UGC NET June 2026 — Solved Paper

Answer: (D) 294. Thursday total = (100−75)% × 1400 = 25% × 1400 = 350. Friday total = 350 × 1.20 = 420. 16 GB share = 7/(7+3) = 7/10, so 16 GB on Friday = 420 × 7/10 = 294.

Answer: (A) 160. 32 GB: Tuesday = 560−240 = 320, Wednesday = 210−130 = 80, together = 400. 16 GB on Tuesday = 240. Difference = 400 − 240 = 160.

Answer: (C) 25. 32 GB on Tue, Wed, Thu = 320, 80, 200; average = 600/3 = 200. 16 GB on Monday = 160. Percentage more = (200−160)/160 × 100 = 25%.

Answer: (A) 80. 32 GB on Monday = 280−160 = 120. 16 GB on Thursday = 680−530 = 150. Required % = 120/150 × 100 = 80%.

Answer: (B) 7 : 5. 16 GB on Wed + Thu = 130 + 150 = 280. 32 GB on Thursday = 350 − 150 = 200. Ratio = 280 : 200 = 7 : 5.

Answer: (B) B and C only. On the square of opposition, B is a particular affirmative (I: “Some S are P”) and C is a particular negative (O: “Some S are not P”). I and O are subcontraries: they cannot both be false (at least one must hold), but they can both be true. Hence B and C.

Answer: (A) Software. Firmware is a specialised class of software permanently stored in a device's non-volatile memory (ROM/flash) to control its hardware. It is software embedded in hardware — not hardware itself, and not a compiler.

Answer: (D) A-II, B-III, C-I, D-IV. Use $\sigma=\sqrt{\frac1N\sum x_i^2-\mu^2}$. A: $\sqrt{178-169}=3$ (II). B: $\sqrt{212-196}=4$ (III). C: $\sqrt{229-225}=2$ (I). D: $\sqrt{281-256}=5$ (IV). So A-II, B-III, C-I, D-IV.

Answer: (D) A and E only. A is true (snooping = secretly monitoring/listening). E is true (cookies store browsing/session information). B is false (a worm is malware, not a firewall), C is false (a keylogger is spyware, not a security system), D is false (HTTP is a protocol, not a cookie). Correct: A and E only.

Answer: (C) A-IV, B-III, C-II, D-I. 1G = analog voice (IV); 2G = voice transmitted in digital form (III); 3G = digital voice + data services (II); 4G = interactive multimedia (I). So A-IV, B-III, C-II, D-I.

Answer: (B) 1512. The rule is $a_n=a_{n-1}\times(n-1)+2(n-1)$: 8→10→24→78→320. Next should be $320\times5+10=1610$, then $1610\times6+12=9672$ (which matches). So the wrong term is 1512 (should be 1610).

Answer: (D) A-III, B-IV, C-I, D-II. Skill India Mission builds the overall skill ecosystem (III); PMKVY provides skill development and vocational training (IV); Skill Hubs act as nodal skill centres (I); Micro-Credentials are short skill-based courses (II). So A-III, B-IV, C-I, D-II.

Answer: (D) Līlāvatī. Līlāvatī, written by Bhāskara II (12th century), is a classic Indian treatise on arithmetic and mathematics. Sāṃkhya Kārikā and Sāṃkhya Kaumudī are philosophical (Sāṃkhya) texts, and Arthasaṃgraha is a work on Mīmāṃsā.

Answer: (C) 6457152. From the codes: C=6, H=4, A=7, I=3, R=2 and E=5, A=7, T=1. So CHEATER = C·H·E·A·T·E·R = 6·4·5·7·1·5·2 = 6457152.

Answer: (C) A, B, C and D only. Interpersonal, mass, intrapersonal and group are all recognised types of communication. “Travelling” is not a type of communication. Correct: A, B, C and D only.

Answer: (A) Pitāmah. The eldest male patriarch who headed the joint family was the Pitāmah (grandsire/patriarch). A Rishi is a sage, a Guru is a teacher, and a Dīkshā Guru is an initiating spiritual teacher.

Answer: (B) E, A, B, D and C. Order of tenure: M. M. Pallam Raju (2012–14) → Smriti Irani (2014–16) → Prakash Javadekar (2016–19) → Ramesh Pokhriyal (2019–21) → Dharmendra Pradhan (2021– ). So E, A, B, D, C.

Answer: (B) C, E, B, D, A. Piaget's developmental sequence by age begins with the Sensorimotor stage (C), then the preconceptual thought sub-stage (E) at the start of the pre-operational stage (B), followed by the Concrete operational stage (D) and finally the (formal) Operational stage (A): C, E, B, D, A.

Answer: (C) B, A, E, C, D. Computer generations: 1st Vacuum Tube (B) → 2nd Transistor (A) → 3rd Integrated Circuit (E) → 4th VLSI (C) → 5th ULSI (D). So B, A, E, C, D.

Answer: (C) Fallacy of Division. A property of the whole (a “big” country) is wrongly transferred to a member of it (a “big” poet). Concluding that what is true of the whole must be true of its parts is the Fallacy of Division.

Answer: (A) Activated Sludge. The activated sludge process uses micro-organisms to biologically (biochemically) break down organic matter. Sedimentation, filtration and the grit chamber are physical (mechanical) treatment stages.

Answer: (D) A, B, D, C. Variance = npq. A: 5(0.4)(0.6)=1.20; B: 6(0.6)(0.4)=1.44; D: 7(0.7)(0.3)=1.47; C: 8(0.3)(0.7)=1.68. Ascending: A < B < D < C.

Answer: (A) C, A, B, D. Greater denotation = wider extension (refers to more things). Science ⊃ Natural Science ⊃ Chemistry ⊃ Organic Chemistry, so decreasing denotation is C, A, B, D.

Answer: (A) Tradition, personal experiences, intuition or authority and all subjective approaches. Field-study, experimental and historical methods are systematic scientific methods. Relying on tradition, personal experience, intuition or authority is a non-scientific (subjective) way of acquiring knowledge.

Answer: (A) It describes social situations, events, systems and structures — the researcher observes and then describes what is found. Descriptive research describes the characteristics of a phenomenon as observed. Explaining causes is explanatory research; studying little-known subjects is exploratory research; common-sense belief is not scientific research.

Answer: (C) A, B and C only. Thermal pollution is a form of water pollution (A), often from power-plant cooling water (B), and it harms aquatic life (C). It decreases dissolved oxygen (warmer water holds less O₂), so D is false. Correct: A, B and C only.

Answer: (D) A, B, C, D and E. NEP 2020 advocates that school curriculum and pedagogy be holistic, integrated, enjoyable, flexible and engaging — i.e. all five qualities.

Answer: (B) Anumāna (Inference). Inference (Anumāna) rests on vyāpti — the invariable concomitance between the reason (hetu) and the inferred object (e.g. smoke and fire). Perception, analogy and verbal testimony do not rely on concomitance.

Answer: (C) Network Interface Card. An Ethernet card is also called a Network Interface Card (NIC) — the hardware that connects a computer to a network.

