Consider the series , where . Which of the following statements is true?
CSIR NET June 2023 Mathematical Sciences — all questions solved
Every Part B and Part C question from this paper, worked out in full — not just the answer key, but why each option holds or fails and which trap it tests. Free to read, no account needed.
Part B
One correct option. 3 marks, −0.75 for a wrong answer.
- A.The series is divergent.
- B.The series is convergent.
- C.The series is conditionally convergent.✓
- D.The series is absolutely convergent.
Solution
n+1−n=1/(n+1+n) decreases to 0, so the alternating series converges (Leibniz). The absolute series behaves like ∑1/(2n), which diverges. Hence conditionally convergent — option 3 is the most precise true statement (official key: 3).
Which of the following assertions is correct?
- A.limsupnexp(cos((nπ+(−1)n⋅2e)/(2n)))>1.
- B.limnexp(loge((nπ2+(−1)ne2)/(7n))) does not exist.
- C.liminfnexp(sin((nπ+(−1)n⋅2e)/(2n)))<π.✓
- D.limnexp(tan((nπ2+(−1)ne2)/(7n))) does not exist.
Solution
All the inner arguments converge: (nπ+(−1)n2e)/(2n)→π/2 and (nπ2+(−1)ne2)/(7n)→π2/7. So the limits in (2) and (4) exist; in (1) the limit is exp(cosπ/2)=1, not > 1; in (3) the limit is exp(sinπ/2)=e<π. Official key: 3.
Which one of the following functions is uniformly continuous on the interval (0, 1)?
- A.f(x) = sin(1/x)
- B.f(x) = e−1/x2✓
- C.f(x) = eˣ cos(1/x)
- D.f(x)=cosx⋅cos(π/x)
Solution
A continuous function on (0,1) is uniformly continuous iff it extends continuously to [0,1]. e−1/x2 → 0 as x → 0⁺, so it extends; the other three oscillate without a limit at 0.
Let A be a 3×3 real matrix whose characteristic polynomial p(T) is divisible by T2. Which of the following statements is true?
- A.The eigenspace of A for the eigenvalue 0 is two-dimensional.
- B.All the eigenvalues of A are real.✓
- C.A3=0.
- D.A is diagonalizable.
Solution
p(T)=T2(T−c) with c real (degree-3 real polynomial). So all eigenvalues are real. The geometric multiplicity of 0 may be 1 (Jordan block), so (1), (4) fail; A need not be nilpotent (c ≠ 0).
Let T be a linear operator on R3. Let f(X)∈R[X] denote its characteristic polynomial. Consider the following statements. (a) Suppose T is non-zero and 0 is an eigenvalue of T. If we write f(X) = X·g(X) in R[X], then the linear operator g(T) is zero. (b) Suppose 0 is an eigenvalue of T with at least two linearly independent eigenvectors. If we write f(X) = X·g(X) in R[X], then the linear operator g(T) is zero. Which of the following is true?
- A.Both (a) and (b) are true.
- B.Both (a) and (b) are false.
- C.(a) is true and (b) is false.
- D.(a) is false and (b) is true.✓
Solution
(a) fails: for the nilpotent Jordan block on R3,f=X3,g=X2, and g(T)=T2=0.(b) If 0 has geometric multiplicity ≥ 2 in dimension 3, the 0-Jordan blocks all have size 1, so the minimal polynomial is X(X − c) (or X if T = 0), which divides g(X) = X(X − c); hence g(T) = 0.
Let x=(x1,…,xn) and y=(y1,…,yn) denote vectors in Rn for a fixed n ≥ 2. Which of the following defines an inner product on Rn?
- A.⟨x, y⟩ =∑i,j=1n xiyj
- B.⟨x, y⟩ =∑i,j=1n (xi2+yj2)
- C.⟨x, y⟩ =∑j=1n j3xjyj✓
- D.⟨x, y⟩ =∑j=1n xjyn₋j₊1
Solution
A weighted sum ∑wjxjyj with positive weights is an inner product. (1) is (∑xi)(∑yj), not positive definite; (2) is not bilinear; (4) is symmetric but ⟨x, x⟩ can be ≤ 0 (e.g. x = (1, −1) for n = 2).
Let p be a prime number. Let G be a group such that for each g ∈ G there exists an n∈N such that gpn = 1. Which of the following statements is FALSE?
