Consider the following subset of {}. Which one of the following statements is true?
CSIR NET December 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.
Part B
One correct option. 3 marks, −0.75 for a wrong answer.
- A.inf U = 5.
- B.inf U = 4.
- C.inf U = 3.✓
- D.inf U = 2.
Solution
x2−9x+18≤0⇔x∈[3,6];x2−7x+12≤0⇔x∈[3,4]. So U = [3, 4] and inf U = 3.
Consider the following infinite series: (a)∑n≥1 sin(nπ/2)/n,(b)∑n≥1 log(1+1/n2). Which one of the following statements is true?
- A.(a) is convergent, but (b) is not convergent.
- B.(a) is not convergent, but (b) is convergent.
- C.Both (a) and (b) are convergent.✓
- D.Neither (a) nor (b) is convergent.
Solution
(a): the non-zero terms are (−1)k/2k+1, an alternating series with terms decreasing to 0 (Leibniz).(b):log(1+1/n2) ~ 1/n2, comparison with a convergent p-series.
Let f(x) be a cubic polynomial with real coefficients. Suppose that f(x) has exactly one real root and that this root is simple. Which one of the following statements holds for ALL antiderivatives F(x) of f(x)?
- A.F(x) has exactly one real root.
- B.F(x) has exactly four real roots.
- C.F(x) has at most two real roots.✓
- D.F(x) has at most one real root.
Solution
F′ = f changes sign exactly once, so F is monotone on each side of one point: at most two real roots (and the constant of integration can realise 0, 1 or 2).
Let X be a non-empty finite set and Y = {f−1(0) : f is a real-valued function on X}. Which one of the following statements is true?
- A.Y is an infinite set.
- B.Y has 2∣X∣ elements.✓
- C.There is a bijective function from X to Y.
- D.There is a surjective function from X to Y.
Solution
Every subset of X is the zero set of its indicator-complement (f = 1 off the subset, 0 on it), so Y is the power set of X, with 2∣X∣ > |X| elements.
Let A be an n × n matrix with complex entries. If n ≥ 4, which one of the following statements is true?
- A.A does not have any non-zero invariant subspace in Cn.
- B.A has an invariant subspace in Cn of dimension n − 3.✓
- C.All eigenvalues of A are real.
- D.A2 does not have any invariant subspace in Cn of dimension n − 1.
Solution
Over C every matrix is triangularisable (Schur), so the span of the first k basis vectors of a triangularising basis is an invariant subspace of every dimension k — in particular n − 3 ≥ 1.
Let A=(aij) be the n × n real matrix with aij= ij for all 1 ≤ i, j ≤ n. If n ≥ 3, which one of the following is an eigenvalue of A?
- A.1
- B.n
- C.n(n+1)/2
- D.n(n+1)(2n+1)/6✓
Solution
A = vvᵀ with v = (1, 2, …, n) has rank 1; its only non-zero eigenvalue is vTv=∑i2=n(n+1)(2n+1)/6.
For a∈R, let A_a = [[2, −1, 0], [−1, 2, −1], [0, −1, a]]. Which one of the following statements is true?
- A.A_a is positive definite for all a < 3.
- B.A_a is positive definite for all a > 3.✓
- C.A_a is positive definite for all a ≥ −2.
- D.A_a is positive definite only for finitely many values of a.
Solution
Leading principal minors: 2, 3 and det A_a = 3a − 2. Positive definite iff a > 2/3, which contains all a > 3.
Let G be any finite group. Which one of the following is necessarily true?
- A.G is a union of proper subgroups.
- B.G is a union of proper subgroups if |G| has at least two distinct prime divisors.
- C.If G is abelian, then G is a union of proper subgroups.
- D.G is a union of proper subgroups if and only if G is not cyclic.✓
Solution
A finite group is the union of its proper subgroups iff it is not cyclic: if it is cyclic a generator lies in no proper subgroup; if not cyclic every element generates a proper subgroup.