Answer: (D) ₹2000. Equal %: 1920 − CP = CP − 1280 ⟹ 2·CP = 3200 ⟹ CP = ₹1600. For 25% profit, SP = 1600 × 1.25 = ₹2000.

Answer: (D) A-III, B-IV, C-II, D-I. A: $(3/8)\div(1/12)=9/2$ (III). B: $x+1.5=1.5x\Rightarrow x=3$ (IV). C: $(5/12)N=375\Rightarrow N=900$ (II). D: $48/7-3/28=189/28=6\tfrac34$ (I). So A-III, B-IV, C-II, D-I.

Answer: (C) A-IV, B-I, C-II, D-III. Intrapersonal = talking with oneself (IV); Interpersonal = face-to-face between two persons (I); Group = theatre/dance/Rama Lila (II); Mass = press, cinema, radio, TV (III). So A-IV, B-I, C-II, D-III.

Answer: (C) Raja Harishchandra. Raja Harishchandra (1913), directed by Dadasaheb Phalke, was India's first full-length feature film. (Alam Ara, 1931, was the first Indian sound film/talkie.)

Answer: (D) A, C and D only. The Modified Mercalli scale grades intensity: what people feel and what happens to buildings and the ground, so A is true. It runs in twelve descriptive steps, I to XII, a small closed scale beside the open-ended Richter magnitude scale, where each step is a tenfold change in wave amplitude, so C is true. Its steps are equal descriptive grades, not powers of ten, so it is a linear scale and D is true. It is not known as a “hazard manager's scale”, so B is false.

Answer: (C) Alam Ara. Alam Ara (1931), directed by Ardeshir Irani, was India's first sound (talkie) feature film.

Answer: (A) A-III, B-IV, C-I, D-II. ICT-based teaching = use of digital tools (III); CBCS = flexibility in course selection (IV); CBT = automated, instant evaluation (I); SWAYAM = MHRD's online education initiative (II). So A-III, B-IV, C-I, D-II.

Answer: (C) Reduce chances of suffering from Alzheimer's and senile dementia. Lifelong learning keeps the brain active and reduces the chances of Alzheimer's and dementia — it does not guarantee complete immunity, so “never” (option 1) is too absolute.

Answer: (B) 280 ppmv. Pre-industrial atmospheric CO₂ was about 280 ppmv (parts per million by volume).

Answer: (B) Fannie Shaftel. The Role-Playing model of teaching (in the social family of models) was developed by Fannie Shaftel (and George Shaftel).

Answer: (D) ₹625. For 2 years, CI − SI = $P\left(\frac{r}{100}\right)^2$. So $1 = P(0.04)^2 = 0.0016P \Rightarrow P = \dfrac{1}{0.0016} = $ ₹625.

Answer: (D) Asking students to examine the material from alternative points of view. Active learning engages students in higher-order thinking. Asking students to examine material from alternative points of view makes them analyse and evaluate actively, rather than passively receiving information.

Answer: (B) B only. “Some animals are birds” is true, so its contradictory “No animals are birds” (B) is false. “No birds are animals” (D) says the same as B, so it is false too. “Some animals are not birds” (A) and “All animals are birds” (C) cannot be decided from the statement: they may be true. Every option except “B only” includes A or C, so B only is the one option made entirely of statements known to be false.

Answer: (A) D, E, C, A, B. Years of establishment: Films Division 1948 (D) → Central Board of Film Certification 1951 (E) → FTII Pune 1960 (C) → Directorate of Film Festivals 1973 (A) → NFDC 1975 (B). So D, E, C, A, B.

Answer: (C) A, B, C and D only. Scientific research is accurate, precise, systematic and objective (A, B, C, D). It is controlled, not uncontrolled, so E is not a characteristic. Correct: A, B, C and D only.

Answer: (D) Methemoglobinemia. Blue baby syndrome is methemoglobinemia, caused by high nitrate levels in drinking water that impair the blood's oxygen-carrying capacity, giving infants a bluish tint.

Answer: (D) In terms of regional languages. The passage opens: “Linguistic diversity divides a nation in terms of regional languages.”

Answer: (C) Cultural. The passage states there must be “psychological, emotional, cultural and economic integration among the masses.” Of the options, only Cultural is mentioned.

Answer: (B) Economic disparities. The passage stresses that removal of economic disparities is vital, and that tension and injustice from such disparities hinder national integration.

Answer: (B) English. The passage explicitly says “English finds no mention in the VIII Schedule of our Constitution.” Tamil, Maithili and Malayalam are all listed there; English is not.

Answer: (D) Feeling of injustice. The passage says “Tension and a sense of injustice among people is bound to hinder the progress of national integration.” Hence a feeling of injustice is harmful.

Answer: (D) $\max\{\nu(A),\nu(B)\}\le \nu(AB)\le \nu(A)+\nu(B)$. Use Sylvester\'s rank inequality $\operatorname{rank}(A)+\operatorname{rank}(B)-n\le \operatorname{rank}(AB)\le \min\{\operatorname{rank}(A),\operatorname{rank}(B)\}$ together with nullity $\nu(M)=n-\operatorname{rank}(M)$. The upper bound $\operatorname{rank}(AB)\le\min\{\operatorname{rank}A,\operatorname{rank}B\}$ becomes $\nu(AB)\ge\max\{\nu(A),\nu(B)\}$, giving the lower bound. The lower bound $\operatorname{rank}(AB)\ge\operatorname{rank}A+\operatorname{rank}B-n$ becomes $\nu(AB)\le\nu(A)+\nu(B)$, giving the upper bound. Hence $\max\{\nu(A),\nu(B)\}\le\nu(AB)\le\nu(A)+\nu(B)$.

Answer: (B) $\left(\tfrac{3}{16},\ \tfrac{81}{128}\right)$. The rejection region is $X>3$, i.e. $X\ge 4$, where $X\sim\text{Bin}(5,p)$. Size $=P(X\ge4\mid p=\tfrac12)=\binom54(\tfrac12)^5+\binom55(\tfrac12)^5=\tfrac{5+1}{32}=\tfrac{6}{32}=\tfrac{3}{16}$. Power $=P(X\ge4\mid p=\tfrac34)=\binom54(\tfrac34)^4(\tfrac14)+(\tfrac34)^5=\tfrac{405}{1024}+\tfrac{243}{1024}=\tfrac{648}{1024}=\tfrac{81}{128}$. Hence $(\tfrac{3}{16},\tfrac{81}{128})$.

Answer: (A) A-II, B-IV, C-III, D-I. The chi-square test checks homogeneity/independence in a contingency table (A–II). The Wilcoxon signed-rank test is the nonparametric test for two related (paired) samples (B–IV). The Mann–Whitney U-test compares two independent samples (C–III). The Kruskal–Wallis test is the nonparametric analogue of one-way ANOVA (D–I). Hence A-II, B-IV, C-III, D-I.