- A.If |G| =p6, then G has a subgroup of index p2.
- B.If |G| =p6, then G has at least five normal subgroups.
- C.Center of G can be infinite.
- D.There exists G with |G| =p6 such that G has exactly six normal subgroups.✓
Solution
A finite p-group has a normal subgroup of every order pᵏ dividing |G|, so |G| =p6 gives at least seven normal subgroups (orders 1,p,…,p6) — 'exactly six' is impossible. (1), (2) follow from the same fact; (3) an infinite abelian p-group (e.g. the Prüfer group) has infinite centre.
The number of solutions of the equation x2=1 in the ring Z/105Z is
- A.0
- B.2
- C.4
- D.8✓
Solution
By CRT, Z/105Z≅Z/3×Z/5×Z/7;x2=1 has exactly 2 solutions in each odd prime field, giving 23=8.
How many real roots does the polynomial x3+3x−2023 have?
- A.0
- B.1✓
- C.2
- D.3
Solution
The derivative 3x2+3>0, so the cubic is strictly increasing: exactly one real root.
Suppose S is an infinite set. Assuming that the axiom of choice holds, which of the following is true?
- A.S is in bijection with the set of rational numbers.
- B.S is in bijection with the set of real numbers.
- C.S is in bijection with S × S.✓
- D.S is in bijection with the power set of S.
Solution
With AC, |S × S| = |S| for every infinite S. (1), (2) fix a specific cardinality; (4) contradicts Cantor's theorem.
Let f(z)=exp(z+1/z),z∈C∖{0}. The residue of f at z = 0 is
- A.∑l=0∞ 1/(l+1)!
- B.∑l=0∞ 1/(l!(l+1))
- C.∑l=0∞ 1/(l!(l+1)!)✓
- D.∑l=0∞ 1/((l2+l)!)
Solution
Multiply the series eᶻ =∑zm/m! and e1/z =∑ z−k/k!; the coefficient of z−1 comes from k=m+1:∑m1/(m!(m+1)!).
Consider the function f defined by f(z)=1/(1−z−z2) for z∈C such that 1−z−z2=0. Which of the following statements is true?
- A.f is an entire function.
- B.f has a simple pole at z = 0.
- C.f has a Taylor series expansion f(z)=∑anzn, where a0=1,a1=0 and an₊2=an+an₊1 for n ≥ 0.
- D.f has a Taylor series expansion f(z)=∑anzn, where a0=1,a1=1 and an₊2=an+an₊1 for n ≥ 0.✓
Solution
(1−z−z2)∑anzn=1 gives a0=1,a1−a0=0⇒a1=1, and an₊2=an₊1+an: the Fibonacci generating function. Poles are at the roots of 1−z−z2, not at 0.
Suppose x(t) is the solution of the initial value problem in R2: ẋ = Ax, x(0)=x0, where A = [[5, 4], [1, 2]]. Which of the following statements is true?
- A.x(t) is a bounded solution for some x0=0.
- B.e−6t|x(t)| → 0 as t→∞, for all x0=0.
- C.e−t|x(t)| →∞ as t→∞, for all x0=0.
- D.e−10t|x(t)| → 0 as t→∞, for all x0=0.✓
Solution
Eigenvalues of A are 6 and 1 (trace 7, det 6), both positive: every non-zero solution grows like e6t or et. So (1) fails, (2) fails along the 6-eigendirection, (3) fails along the 1-eigendirection, and (4) holds since 10 > 6.
Let u(x, y) be the solution of the Cauchy problem u⋅ux+uy=0 for x∈R,y>0, with u(x, 0) = x for x∈R. Which of the following is the value of u(2, 3)?
- A.2
- B.3
- C.1/2✓
- D.1/3
Solution
Characteristics: dx/dy = u with u constant along them, so x=x0+u⋅y and u=x0. Hence u = x/(1 + y), and u(2, 3) = 1/2.
Consider the variational problem (P):J(y)=∫01[(y′)2−y|y|·y′ + xy] dx, y(0) = 0, y(1) = 0. Which of the following statements is correct?
- A.(P) has no stationary function (extremal).
- B.y ≡ 0 is the only stationary function (extremal) for (P).
