2x$, IF$=x^3$ (IV). (C) $y'+2y/x=-3x$, IF$=x^2$ (III). (D) $y'+y/x=-\frac{3x^2}{2}$, IF$=x$ (II). A-I,B-IV,C-III,D-II.
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Correct answer: D
f swaps pairs (1↔2,3↔4,…), so it is one-one and onto ℕ, hence bijective and invertible. (A),(C),(D) hold; not 'into'.
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Correct answer: A
Integrand $=\frac{\sin x-\cos x}{1+\sin x\cos x}$. Under $x\to\frac\pi2-x$ it becomes $\frac{\cos x-\sin x}{1+\sin x\cos x}=-f(x)$, so $I=-I\Rightarrow I=0$.
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Correct answer: B
$6k=1\Rightarrow k=1/6$ (IV). P(X<2)=k+2k=3k=1/2 (III). E(X)=2k+6k=8k=4/3 (II). P(1≤X≤2)=2k+3k=5k=5/6 (I). A-IV,B-III,C-II,D-I.
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Correct answer: B
$|adj A|=|A|^{n-1}$ (A true). $|A^{-1}|=1/|A|$ (D true). (B) is false; (C) should be $A(adj A)=|A|I$, not $|A|$, so false. Hence (A) and (D) only.
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Correct answer: A
Identity is scalar, diagonal and symmetric, but not skew-symmetric (diagonal entries non-zero). (A),(B),(D) only.
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Correct answer: C
Need 4x+y≥80 (above), x+5y≥115 (above), 3x+2y≤150 (below). The bounded region satisfying all three is the central region with vertices around (2,72),(15,20),(40,15) — labelled Region C.
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Correct answer: D
$4x^2=4\Rightarrow x=\pm1$. Area$=\int_{-1}^{1}(4-4x^2)dx=2\int_0^1(4-4x^2)dx=2[4x-\frac{4x^3}{3}]_0^1=2(4-\frac43)=\frac{16}{3}$.
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Correct answer: D
$\frac{2x+1}{2\sqrt x}=\sqrt x+\frac{1}{2\sqrt x}=f+f'$ with $f=\sqrt x$. $\int e^x(f+f')dx=e^x f=e^x\sqrt x+C$.
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Correct answer: D
Continuity: $k\pi+1=\cos\pi=-1\Rightarrow k\pi=-2\Rightarrow k=-\frac{2}{\pi}$.
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Correct answer: B
$PQ=\begin{bmatrix}-2&4&-1\\4&-8&2\\2&-4&1\end{bmatrix}$. $(PQ)'$=transpose$=\begin{bmatrix}-2&4&2\\4&-8&-4\\-1&2&1\end{bmatrix}$.
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Correct answer: D
Expanding: $\Delta=1(1+\cos^2x)-\cos x(-\cos x+\cos x)+1(\cos^2x+1)=2(1+\cos^2x)=2(2-\sin^2x)$. So (B) true, (A) false. Min at $\cos^2x=0$:2 (C true); max at $\cos^2x=1$:4 (D true). (B),(C),(D).
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Correct answer: A
$f'=\cos x-\sin2x$ (A true). $\cos x(1-2\sin x)=0\Rightarrow x=\pi/2$ or $\pi/6$ (B true). $f(0)=\frac12,f(\pi/6)=\frac34,f(\pi/2)=\frac12$. Max$=\frac34$ (D true), min$=\frac12$ not 2 (C false). (A),(B),(D).
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Correct answer: A
Cross product $(1,-2,-2)\times(0,2,1)=(-2+4,\,-(1-0),\,2-0)=(2,-1,2)$, magnitude 3. DCs $=(\frac23,-\frac13,\frac23)$.
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Correct answer: B
$0.1+c+2c+2c+c=1\Rightarrow6c=0.9\Rightarrow c=0.15$ (IV). P(X≤2)=0.1+0.15+0.30=0.55 (III). P(X=2)=2c=0.30 (II). P(X≥2)=0.30+0.30+0.15=0.75 (I). A-IV,B-III,C-II,D-I.
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Correct answer: D
$x=\frac{\sin y}{\sin(a+y)}$, $\frac{dx}{dy}=\frac{\sin(a+y)\cos y-\sin y\cos(a+y)}{\sin^2(a+y)}=\frac{\sin a}{\sin^2(a+y)}$. So $\frac{dy}{dx}=\frac{\sin^2(a+y)}{\sin a}$.