Answer: (D) A, C, D only. $\Sigma$ is an equicorrelation matrix (diagonal 1, off-diagonal $\rho=\tfrac12$) of order 3. Its eigenvalues are $1+(n-1)\rho=1+2(\tfrac12)=2$ (eigenvector $\tfrac{1}{\sqrt3}(1,1,1)$) and $1-\rho=\tfrac12$ (multiplicity 2). So A is true (first PC is the equally-weighted average direction), C is true ($2/\text{trace}=2/3\approx66.7\%$) and D is true (smallest eigenvalue $=\tfrac12$). B is false: the largest eigenvalue is 2, not 3. Hence A, C, D only.

Answer: (C) 66.7365. $F=\dfrac{\text{MS}_{\text{treatment}}}{\text{MS}_{\text{error}}}=\dfrac{66870/3}{5344/16}=\dfrac{22290}{334}\approx 66.74$.

Answer: (C) 7.0. Sample mean $\bar X=(3,3)$, so $d=\bar X-\mu_0=(-0.5,\,0.5)$. The sample covariance (divisor $n-1=2$) is $S=\begin{pmatrix}3&1.5\\1.5&1\end{pmatrix}$, with $\det S=0.75$ and $S^{-1}=\begin{pmatrix}4/3&-2\\-2&4\end{pmatrix}$. Then $d\'S^{-1}d=(-0.5)(-\tfrac53)+(0.5)(3)=\tfrac56+\tfrac32=\tfrac73$, and $T^2=n\,d\'S^{-1}d=3\cdot\tfrac73=7.0$.

Answer: (A) A, B only. This is a shifted (location) exponential with $E(X)=\theta+1$ and $V(X)=1$. A is true: by the WLLN $\bar X_n\xrightarrow{P}E(X)=1+\theta$. B is true: the minimum of $n$ iid unit-rate exponentials shifted by $\theta$ has rate $n$, so $f_Y(y)=n\,e^{-n(y-\theta)},\ y\ge\theta$. C is false: $V(\bar X_n)=V(X)/n=1/n$, not $n$. D is false: $\bar X_n$ converges to $1+\theta$, not $\theta$. Hence A, B only.

Answer: (D) $\sigma^2(I_n-P)$. $P$ is the (symmetric, idempotent) hat matrix, so $M=I_n-P$ is also symmetric and idempotent with $MX=0$. Hence $e=My=M(X\beta+u)=Mu$, and $E(ee\')=M\,E(uu\')\,M\'=\sigma^2 MM=\sigma^2 M=\sigma^2(I_n-P)$.

Answer: (D) E, C, A, B, D. A valid session must first create the objects, then operate on them, then print output. First define the matrix (E), then the vector (C); compute the element-wise product $y=A*x$ (A); extract its diagonal with diag(y) (B); the console then prints the result (D). Hence the order is E, C, A, B, D.

Answer: (B) 9. Inside the function the assignment x <- 5 creates a local variable that shadows the global x = 3. The last evaluated expression x + y uses this local value, so f(4) returns $5+4=9$.

Answer: (D) $1-\dfrac{1}{2^{n(n+1)/2}}$. $P(\text{at least one})=1-\prod_{i=1}^{n}P(A_i^c)=1-\prod_{i=1}^{n}\dfrac{1}{2^i}=1-\dfrac{1}{2^{1+2+\cdots+n}}=1-\dfrac{1}{2^{n(n+1)/2}}$.

Answer: (C) $r_k=(-0.4)^k$. The AR polynomial factors as $1-0.6B-0.4B^2=(1-B)(1+0.4B)$, which contains a unit root. So $(1+0.4B)(1-B)y_t=\epsilon_t$, i.e. $(1+0.4B)\Delta y_t=\epsilon_t$. Hence $\Delta y_t=-0.4\,\Delta y_{t-1}+\epsilon_t$ is an AR(1) with parameter $\phi=-0.4$, whose autocorrelation function is $r_k=\phi^k=(-0.4)^k$.

Answer: (A) B and C only. A is false: SRSWOR is more efficient than SRSWR (the finite population correction reduces the variance). B is true: the ratio estimator is biased, with bias of order $1/n$. C is true: cluster sampling beats SRSWOR when the intraclass correlation is negative (elements within a cluster are dissimilar). D is false: systematic sampling is superior only when the intraclass correlation is less than $-1/(kn-1)$, not greater. Hence B and C only.

Answer: (D) A-III, B-IV, C-II, D-I. Simple random sampling works best for a homogeneous population (A–III). Stratified sampling is used for a heterogeneous population, splitting it into homogeneous strata (B–IV). Cluster sampling is adopted when a complete sampling frame of elements is unavailable (C–II). Sampling with varying probabilities (PPS) gives due weightage to larger units (D–I). Hence A-III, B-IV, C-II, D-I.

Answer: (D) $T_1+(n-r)T_2$. Under Type-II censoring at the $r$th failure, the total time on test $S=\sum_{i=1}^{r}X_{(i)}+(n-r)X_{(r)}$ satisfies $E(S)=r\theta$, so $S/r$ is unbiased for $\theta$. Now $S/r=\dfrac{1}{r}\sum X_{(i)}+(n-r)\dfrac{X_{(r)}}{r}=T_1+(n-r)T_2$.

Answer: (D) C, A, B, D, E. A test proceeds as: first form the hypotheses (C), then check the assumptions of the chosen test (A), compute the test statistic (B), compare with the tabulated critical value (D), and finally draw the inference (E). Hence C, A, B, D, E.

Answer: (D) 48. $M_X(t)=e^{2t+2t^2}$ matches the normal MGF $e^{\mu t+\sigma^2 t^2/2}$ with $\mu=2$ and $\sigma^2=4$. For a normal distribution the fourth central moment is $\mu_4=3\sigma^4=3(4)^2=48$.

Answer: (D) B, D only. At $x=1$: left value $1^2+1=2$, right value $3-1=2$, so $f$ is continuous. But the left derivative $2x\big|_{1}=2$ and right derivative $-1$ differ, so $f$ is not differentiable at $x=1$ (B true, A false). Rolle\'s theorem needs differentiability on the whole open interval; since it fails at $x=1\in(0,2)$, Rolle\'s theorem is not applicable (D true, C false) — even though $f(0)=f(2)=1$. Hence B, D only.

Answer: (B) $\dfrac16(y_1+2y_2+y_3),\ \dfrac12(y_1-y_3)$. With design matrix $X=\begin{pmatrix}1&1\\2&0\\1&-1\end{pmatrix}$, the errors are homoscedastic and uncorrelated, so the BLUE is OLS: $\hat\beta=(X\'X)^{-1}X\'y$. Here $X\'X=\begin{pmatrix}6&0\\0&2\end{pmatrix}$ and $X\'y=\begin{pmatrix}y_1+2y_2+y_3\\ y_1-y_3\end{pmatrix}$, giving $\hat\beta_1=\tfrac16(y_1+2y_2+y_3)$ and $\hat\beta_2=\tfrac12(y_1-y_3)$.