- C.(P) has a unique stationary function (extremal) y not identically equal to 0.✓
- D.(P) has infinitely many stationary functions (extremals).
Solution
The term y|y|y′ is an exact derivative ((|y|3/3)′ up to sign), so it does not affect the Euler–Lagrange equation, which reduces to 2y″ = x. With y(0) = y(1) = 0 this gives the unique extremal y=(x3−x)/12, not identically zero.
For the unknown y:[0,1]→R, consider the boundary value problem y″(x) + 2y(x) = 0 for x ∈ (0,1), y(0) = y(1) = 0. It is given that it corresponds to the integral equation y(x)=2∫01K(x,t)y(t) dt. Which of the following is the kernel K(x, t)?
- A.K(x,t) = t(1 − x) for t < x; x(1 − t) for t > x✓
- B.K(x,t)=t2(1−x) for t<x;x2(1−t) for t > x
- C.K(x,t)=t(1−x) for t<x;x(1−t) for t > x
- D.K(x,t)=t3(1−x) for t<x;x3(1−t) for t > x
Solution
The Green's function of −y″ with Dirichlet conditions on [0,1] is G(x,t) = t(1 − x) for t < x and x(1 − t) for t > x; then y=2∫Gy.
Let X1,X2,X3,X4 be i.i.d. Bernoulli(1/3) and X(1)≤⋯≤X(4) the order statistics. Which of the following is true?
- A.X(1) and X(4) are independent.
- B.Expectation of X(2) is 1/2.
- C.Variance of X(2) is 8/81.✓
- D.X(4) is a degenerate random variable.
Solution
X(2)=1 iff at least three of the four are 1:P=4(1/3)3(2/3)+(1/3)4=9/81=1/9.A Bernoulli(1/9) has variance (1/9)(8/9) = 8/81.
Let X be a Poisson random variable with mean λ. Which of the following parametric functions is not estimable?
- A.λ⁻1✓
- B.λ
- C.λ2
- D.e−λ
Solution
An unbiased estimator g(X) must satisfy ∑g(k)λke−λ/k! =h(λ); the left side is e−λ times a power series in λ, which can never equal λ⁻1(unbounded at 0).λ,λ2 and e−λ = P(X = 0) are all estimable.
Let X1,…,X7 and Y1,…,Y9 be independent random samples from continuous CDFs F and G. For the Wald–Wolfowitz run test of H0:F=G vs H1:F=G, let R be the total number of runs in the combined ordered arrangement. Which of the following is true?
- A.PH0(R = 6) = 28/286, PH0(R = 9) = 28/143.
- B.PH0(R = 6) = 21/286, PH0(R = 9) = 15/286.
- C.PH0(R = 6) = 21/286, PH0(R = 9) = 28/143.✓
- D.PH0(R = 6) = 21/286, PH0(R = 9) = 15/286.
Solution
With m = 7, n = 9 and C(16,7) = 11440 arrangements: P(R = 2k) = 2·C(m−1,k−1)·C(n−1,k−1)/C(16,7) gives P(R=6) = 2·15·28/11440 = 21/286; P(R = 2k+1) = [C(m−1,k)C(n−1,k−1) + C(m−1,k−1)C(n−1,k)]/C(16,7) gives P(R=9) = (15·56 + 20·70)/11440 = 28/143.
Consider the simple linear regression model Yi=βxi+εi,i=1,…,n, where E(εi)=0,Cov(εi,εk)=0 for i ≠ k and Var(εi)=xi2σ2. The best linear unbiased estimator of β is
- A.∑Yixi/∑xi2
- B.∑Yi/∑xi
- C.(1/n)∑Yi/xi✓
- D.(1/n)∑Yixi/xi2
Solution
Heteroscedastic errors: divide by xi to get Yi/xi=β+εi/xi with constant variance σ2; the BLUE is the plain mean of Yi/xi(weighted least squares with weights 1/xi2).
Let X=(X1,X2)T be bivariate normal with mean (0, 0)ᵀ and covariance ∑=[[5,−3],[−3,10]]. The mean vector and covariance matrix of Y=(X1,5−2X2)T are
- A.(0, 5)ᵀ, [[5, −3], [−3, 40]]
- B.(0, 5)ᵀ, [[5, −6], [−6, 20]]
- C.(0, 5)ᵀ, [[5, 3], [3, 20]]
- D.(0, 5)ᵀ, [[5, 6], [6, 40]]✓
Solution
E[5−2X2]=5;Var(5−2X2)=4⋅10=40;Cov(X1,5−2X2)=−2⋅(−3)=6.