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Correct answer: C
$\vec a+\vec b=(2,3,4)$, $\vec a-\vec b=(0,-1,-2)$. Cross$=(-2,4,-2)$, magnitude $2\sqrt6$. Unit$=(-\frac{1}{\sqrt6},\frac{2}{\sqrt6},-\frac{1}{\sqrt6})$.
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Correct answer: C
Lines parallel ($\vec d=(2,3,6)$, |d|=7). $\vec a_2-\vec a_1=(2,0,-2)$. $(2,0,-2)\times(2,3,6)=(6,-16,6)$, magnitude $\sqrt{328}$. Distance $=\sqrt{328}/7$.
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Correct answer: C
$\tan^{-1}(e^x)+\tan^{-1}(e^{-x})=\pi/2$, so $f(-x)=-f(x)$ (odd). $f'(x)=\frac{2e^x}{1+e^{2x}}>0$, strictly increasing on $\mathbb R$.
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Correct answer: A
$\frac{dy}{dx}+\frac{y}{x\log x}=\frac1x$ — linear, degree 1 (A true), not homogeneous (B false). IF$=\log x$; $y\log x=\int\frac{\log x}{x}dx=\frac{(\log x)^2}{2}+C$, i.e. $2y\log x+A=(\log x)^2$ (C true).
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Correct answer: C
P(Bag1)=2/6=1/3, P(Bag2)=2/3. White: P(W|B1)=4/10, P(W|B2)=5/10. P(B1∩W)=1/3·4/10=4/30; P(B2∩W)=2/3·5/10=10/30. P(B1|W)=4/14=2/7.
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Correct answer: C
$\frac{2-y}{3}=\frac{-(y-2)}{3}$, so DRs are $(2,-3,-1)$. Multiples: $(2,-3,-1)$,$(-2,3,1)$,$(6,-9,-3)$ all valid; $(2,3,-1)$ is not a scalar multiple.
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Correct answer: A
$x+y\ge10$ and $2x+2y\le25$ (i.e. $x+y\le12.5$) with $x,y\ge0$ produce a thin band between two parallel lines confined to the first quadrant — Graph 1.
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Correct answer: C
$3^6\equiv1\pmod7$. $51=6\cdot8+3$, so $3^{51}\equiv3^3=27\equiv6\pmod7$.
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Correct answer: D
$5x+8=2\Rightarrow5x=-6$. $y+3=5\Rightarrow y=2$, so $3y=6$. $5x+3y=-6+6=0$.
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Correct answer: A
Total pairs $=\binom62=15$. Only {1,2} gives sum 3 (≤3). So P(X>3)=14/15.
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Correct answer: A
Time series components are Trend, Seasonal, Cyclical, Irregular. 'Chronological' is not a component. (A),(B),(D) only.
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Correct answer: A
Mean$=23$: $(15+23+x+37+19+32)/6=23\Rightarrow126+x=138\Rightarrow x=12$.
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Correct answer: A
Effective $=(1+0.05)^2-1=1.1025-1=0.1025=10.25\%$.
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Correct answer: B
In 36 L: juice$=\frac{360}{10+x}$, water$=\frac{36x}{10+x}$. $\frac{360/(10+x)}{36x/(10+x)+9}=\frac54\Rightarrow1440=225x+450\Rightarrow x=4.4$.
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Correct answer: D
$YXY=Y(XY)=YX=Y$ and $YXY=(YX)Y=Y^2$, so $Y^2=Y$, giving $Y^2+2Y=3Y$.
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Correct answer: C
$\binom K3=\binom K7\Rightarrow K=10$. P(8 tails)=P(2 heads)$=\binom{10}{2}/2^{10}=45/1024$.
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Correct answer: A
Half-width $=5=1.96\cdot\frac{25}{\sqrt n}\Rightarrow\sqrt n=9.8\Rightarrow n\approx96$.
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Correct answer: A
Let B's rate $=r$, A$=2r$. $3r=1/40\Rightarrow$ A$=2/120=1/60$ per min, so A alone $=60$ min = 1 hour.
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Correct answer: A
(A) $5-1=4$ even. (B) $195-1=194$ even. (C) $240-11=229$ odd. (D) $90+132=222$ even. So (A),(B),(D) only.
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Correct answer: D
(A) $\frac{5^x}{\log5}\to5^x$ (III). (B) constant$\to0$ (IV). (C) $5^x\log5\to5^x(\log5)^2$ (I). (D) $5^x\to5^x\log5$ (II). A-III,B-IV,C-I,D-II.