Answer: (D) $\dfrac{\lambda+1}{\lambda^2}$. Using the law of total variance with $X\sim\text{Exp}$ (mean $1/\lambda$, variance $1/\lambda^2$) and $Y\mid X\sim\text{Poisson}(X)$: $\mathrm{Var}(Y)=E[\mathrm{Var}(Y\mid X)]+\mathrm{Var}[E(Y\mid X)]=E(X)+\mathrm{Var}(X)=\dfrac1\lambda+\dfrac{1}{\lambda^2}=\dfrac{\lambda+1}{\lambda^2}$.

Answer: (B) $\begin{pmatrix}-3\\0\end{pmatrix},\ \begin{pmatrix}2&1\\1&1\end{pmatrix}$. Partition with $\mu_1=2,\ \mu_2=(-2,1)\',\ \Sigma_{11}=1,\ \Sigma_{21}=(1,1)\',\ \Sigma_{22}=\begin{pmatrix}3&2\\2&2\end{pmatrix}$. Conditional mean $=\mu_2+\Sigma_{21}\Sigma_{11}^{-1}(x_1-\mu_1)=(-2,1)\'+(1,1)\'(1-2)=(-3,0)\'$. Conditional variance $=\Sigma_{22}-\Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12}=\begin{pmatrix}3&2\\2&2\end{pmatrix}-\begin{pmatrix}1&1\\1&1\end{pmatrix}=\begin{pmatrix}2&1\\1&1\end{pmatrix}$.

Answer: (C) A-III, B-IV, C-I, D-II. x just echoes the vector (A–III). sort returns the values in ascending order $-51,-8,4,12,23,54$ (B–IV). rank gives each element\'s ascending rank: $4,3,2,6,5,1$ (C–I). order gives the indices that sort the vector: $6,3,2,1,5,4$ (D–II). Hence A-III, B-IV, C-I, D-II.

Answer: (C) A, B and D. For a non-negative random variable the tail-integral identities give $E(X)=\int_0^\infty[1-F(x)]dx$ (B true) and $E(X^2)=\int_0^\infty 2x[1-F(x)]dx$ (A true). By Jensen\'s inequality with the convex function $1/x$, $E(1/X)\ge 1/E(X)$, so C (with $\le$) is false. With the concave function $\log$, $E(\log X)\le\log E(X)$, so D is true. Hence A, B and D.

Answer: (B) $\dfrac{2^5}{3^4}$. Given the total $N_1+N_2=n$, $N_1$ is Binomial$(n,p)$ with $p=\dfrac{\lambda}{\lambda+2\lambda}=\tfrac13$. Hence $P(N_1=1\mid \text{sum}=4)=\binom41(\tfrac13)(\tfrac23)^3=4\cdot\tfrac13\cdot\tfrac{8}{27}=\tfrac{32}{81}=\dfrac{2^5}{3^4}$.

Answer: (B) 0.345. Ranking each series and taking rank differences $d$ gives $\sum d^2=108$ (the year-by-year $d^2$ values are 4, 25, 16, 9, 16, 4, 25, 9, 0, 0). Then $\rho=1-\dfrac{6\sum d^2}{n(n^2-1)}=1-\dfrac{6\times108}{10\times99}=1-0.6545\approx0.345$.

Answer: (A) $\mu$ when $\sigma^2$ is known. $\dfrac{\bar X-\mu}{\sigma/\sqrt n}\sim N(0,1)$, whose distribution is free of unknown parameters. It can serve as a pivot only when $\sigma$ is a known constant; then it depends on the data and on the single unknown $\mu$, giving a confidence interval for $\mu$. Hence it is a pivot for $\mu$ when $\sigma^2$ is known.

Answer: (C) A-IV, B-III, C-I, D-II. Two means with known variances → Z-test (A–IV). Ratio of two normal variances → F-test (B–III). Equality of several proportions → $\chi^2$-test (C–I). Single mean with unknown variance → t-test (D–II). Hence A-IV, B-III, C-I, D-II.

Answer: (D) $2^{1/4}$. Since $2^{u}$ is increasing, maximize the exponent $g(x)=x(1-x)=x-x^2$. Setting $g\'(x)=1-2x=0$ gives $x=\tfrac12$, where $g=\tfrac14$. Hence the maximum of $f$ is $2^{1/4}$.

Answer: (D) No UMP test exists. The exponential family has monotone likelihood ratio, so a UMP test exists for a one-sided alternative. But for the two-sided alternative $H_1:\theta\ne\theta_0$ the best rejection region differs on the two sides, so no single test is uniformly most powerful. Hence no UMP test exists (only a UMP unbiased test does).

Answer: (C) (4, 6). With $t=|x-5|$ and $a_n=\dfrac{1}{(n+1)(n+2)}$, the ratio test gives $\lim\left|\dfrac{a_{n+1}}{a_n}\right|=\lim\dfrac{(n+1)(n+2)}{(n+2)(n+3)}=1$, so the radius of convergence is 1. Thus the series converges for $|x-5|<1$, i.e. $4<x<6$.

Answer: (B) $\lambda^{-1}$ is an eigenvalue of $A^{-1}$ and the corresponding eigenvector is $x$. From $Ax=\lambda x$ (with $\lambda\ne0$ since $A$ is invertible), multiply by $A^{-1}$: $x=\lambda A^{-1}x$, so $A^{-1}x=\tfrac{1}{\lambda}x$. Thus $\lambda^{-1}$ is an eigenvalue of $A^{-1}$ with the same eigenvector $x$.

Answer: (D) A, B, D only. By the Rao–Blackwell theorem, $E(h\mid T)$ is unbiased (A) and has variance no larger than $h$ (B). Because $T$ is sufficient, the conditional distribution given $T$ is free of $\theta$, so $E(h\mid T)$ is a genuine statistic not depending on $\theta$ (D). Statement C is false: uniqueness/UMVUE status requires $T$ to be complete sufficient (Lehmann–Scheffé), which is not assumed here. Hence A, B, D only.

Answer: (D) B, A, D, C. First create the vector $x=(1,-1,3.5,2)$ (B), copy it to $y$ (A), then evaluate $y^2$ (D); the console prints $(1,1,12.25,4)$ (C). Hence B, A, D, C.

Answer: (B) $C_1=C_2$. Communication of states is an equivalence relation (reflexive, symmetric, transitive). Since $k\leftrightarrow i$ and $k\leftrightarrow j$, transitivity gives $i\leftrightarrow j$, so $i$ and $j$ lie in the same communicating class. Hence $C_1=C_2$ (and $k$ belongs to it, states in a class share the same period).

Answer: (C) A-IV, B-III, C-I, D-II. P. C. Mahalanobis — the Mahalanobis distance in multivariate analysis (A–IV). P. V. Sukhatme — sampling theory, including ordered estimates in PPS sampling without replacement (B–III). R. C. Bose — partially balanced incomplete block designs (C–I). S. N. Roy — the union–intersection principle and largest-root test in multivariate hypothesis testing (D–II). Hence A-IV, B-III, C-I, D-II.