Suppose X=(X1,X2,X3,X4)T ~ N4(0,I2 ⊗ ∑) with ∑=[[2,−1],[−1,2]]. Define Z=[[X1,X2],[X3,X4]] and Q = ZᵀZ. If Wm(n,∑) denotes a Wishart distribution of order m with n degrees of freedom, the distribution of Q11+Q12+Q21+Q22 is
- A.W1(2,2)✓
- B.W1(1,2)
- C.W1(2,1)
- D.2χ42
Solution
The rows of Z are i.i.d.N2(0,∑), so Q ~ W2(2,∑). The sum of all entries of Q is aᵀQa with a = (1, 1)ᵀ, which is W1(2,aT∑a)=W1(2,2).
Part C
One or more correct options. 4.75 marks, no negative marking, and credit only for exactly the right set.
Under which of the following conditions is the sequence {xn} of real numbers convergent?
- A.The subsequences {x2n₊1}, {x2n} and {x3n} are convergent and have the same limit.✓
- B.The subsequences {x2n₊1}, {x2n} and {x3n} are convergent.✓
- C.The subsequences {xkn}n are convergent for every k ≥ 2.
- D.lim |xn₊1−xn| = 0.
Solution
(1) Even and odd subsequences with a common limit ⇒ convergent. (2) {x6n} lies in both {x2n} and {x3n}, so those limits agree; {x3(2n₊1)} lies in both {x2n₊1} and {x3n}, so all three limits agree ⇒ convergent. (3) Fails: xn=1 if n is prime, 0 otherwise — every {xkn} is eventually 0 but xn diverges. (4) Fails: xn=n.
Let f:R→R be defined as f(x)=1/4+x−x2. Given a∈R, define the sequence {xn} by x0=a and xn=f(xn₋1) for n ≥ 1. Which of the following statements are true?
- A.If a = 0, then the sequence {xn} converges to 1/2.✓
- B.If a = 0, then the sequence {xn} converges to −1/2.
- C.The sequence {xn} converges for every a ∈ (−1/2, 3/2), and it converges to 1/2.✓
- D.If a = 0, then the sequence {xn} does not converge.
Solution
Write f(x)=1/2−(x−1/2)2. Fixed points are ±1/2. For |a − 1/2| < 1 the distance dn=1/2−xn satisfies dn₊1=dn2, which tends to 0, so xn→1/2 for all a ∈ (−1/2, 3/2); a = 0 is included.
Consider the function f:R2→R defined by f(x,y)=x2−y3. Which of the following statements are true?
- A.There is no continuous real-valued function g defined on any interval of R containing 0 such that f(x, g(x)) = 0.
- B.There is exactly one continuous real-valued function g defined on an interval of R containing 0 such that f(x, g(x)) = 0.✓
- C.There is exactly one differentiable real-valued function g defined on an interval of R containing 0 such that f(x, g(x)) = 0.
- D.There are two distinct differentiable real-valued functions g on an interval of R containing 0 such that f(x, g(x)) = 0.
Solution
g(x)3=x2 forces g(x) = x2/3 (the real cube root is a bijection), so there is exactly one continuous solution. It is not differentiable at 0, so no differentiable g exists. The implicit function theorem does not apply since f_y(0,0) = 0.
Consider the following quadratic forms over R:(a)6X2−13XY +6Y2,(b)X2− XY +2Y2,(c)X2− XY −2Y2. Which of the following statements are true?
- A.Quadratic forms (a) and (b) are equivalent.
- B.Quadratic forms (a) and (c) are equivalent.✓
- C.Quadratic form (b) is positive definite.✓
- D.Quadratic form (c) is positive definite.
Solution
Discriminants b2−4ac: (a) 169 − 144 > 0 indefinite; (b) 1 − 8 < 0 with a > 0 ⇒ positive definite; (c) 1 + 8 > 0 indefinite. Over R, non-degenerate binary forms are equivalent iff they have the same signature, so (a) ~ (c).