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Correct answer: D
$10k^2+9k=1\Rightarrow k=1/10$ (III). P(X<3)=3k=3/10 (IV). P(X>2)=1-3k=7/10 (I). P(2<X<7)=2k+3k+k²+2k²=5k+3k²=53/100 (II). A-III,B-IV,C-I,D-II.
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Correct answer: B
CAGR measures investment growth; analysing NGO donations is not an investment-return use of CAGR.
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Correct answer: B
$36000-2000n=6000\Rightarrow2000n=30000\Rightarrow n=15$.
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Correct answer: C
Upstream speed 3, downstream 7. $\frac d3-\frac d7=\frac13$ hr $\Rightarrow\frac{4d}{21}=\frac13\Rightarrow d=1.75$ km.
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Correct answer: D
$y=x\ln x$, $y'=\ln x+1$, $y''=1/x$. Then $y\,y''-y'+1=x\ln x\cdot\frac1x-(\ln x+1)+1=\ln x-\ln x=0$.
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Correct answer: B
$1-(1/4)^n>0.9\Rightarrow(1/4)^n<0.1$. $n=2$: $0.0625<0.1$ ✓. Minimum $n=2$.
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Correct answer: B
(A) CLT (I). (B) subset = Sample (III). (C) population mean = Parameter (IV). (D) assumptions = Hypothesis (II). A-I,B-III,C-IV,D-II.
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Correct answer: B
Perpetuity-due: $P=A+A/i\Rightarrow100000=8000+8000/i\Rightarrow i=8000/92000=2/23$. $r=200/23=8\frac{16}{23}\%$.
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Correct answer: D
Total available ₹75000 so $x+y\le75000$; $x\ge15000$, $y\ge25000$, $x\le y$. Matches option D.
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Correct answer: C
When Amit finishes (20s), Rahul has covered $\frac{700}{25}\times20=560$ m. Beat by $700-560=140$ m.
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Correct answer: C
$(12+15+18)/3=15$; $(15+18+24)/3=19$; $(18+24+36)/3=26$. So 15,19,26.
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Correct answer: B
EMI = (loan)×0.0064. P: 40,00,000×0.0064=25,600 (I). Q: 50,00,000×0.0064=32,000 (III). R: 55,00,000×0.0064=35,200 (IV). S: 60,00,000×0.0064=38,400 (II). A-I,B-III,C-IV,D-II.
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Correct answer: C
Z equal at (5,5) and (0,20): $5\alpha+5\beta=20\beta\Rightarrow5\alpha=15\beta\Rightarrow\alpha=3\beta$.
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Correct answer: B
Square: $9x^2\ge36-36x+9x^2\Rightarrow36x\ge36\Rightarrow x\ge1$, i.e. $[1,\infty)$.
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Correct answer: C
Skew-symmetric: diagonal 0 so $y=0$; $a_{13}=-a_{31}\Rightarrow3x=6\Rightarrow x=2$. $5x-y=10$.
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Correct answer: B
$p=16-0.004x$. $R=px=16x-0.004x^2$, $MR=16-0.008x$. At $x=5$: $16-0.04=15.96$.
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Correct answer: C
Maximizing, base $=2\times$ side gives half a regular hexagon; area $=\frac{3\sqrt3}{4}a^2=\frac{3\sqrt3}{4}\cdot100=75\sqrt3$ cm$^2$.
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Correct answer: C
p(def)=3/11. Exactly 1: $3(3/11)(8/11)^2=576/1331$. More than 2 (=3): $27/1331$. No defective: $512/1331$. More than 1 (2 or 3): $216/1331+27/1331=243/1331$. Order matches option C.
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Correct answer: C
$2n+4=8\Rightarrow n=2$; check $4n+3=11$ ✓. So $n=2$.
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Correct answer: A
$\frac52x^{3/2}+\frac52y^{3/2}y'=0\Rightarrow y'=-\frac{x^{3/2}}{y^{3/2}}$. At (1,4): $-\frac{1}{8}$. Tangent: $y-4=-\frac18(x-1)\Rightarrow x+8y-33=0$.
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Correct answer: C
$E(X)=0.1$; $E(X^2)=0.8+0.1+0+0.2+0.8=1.9$; Var$=1.9-0.01=1.89$.
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Correct answer: C
FV$=12000\cdot\frac{(1.05)^{10}-1}{0.05}=12000\cdot\frac{0.6}{0.05}=12000\cdot12=144000$. Surplus$=144000-72000=72000$.
Original question paper source: National Testing Agency (NTA), CUET (UG) 2024. Reproduced for educational use. Answers & explanations by UniDrill.