Answer: (D) 120. The loop multiplies $x$ by $i$ for $i=1,2,3,4,5$, computing the factorial $5!=1\cdot2\cdot3\cdot4\cdot5=120$. Hence f(5) returns 120.

Answer: (C) B, C only. For jointly normal variables, independence $\Leftrightarrow$ zero covariance. A: $\mathrm{Cov}(X_1,X_2)=-1\ne0$ → dependent. B: $\mathrm{Cov}(X_3,X_1)=\mathrm{Cov}(X_3,X_2)=0$, so $X_3$ is independent of $(X_1,X_2)$ → true. C: $\mathrm{Cov}(X_2,X_2+2X_1)=\mathrm{Var}(X_2)+2\mathrm{Cov}(X_2,X_1)=2+2(-1)=0$ → independent. D: $\tfrac14[\mathrm{Cov}(X_2,X_1)+\mathrm{Var}(X_2)+\mathrm{Cov}(X_3,X_1)+\mathrm{Cov}(X_3,X_2)]=\tfrac14(-1+2+0+0)=\tfrac14\ne0$ → dependent. Hence B, C only.

Answer: (C) Estimates are biased. Near multicollinearity inflates the variances of the OLS estimates — producing unstable, large-magnitude coefficients with high standard errors, insignificant $t$-values despite a high $R^2$. However, OLS estimates remain unbiased under multicollinearity; only their precision suffers. Hence “estimates are biased” is not a consequence.

Answer: (B) $Z\sim W_4(5,I_4)$. $XX\'=\sum_{i=1}^{2}X_iX_i\'\sim W_4(2,\Sigma_1)$, so $\Sigma_1^{-1/2}XX\'\Sigma_1^{-1/2}\sim W_4(2,I_4)$. Likewise $\Sigma_2^{-1/2}YY\'\Sigma_2^{-1/2}\sim W_4(3,I_4)$. Independent Wisharts with the same scale matrix add their degrees of freedom, giving $Z\sim W_4(2+3,I_4)=W_4(5,I_4)$.

Answer: (B) B, C, D only. Marginals: $f_X(x)=\int_{-1}^1\tfrac14(1+cxy)\,dy=\tfrac12$ (and similarly $f_Y(y)=\tfrac12$), so B is true. But $f_X f_Y=\tfrac14\ne\tfrac14(1+cxy)$ (for $c\ne0$), so $X,Y$ are dependent — A is false. Summing the density over the four sign-combinations of $(x,y)$ shows $(|X|,|Y|)$ is uniform on $(0,1)^2$, hence $X^2$ and $Y^2$ are independent (C true). Finally $E(X^2Y^2)=\tfrac14\big[(\int x^2)(\int y^2)+c(\int x^3)(\int y^3)\big]=\tfrac14\cdot\tfrac49=\tfrac19$ (D true). Hence B, C, D only.

Answer: (D) A, B, C, D only. $f$ is a polynomial, hence continuous on $[0,3]$ (A) and differentiable on $(0,3)$ (B). Also $f(0)=0=f(3)$, so all three hypotheses of Rolle\'s theorem hold (D). Indeed $f\'(x)=3(x-1)(x-3)$ vanishes at $C=1\in(0,3)$ (C). Hence all of A, B, C, D are correct.

Answer: (D) A, B, D only. Here every simple correlation is $\tfrac12$. Partial correlation $r_{13.2}=\dfrac{r_{13}-r_{12}r_{23}}{\sqrt{(1-r_{12}^2)(1-r_{23}^2)}}=\dfrac{\tfrac12-\tfrac14}{\tfrac34}=\tfrac13$ (B true); by symmetry $r_{23.1}=\tfrac13=r_{13.2}$ (D true) and the multiple correlations $R_{3.12}=R_{2.13}$ are equal (A true). For C, $R^2_{1.23}=\dfrac{r_{12}^2+r_{13}^2-2r_{12}r_{13}r_{23}}{1-r_{23}^2}=\dfrac{1/4}{3/4}=\tfrac13$, so $R_{1.23}=1/\sqrt3\approx0.577\ne\tfrac23$ (C false; note it is $1-R^2_{1.23}=\tfrac23$). Hence A, B, D only.

Answer: (D) $t\mu,\ t\sigma^2_\epsilon$. Iterating from $y_0=0$ gives $y_t=\mu t+\sum_{i=1}^{t}\epsilon_i$. Hence $E(y_t)=t\mu$ and, since the $\epsilon_i$ are uncorrelated, $\mathrm{Var}(y_t)=t\sigma^2_\epsilon$ — both grow with $t$, reflecting the non-stationarity of a random walk.

Answer: (C) A, C, D only. Under $\rho=0$ the null density is $f(r)\propto(1-r^2)^{(n-4)/2},\ -1<r<1$. For $n=3$ the exponent is $-\tfrac12$, giving the U-shaped arcsine density (C true); for $n=4$ the exponent is 0, giving the uniform $U(-1,1)$ density (D true). The sampling distribution of $r$ also becomes increasingly skewed as the true correlation approaches $\pm1$ (A true). Statement B is the odd one out — $r$ is supported on the open interval and the blanket “non-degenerate on $[-1,1]$” claim is not the accepted correct statement. Hence A, C, D only.

Answer: (C) A, B, C only. The log-likelihood $n\ln\theta-\theta\sum x_i$ is maximised at $\hat\theta=n/\sum x_i$ (A true). By factorization $\sum x_i$ is sufficient (C true), and it is complete, with $\sum x_i\sim\text{Gamma(shape }n,\text{ rate }\theta)$ — not $(1,\theta)$ — so D is false. Since $E(1/\sum x_i)=\theta/(n-1)$, the estimator $(n-1)/\sum x_i$ is unbiased and, being a function of the complete sufficient statistic, is the UMVUE (B true). Hence A, B, C only.

Answer: (C) $\dfrac{X_i}{N\bar X}$. In PPS (probability proportional to size) selection, unit $i$ is chosen with probability $\dfrac{X_i}{\sum_{j=1}^N X_j}$. Since $\sum_j X_j=N\bar X$, this equals $\dfrac{X_i}{N\bar X}$. Lahiri\'s method is precisely a mechanism that realizes these selection probabilities.

Answer: (B) 2. Every row equals $R_1+k(1,1,0,0)$ for $k=0,1,2,3$, so only two rows are linearly independent (equivalently, column 3 $=2\times$column 4 and column 2 $=$ column 1 $+$ column 3). Hence the rank is 2.

Answer: (D) A, B, D only. For Poisson$(\lambda)$ both the mean and the variance equal $\lambda$, so $E(\bar X)=\lambda$ (A) and $E(S^2)=\lambda$ (B). Any weighted average $a\bar X+(1-a)S^2$ with weights summing to 1 is therefore unbiased: $0.3+0.7=1$ gives $E=\lambda$ (D true). But $\bar X+S^2$ has weights summing to 2, so $E=2\lambda\ne\lambda$ (C false). Hence A, B, D only.