Let B be a 3×5 matrix with entries from Q. Assume that {v∈R5 | Bv = 0} is a three-dimensional real vector space. Which of the following statements are true?
- A.{v∈Q5 | Bv = 0} is a three-dimensional vector space over Q.✓
- B.The linear transformation T:Q3→Q5 given by T(v) = Bᵗv is injective.
- C.The column span of B is two-dimensional.✓
- D.The linear transformation T:Q3→Q3 given by T(v) = BBᵗv is injective.
Solution
Rank does not change under field extension, so rank B = 5 − 3 = 2 over Q as well: nullity over Q is 3 and the column span is 2-dimensional. Bᵗ : Q3→Q5 has rank 2 < 3, not injective; BBᵗ has rank 2 < 3, not injective.
Let V be a finite dimensional real vector space and T1,T2 be two nilpotent operators on V. Let W1= {v∈V:T1(v)=0} and W2= {v∈V:T2(v)=0}. Which of the following statements are FALSE?
- A.If T1 and T2 are similar, then W1 and W2 are isomorphic vector spaces.
- B.If W1 and W2 are isomorphic vector spaces, then T1 and T2 have the same minimal polynomial.✓
- C.If W1=W2=V, then T1 and T2 are similar.
- D.If W1 and W2 are isomorphic, then T1 and T2 have the same characteristic polynomial.
Solution
dim W = number of Jordan blocks. Equal numbers of blocks do not fix the largest block (minimal polynomial): blocks {2,2} vs {3,1} in dimension 4. (1) similar ⇒ equal kernel dimension. (3) W = V ⇒ T = 0. (4) nilpotent ⇒ characteristic polynomial xn always.
Let G be a group of order 2023. Which of the following statements are true?
- A.G is an Abelian group.✓
- B.G is a cyclic group.
- C.G is a simple group.
- D.G is not a simple group.✓
Solution
2023=7⋅172.n7≡1(mod7) and n7 | 289⇒n7=1;n17 | 7 and ≡1(mod17)⇒n17=1. So G≅Z7×P with P of order 172 abelian: G is abelian and not simple. P may be Z17×Z17, so G need not be cyclic.
Let f(z) be an entire function on C. Which of the following statements are true?
- A.f(z̄) is an entire function.
- B.conj(f(z)) is an entire function.
- C.conj(f(z̄)) is an entire function.✓
- D.conj(f(z̄)) + f(z̄) is an entire function.
Solution
If f=∑anzn then conj(f(zˉ))=∑ ānzn is entire. z ↦ f(z̄) and z ↦ conj(f(z)) are anti-holomorphic (entire only if f is constant). (4) is entire plus non-entire ⇒ not entire in general.
Let f∈C1(R) be bounded. Consider the initial-value problem (P): x′(t) = f(x(t)), t > 0, x(0) = 0. Which of the following statements are true?
- A.(P) has solution(s) defined for all t > 0.✓
- B.(P) has a unique solution.✓
- C.(P) has infinitely many solutions.
- D.The solution(s) of (P) is/are Lipschitz.✓
Solution
f∈C1⇒ locally Lipschitz ⇒ unique local solution (Picard). |x′| ≤ sup|f| <∞⇒ the solution cannot blow up, so it is global, and it is Lipschitz with constant sup|f|.
Let u:R2→R solve ∂xu+2∂yu=0 on R2 with u(x, y) = sin x on the line y = 3x + 1, and let v:R2→R solve ∂xv+2∂yv=0 with v(x, 0) = sin x. Let S = [0,1] × [0,1]. Which of the following statements are true?
- A.u changes sign in the interior of S.
- B.u(x, y) = v(x, y) along a line in S.✓
- C.v changes sign in the interior of S.✓
- D.v vanishes along a line in S.✓
Solution
Both are constant on lines y − 2x = c. v = sin(x − y/2), which vanishes on x = y/2 and changes sign across it inside S. For u, the data line y = 3x + 1 meets the characteristic through (x, y) at x0=y−2x−1, so u = sin(y − 2x − 1); on S the argument lies in [−3,0]⊂(−π,0], so u ≤ 0 and does not change sign. u = v where y − 2x − 1 = x − y/2, i.e. on the line y = 2x + 2/3, which crosses S.
Let y(x) and z(x) be the stationary functions (extremals) of J(y,z)=∫01[(y′)2+(z′)2+y′z′] dx subject to y(0) = 1, y(1) = 0, z(0) = −1, z(1) = 2. Which of the following statements are correct?
- A.z(x) + 3y(x) = 2 for x ∈ [0,1].✓
- B.3z(x) + y(x) = 2 for x ∈ [0,1].
- C.y(x) + z(x) = 2x for x ∈ [0,1].✓
- D.y(x) + z(x) = x for x ∈ [0,1].
Solution
Euler–Lagrange: 2y″ + z″ = 0 and 2z″ + y″ = 0 ⇒ y″ = z″ = 0. So y = 1 − x and z = −1 + 3x. Then y + z = 2x and z + 3y = 2.
Let A, B be two events in a discrete probability space with P(A) > 0 and P(B) > 0. Which of the following are necessarily true?
- A.If P(A | B) = 0 then P(B | A) = 0.✓
- B.If P(A | B) = 1 then P(B | A) = 1.
- C.If P(A | B) > P(A) then P(B | A) > P(B).✓
- D.If P(A | B) > P(B) then P(B | A) > P(A).
Solution
(1) P(A∩B) = 0 is symmetric. (3) P(A|B) > P(A) ⇔ P(A∩B) > P(A)P(B), symmetric in A, B. (2) A ⊇ B (a.s.) does not give B ⊇ A. (4) Take B ⊂ A with P(B) small.
Suppose X1,X2,… are independent and identically distributed N(0,1) random variables and Yn=X14+X24+⋯+Xn4. Which of the following probabilities converge to 1/2 as n→∞?
- A.P{Yn∈[0,2n]}
- B.P{Yn∈[n,3n]}✓
- C.P{Yn∈[2n,4n]}
- D.P{Yn∈[3n,5n]}✓
Solution
E[X4]=3, so Yn/n→3(LLN) and (Yn−3n)/n is asymptotically normal (CLT). Intervals with 3n as an endpoint have probability → 1/2; [2n, 4n] contains 3n in its interior (→ 1); [0, 2n] excludes it (→ 0).
Let X1 and X2 be independent, X1 gamma with mean 10 and variance 10, and X2 ~ N(3, 4). Let f1,f2 be their densities. Define Y with density f(y)=0.4f1(y)+qf2(y). Which of the following are true?
- A.q = 0.6✓
- B.E[Y] = 5.8✓
- C.Var(Y) = 3.04
- D.Y=0.4X1+ qX2
Solution
Densities integrate to 1 ⇒ q = 0.6. Mixture mean 0.4⋅10+0.6⋅3=5.8.E[Y2]=0.4(10+100)+0.6(4+9)=51.8, so Var(Y)=51.8−5.82=18.16.A mixture is not a linear combination of the variables.
Let {Xi:1≤i≤2n} be i.i.d. normal with mean μ and variance 1, independent of a standard Cauchy random variable W. Which of the following statistics are consistent for μ?
- A.n⁻1∑i=1n Xi✓
- B.n⁻1∑i=12n Xi
- C.n⁻1∑i=1n X2i₋1✓
- D.n⁻1(∑i=1n Xi+W)✓
Solution
(1), (3) are sample means of n i.i.d. terms. (2) converges to 2μ.(4)W/n→0 in probability since W is a fixed random variable, so consistency is preserved.
Under H: X ~ p with p(x) = 1/20, and under K: X ~ q with q(x) = x/210, x ∈ {1, …, 20}. Define test functions φ(x)=1 if x ≤ 2 (else 0) and ψ(x)=1 if x ≥ 19 (else 0). Which of the following statements are true?
- A.Size of the test φ is 0.1.✓
- B.Size of the test ψ is 0.05.
- C.(Power of the test ψ)>0.05.✓
- D.(Power of the test ψ)>(Power of the test φ).✓
Solution
Sizes: P_H(X ≤ 2) = 2/20 = 0.1 and P_H(X ≥ 19) = 0.1. Powers: P_K(X ≥ 19) = 39/210 ≈ 0.186 and PK(X≤2)=3/210≈0.014.ψ rejects where the likelihood ratio is largest — the Neyman–Pearson direction.
Keep going
Drill these by trap type rather than by paper, sit a full timed paper with the real attempt limits, or work the syllabus topic by topic with curated lectures.