Answer: (B) $\dfrac32$. Convert covariances to correlations: $r_{12}=r_{13}=\dfrac{1}{\sqrt{4\cdot2}}=\dfrac{1}{2\sqrt2}$ (so $r_{12}^2=\tfrac18$) and $r_{23}=\dfrac{\alpha}{2}$. Then $r_{23.1}=\dfrac{r_{23}-r_{12}r_{13}}{\sqrt{(1-r_{12}^2)(1-r_{13}^2)}}=\dfrac{\alpha/2-1/8}{7/8}=\dfrac{4\alpha-1}{7}$. Setting this equal to $\tfrac57$ gives $4\alpha-1=5$, i.e. $\alpha=\tfrac32$.

Answer: (B) 0.25, 1.50. $\hat\rho=\dfrac{\sum e_t e_{t-1}}{\sum e_t^2}=\dfrac{1.32}{5.28}=0.25$. The Durbin–Watson statistic $d\approx2(1-\hat\rho)=2(1-0.25)=1.50$.

Answer: (B) $\dfrac{R^2/(k-1)}{(1-R^2)/(n-k)}$. With $k$ parameters (including the intercept) there are $k-1$ regressors. The overall F-test compares explained to unexplained variation: $F=\dfrac{R^2/(k-1)}{(1-R^2)/(n-k)}$, with $(k-1,\,n-k)$ degrees of freedom.

Answer: (C) $e^{-x_0/\theta}$. $T(X_1)$ is unbiased for $E[T(X_1)]=P(X_1>X_0)=e^{-X_0/\theta}$ (the exponential survival probability). Since $T_S=\sum X_i$ is a complete sufficient statistic, by the Rao–Blackwell/Lehmann–Scheffé theorems $E[T(X_1)\mid T_S]$ is the UMVUE of $e^{-x_0/\theta}$.

Answer: (C) A, D only. A gives the correct definition of $\lambda$ (A true). Since the numerator is a supremum over a subset of the denominator\'s set, $0\le\lambda\le1$ — not $0<\lambda<\infty$, so B is false. By Wilks\' theorem $-2\log_e\lambda$ is asymptotically $\chi^2$ (not $N(0,1)$), so C is false. Because $\lambda$ depends on the data only through the likelihood, it is a function of the sufficient statistic (D true). Hence A, D only.

Answer: (B) $\dfrac1n\sum_{i=1}^n x_i^2\xrightarrow{P}4$. Here $E(X)=1,\ \mathrm{Var}(X)=4$, so $E(X^2)=4+1^2=5$. Option 2 is incorrect: by the WLLN $\tfrac1n\sum x_i^2\xrightarrow{P}E(X^2)=5$, not 4. The others are correct: $\tfrac1n\sum x_i\xrightarrow{P}1$ (WLLN), $\tfrac{\sqrt n}{2}(\bar X_n-1)\xrightarrow{d}N(0,1)$ (CLT with $\sigma=2$), and $E(S^2)=4$ (unbiasedness of the sample variance).

Answer: (A) $\dfrac12$. For a sample of size $n=5$ from $U[0,1]$, the median is the 3rd order statistic. In general $E(X_{(k)})=\dfrac{k}{n+1}$, so $E(X_{(3)})=\dfrac{3}{6}=\dfrac12$.

Answer: (B) $e^{-4}$. Interarrival times are exponential with rate $\lambda=2$. By the memoryless property, $P(T>6\mid T>4)=P(T>2)=e^{-\lambda\cdot2}=e^{-4}$.

Answer: (B) 0.51. With $n=50$ observations and $k=5$ parameters, $\bar R^2=1-(1-R^2)\dfrac{n-1}{n-k}=1-(0.45)\dfrac{49}{45}=1-0.49=0.51$.

Answer: (B) Irreducible and periodic. All three states communicate ($1\to2\to3$ and back), so the chain is irreducible. Returns to any state occur only in an even number of steps ($P^3=P$ means odd powers equal $P$ and even powers equal $P^2$), so the period is 2 — the chain is periodic.

Answer: (C) $n-k$ eigenvalues of $P$ are 1 and $k$ eigenvalues are 0.. $P=I_n-H$ where $H=X(X\'X)^{-1}X\'$ is the hat matrix — a symmetric idempotent projection of rank $k$. Thus $P$ is also symmetric and idempotent (a projection onto the orthogonal complement) with rank $n-k$. An idempotent matrix has eigenvalues 0 and 1 only; $P$ has eigenvalue 1 with multiplicity $n-k$ and 0 with multiplicity $k$.

Answer: (A) 32.5. Ranking all 30 scores together (averaging ties) gives the ES rank-sum $R_{ES}=152.5$. Then $U_{ES}=R_{ES}-\dfrac{n_1(n_1+1)}{2}=152.5-\dfrac{15\cdot16}{2}=152.5-120=32.5$, and $U_{LC}=n_1n_2-U_{ES}=225-32.5=192.5$. The Mann–Whitney statistic is the smaller value, $U=32.5$ (reflecting that ES scores are generally lower).

Answer: (C) B and C only. The Horvitz–Thompson estimator is unbiased (so D is false) and is an unordered estimator — the ordered estimator is Des Raj\'s (so A is false). Its standard (Horvitz–Thompson form) variance estimator has two well-known drawbacks: it can take negative values for some samples (C true) and, unlike the Yates–Grundy–Sen form, it does not reduce to zero even when all $y_i/\pi_i$ are equal (B true). Hence B and C only.

Answer: (C) Rangarajan commission. The National Statistical Commission was set up (constituted in 2005) based on the recommendations of the Rangarajan Commission on Statistics, chaired by Dr. C. Rangarajan.

Answer: (B) B, C only. $f\'(x)=3x^2-12x+9=3(x-1)(x-3)$ vanishes at $x=1,3$. With $f\'\'(x)=6x-12$: $f\'\'(1)=-6<0$, so $x=1$ is a local maximum (B); $f\'\'(3)=6>0$, so $x=3$ is a local minimum (C). Hence B, C only.

Answer: (B) 0.63. With $\Sigma_{11}=\Sigma_{22}=I_2$, $\Sigma_{12}=\begin{pmatrix}0.5&0\\0.3&0.4\end{pmatrix}$, the squared canonical correlations are the eigenvalues of $\Sigma_{12}\Sigma_{21}=\begin{pmatrix}0.25&0.15\\0.15&0.25\end{pmatrix}$, namely $0.25\pm0.15=0.4$ and $0.1$. The maximum canonical correlation is $\sqrt{0.4}\approx0.63$.

Answer: (D) B, A, D, E, C. The National Sample Survey began in 1950 (B), the Central Statistics Office in 1951 (A), MOSPI was formed in 1999 (D), the National Statistical Commission was constituted in 2005 (E), and the National Statistical Office (merging CSO and NSSO) in 2019 (C). Hence B, A, D, E, C.

Answer: (C) B and C only. CRD uses only randomization and replication — it has no local control. The Randomized Block Design (blocking in one direction) and the Latin Square Design (blocking by rows and columns) both employ all three principles — randomization, replication and local control. Hence B and C only.

Answer: (A) $A\ge B$. Model (I) is nested within model (II) (set $\gamma=0$). Least squares chooses coefficients to minimize the residual sum of squares, and adding a regressor can never increase the minimized RSS. Hence $B\le A$, i.e. $A\ge B$ always. (A residual sum of squares can never be negative, so options with $B\le0$ are impossible.)

Answer: (D) Stratified multi-stage sampling. NSSO large-scale surveys use a stratified multi-stage sampling design — e.g. villages/urban blocks as first-stage units (after stratification) and households as second-stage units — which balances cost and precision over a vast, heterogeneous population.

Answer: (A) $\dfrac{-7}{18}$. The Yule–Walker equations $r_1=\Phi_1+\Phi_2 r_1$ and $r_2=\Phi_1 r_1+\Phi_2$ give $\Phi_2=\dfrac{r_2-r_1^2}{1-r_1^2}=\dfrac{0.5-0.64}{1-0.64}=\dfrac{-0.14}{0.36}=-\dfrac{7}{18}$.

Answer: (A) A-III, B-IV, C-I, D-II. Number of successes in a fixed number of trials → Binomial (A–III). Trials until the first success → Geometric (B–IV). Number of failures before the $r$th success → Negative Binomial (C–I). The limiting case of many trials with a tiny success probability → Poisson (D–II). Hence A-III, B-IV, C-I, D-II.

Answer: (D) Periodogram. The periodogram — based on the squared magnitude of the finite Fourier transform of the series — is the classical estimator of the spectral density function (its smoothed versions give consistent spectral estimates).

Answer: (D) $(1/3)^{100}$. Against an effectively infinitely rich casino, the classical gambler\'s-ruin probability with win probability $p>q$ is $\left(\dfrac{q}{p}\right)^{i}$ for initial capital $i$. Here $p=\tfrac34,\ q=\tfrac14$, so $q/p=\tfrac13$ and the ruin probability is $(1/3)^{100}$.

Answer: (A) 1, 1. For the Laplace density, $|X|$ has density $f_{|X|}(y)=2\cdot\tfrac12 e^{-y}=e^{-y},\ y>0$, i.e. $|X|\sim\text{Exponential(rate 1)}$. Hence $E(|X|)=1$ and $\mathrm{Var}(|X|)=1$.

Answer: (C) 12.725. Under Binomial$(4,\tfrac12)$ with $N=160$, the expected frequencies are $10,40,60,40,10$. Then $\chi^2=\dfrac{(17-10)^2}{10}+\dfrac{(52-40)^2}{40}+\dfrac{(54-60)^2}{60}+\dfrac{(31-40)^2}{40}+\dfrac{(6-10)^2}{10}=4.9+3.6+0.6+2.025+1.6=12.725$.

Answer: (D) $(X\'X)^{-1}\delta$. $b=(X\'X)^{-1}X\'y=\beta+(X\'X)^{-1}X\'u$, so $b-\beta=(X\'X)^{-1}X\'u$ and $E(b-\beta)=(X\'X)^{-1}E(X\'u)=(X\'X)^{-1}\delta$. (When $\delta=0$ this reduces to the usual unbiasedness; here $\delta\ne0$ gives the endogeneity bias.)

Answer: (C) 6.25. For a stationary AR(1), $\mathrm{Var}(y_t)=\dfrac{\sigma^2_\epsilon}{1-\phi^2}=\dfrac{4}{1-0.6^2}=\dfrac{4}{0.64}=6.25$.

Answer: (A) 7. $SS_{error}=SS_{total}-SS_{treat}-SS_{blocks}=477-177-195=105$. The error degrees of freedom are $(t-1)(b-1)=(4-1)(6-1)=15$. Hence $MSE=105/15=7$.

Answer: (C) (ii) is identified but (i) is not. The system has 2 predetermined variables ($x_1,x_2$). Order condition: (excluded predetermined) $\ge$ (included endogenous $-1$). Equation (i) excludes no exogenous variable ($0\ge1$ fails) → not identified. Equation (ii) excludes $x_1$ ($1\ge1$ holds) → identified (just-identified). Hence (ii) is identified but (i) is not.

Answer: (C) 35. Neyman allocation: $n_h=n\dfrac{N_h S_h}{\sum N_k S_k}$. Since $S_1=S_2$, this reduces to proportional allocation: $n_2=50\times\dfrac{350}{150+350}=50\times0.7=35$.

Answer: (A) $\left(-\infty,\ \bar X+t_{n-1,\alpha}\,S/\sqrt n\right)$. With $\sigma$ unknown, $\dfrac{\bar X-\mu}{S/\sqrt n}\sim t_{n-1}$. A one-sided $(1-\alpha)$ upper bound satisfies $P\!\left(\mu\le\bar X+t_{n-1,\alpha}\,S/\sqrt n\right)=1-\alpha$, giving the interval $\left(-\infty,\ \bar X+t_{n-1,\alpha}\,S/\sqrt n\right)$ (note the $n-1$ degrees of freedom and the $+$ sign for an upper bound).

Answer: (B) A-III, B-II, C-I, D-IV. min returns $-154$ (A–III); max returns 6 (B–II); range returns the pair $(-154,\ 6)$ (C–I); which(...<0) returns the positions of the negative entries $-14$ and $-154$, namely 3 and 4 (D–IV). Hence A-III, B-II, C-I, D-IV.

Answer: (A) 0. $PQ=\left(I_n-\dfrac{uv\'}{u\'v}\right)uv\'=uv\'-\dfrac{u(v\'u)v\'}{u\'v}$. Since the scalar $v\'u=u\'v$, the second term is $uv\'$, so $PQ=uv\'-uv\'=0$. ($P$ is the oblique projector annihilating the column space of $Q$.)

Answer: (C) Uniform$(0,2)$. For two iid exponential variables, the conditional density of $X$ given $X+Y=s$ is $\dfrac{f_X(x)f_Y(s-x)}{f_{X+Y}(s)}=\dfrac{\lambda^2 e^{-\lambda s}}{\lambda^2 s\,e^{-\lambda s}}=\dfrac1s$ for $0<x<s$ — i.e. Uniform$(0,s)$. With $s=2$, $X\mid(X+Y=2)\sim\text{Uniform}(0,2)$.

Answer: (B) 7. Total observations $=2^3\times2=16$, so total df $=15$. The 7 factorial effects (A, B, C, AB, AC, BC, ABC) take 7 df, and the two replications, laid out as blocks as a replicated factorial normally is, take $r-1=1$. Hence error df $=15-7-1=(r-1)(2^3-1)=7$. (Only if the replications were not treated as blocks, a completely randomized layout, would the error df be $2^3(r-1)=8$.)

Answer: (B) $rk(r-\lambda)^{v-1}$. For a BIBD, $NN\'=(r-\lambda)I_v+\lambda J_v$. Its determinant is $(r-\lambda)^{v-1}\big[r+(v-1)\lambda\big]$. Using $\lambda(v-1)=r(k-1)$, we get $r+(v-1)\lambda=rk$, so $|NN\'|=rk\,(r-\lambda)^{v-1}$.

Answer: (C) $(1-e^{-x})^7$. The CDF of a single observation is $F(x)=1-e^{-x}$. The maximum $X_{(n)}\le x$ iff all $n$ observations are $\le x$, so $F_{X_{(7)}}(x)=[F(x)]^7=(1-e^{-x})^7$.

Answer: (B) 0. $Cy=X\beta+Cu$ (fitted values) and $(I_n-C)y=(I_n-C)u$ (residuals). Then $E[(Cy)\{(I_n-C)y\}\']=X\beta\,E(u\')(I_n-C)+C\,E(uu\')(I_n-C)=0+\sigma^2 C(I_n-C)=\sigma^2(C-C^2)=0$, since $C$ is idempotent. Fitted values and residuals are uncorrelated.

Answer: (B) 0.75. Power $=P(X\ge0.5\mid\theta=2)$. Under $H_1$, $X\sim U[0,2]$ with density $\tfrac12$, so $P(X\ge0.5)=\int_{0.5}^{2}\tfrac12\,dx=\tfrac12(1.5)=0.75$.

Answer: (B) B, C, D only. In the SPRT both error probabilities $\alpha$ and $\beta$ are fixed in advance (B true, so A is false), and sampling continues until a boundary is crossed, making the sample size $N$ a random variable (C true). Wald proved the SPRT terminates with probability 1 (D true). Hence B, C, D only.

Answer: (D) $\begin{pmatrix}3\\3\end{pmatrix},\ \begin{pmatrix}2&1\\1&2/3\end{pmatrix}$. $\hat\mu=\bar X=(3,3)\'$. The MLE of $\Sigma$ uses divisor $n$: with deviations $(-2,-1),(1,1),(1,0)$, $\sum(X_i-\bar X)(X_i-\bar X)\'=\begin{pmatrix}6&3\\3&2\end{pmatrix}$, so $\hat\Sigma=\tfrac13\begin{pmatrix}6&3\\3&2\end{pmatrix}=\begin{pmatrix}2&1\\1&2/3\end{pmatrix}$.

Answer: (B) $pq(1+p)$. Computing $P^2$ gives first row $(q^2+pq,\ pq,\ p^2)$. Then $p_{12}^{(3)}=(P^2)_{11}P_{12}+(P^2)_{12}P_{22}+(P^2)_{13}P_{32}=(q^2+pq)p+pq\cdot0+p^2\cdot q=pq^2+2p^2q=pq(q+2p)$. Since $q+2p=(q+p)+p=1+p$, this equals $pq(1+p)$.

Answer: (C) $0$. A first return to state 1 at step 3 must avoid state 1 at steps 1 and 2: the only feasible path is $1\to2\to3\to1$ (from 1 only $1\to2$ has positive probability among non-1 moves, then $2\to3$). But $P_{31}=0$, so the last step $3\to1$ is impossible. Hence $f_{11}^{(3)}=p\cdot p\cdot0=0$.

Answer: (D) $p^{-1}$. The mean first-passage time satisfies $\mu_{12}=1+\sum_{k\ne2}P_{1k}\mu_{k2}=1+P_{11}\mu_{12}+P_{13}\mu_{32}=1+q\,\mu_{12}$ (since $P_{13}=0$). Hence $\mu_{12}(1-q)=1$, giving $\mu_{12}=\dfrac{1}{1-q}=\dfrac1p=p^{-1}$.

Answer: (A) $\left(\dfrac{q^2}{1-pq},\ \dfrac{pq}{1-pq},\ \dfrac{p^2}{1-pq}\right)$. Solving $\pi P=\pi$ gives $\pi_2=\tfrac pq\pi_1$ and $\pi_3=\tfrac{p^2}{q^2}\pi_1$, so $\pi\propto(q^2,\,pq,\,p^2)$. Direct check confirms $(q^2,pq,p^2)P=(q^2,pq,p^2)$ using $p+q=1$. Normalizing by $q^2+pq+p^2=(p+q)^2-pq=1-pq$ gives $\pi=\left(\dfrac{q^2}{1-pq},\dfrac{pq}{1-pq},\dfrac{p^2}{1-pq}\right)$. (Only when $p=q=\tfrac12$ does this reduce to $(\tfrac13,\tfrac13,\tfrac13)$.)

Answer: (A) $q^4 p$. A first passage from 1 to 2 in 5 steps must avoid state 2 for the first 4 steps. Among the non-2 states, from state 1 only the self-loop $1\to1$ has positive probability ($P_{13}=0$), and state 3 is unreachable from 1 without passing through 2. So the walk stays at state 1 for four steps ($q$ each) and then moves $1\to2$ (probability $p$): $f_{12}^{(5)}=q^4 p$.

Answer: (D) ARIMA(1, 1, 1) process. The AR polynomial $1-0.6B-0.4B^2=(1-B)(1+0.4B)$ has a unit root, so the process is non-stationary and needs one difference ($d=1$). After differencing, the stationary AR part is order 1 and the MA part is order 1 (only $\theta_1\ne0$). Hence it is an ARIMA(1, 1, 1) process.

Answer: (C) $\alpha(h)=0$ for all $h\ge3$. With $\theta_1=\theta_2=0$ the model is a pure AR(2) process. The partial autocorrelation function of an AR($p$) process cuts off after lag $p$, i.e. $\alpha(h)=0$ for all $h>p$. Here $p=2$, so $\alpha(h)=0$ for all $h\ge3$.

Answer: (B) Stationary but not invertible. AR part: with $\Phi_1+\Phi_2=0.7<1,\ \Phi_2-\Phi_1=-0.1<1,\ |\Phi_2|<1$, all stationarity conditions hold — the process is stationary. MA part: $1-0.6B-0.4B^2=(1-B)(1+0.4B)$ has a unit root ($\theta_1+\theta_2=1$), so the MA polynomial is not invertible. Hence stationary but not invertible.

Answer: (C) $\dfrac{(0.6)^h}{0.64}$. The model reduces to the AR(1) process $y_t=0.6\,y_{t-1}+\epsilon_t$ with $\sigma^2_\epsilon=1$. Its autocovariance is $r(h)=\phi^{h}\dfrac{\sigma^2_\epsilon}{1-\phi^2}=\dfrac{(0.6)^h}{1-0.36}=\dfrac{(0.6)^h}{0.64}$. (Note $(0.6)^h$ alone is the autocorrelation, not the autocovariance.)

Answer: (D) $[1.49-1.4\cos(2\pi\omega)]^{-1}$. The model reduces to the MA(1) process $y_t=\epsilon_t-0.7\,\epsilon_{t-1}$ with $\sigma^2=1$. Its spectral density is proportional to $|1-0.7e^{-2\pi i\omega}|^2=1+0.49-1.4\cos(2\pi\omega)=1.49-1.4\cos(2\pi\omega)$. Option D is the only one with these coefficients. Its printed $-1$ exponent is a misprint: an MA spectrum is a polynomial in $\cos(2\pi\omega)$, while the reciprocal form belongs to an AR(1) process.