Which one of the following is equal to 137+237+337+⋯+8837 in Z/89Z?
- A.88
- B.−88
- C.−2
- D.0✓
Solution
89 is prime and (Z/89)∗ is cyclic of order 88. For a generator g,∑k k37=∑j g37j is a geometric sum equal to 0 because 88 ∤ 37 (so g37=1).
Let f:C→C be a real-differentiable function and define u(x, y) = Re f(x + iy), v(x, y) = Im f(x + iy). Let ∇u=(ux,uy) denote the gradient. Which one of the following is necessarily true?
- A.For c1,c2∈C, the level curves u=c1 and v=c2 are orthogonal wherever they intersect.
- B.∇u⋅∇v=0 at every point.
- C.If f is an entire function, then ∇u⋅∇v=0 at every point.✓
- D.If ∇u⋅∇v=0 at every point, then f is an entire function.
Solution
For entire f the Cauchy–Riemann equations give ∇u⋅∇v=uxvx+uyvy=ux(−uy)+uyux=0. Without holomorphy there is no such relation, and orthogonal gradients alone do not force CR (e.g. f = z̄ has ∇u⋅∇v=0).
How many roots does the polynomial z100−50z30+40z10+6z+1 have in the open disc {z∈C : |z| < 1}?
- A.100
- B.50
- C.30✓
- D.0
Solution
On |z| = 1, |−50z30| = 50 > 1 + 40 + 6 + 1 = 48 ≥ |z100+40z10+6z+1|, so by Rouché the polynomial has as many zeros inside as −50z30: thirty.
The partial differential equation x2uxx−2xy uxᵧ −3y2uyy+ux−uy=0 is
- A.elliptic in {(x, y) : y > 0}
- B.parabolic in {(x, y) : x > 0, y > 0}
- C.hyperbolic in {(x, y) : xy ≠ 0}✓
- D.parabolic in {(x, y) : xy ≠ 0}
Solution
B2−4AC =4x2y2+12x2y2=16x2y2>0 wherever xy ≠ 0: hyperbolic.
Using Euler's method with step size 0.05, the approximate value of the solution of dy/dx =3x+2y+1,y(1)=1, at x = 1.1 (rounded to two decimal places) is
- A.1.50
- B.1.65
- C.1.25✓
- D.1.15
Solution
y1=1+0.056≈1.1225;y2=1.1225+0.053.15+2.245+1≈1.1225+0.05⋅2.529≈1.249≈1.25.
The value of λ for which the integral equation y(x)=λ∫01x2 ex+t y(t) dt has a non-zero solution is
- A.4/(1+e2)
- B.2/(1+e2)
- C.4/(e2−1)✓
- D.2/(e2−1)
Solution
Separable kernel: y=cx2eˣ with c=λc∫01t2 e2t dt =λc(e2−1)/4, so λ=4/(e2−1).
Suppose X ~ Poisson(3/4). Then which of the following statements is true?
- A.P(X > 9) ≥ 11/12
- B.P(X < 9) ≥ 11/12✓
- C.E(X−3/4)2≥11/12
- D.(11/9)X ~ Poisson(11/12)
Solution
Markov: P(X ≥ 9) ≤ E[X]/9 = 1/12, so P(X < 9) ≥ 11/12. The variance is 3/4 < 11/12, and a scaled Poisson is not Poisson.
Let X1,X2,… be i.i.d. with CDF F(x) = 0 for x < 5 and 1 − e5−x for x ≥ 5. Define Yn=min{X1,…,Xn} and Zn=n(Yn−5), and let Z be standard normal. Which one of the following statements is true?
- A.limP(1/2<Yn<3/2)=1
- B.Yn→5 in probability as n→∞✓
- C.Zn→Z in distribution as n→∞
- D.limP(1<Zn<2)=Φ(2)−Φ(1)
Solution
Yn−5 ~ Exp(n), so Yn→5 in probability while n(Yn−5)→0 in probability (the right scaling is n, not n).
The probability of a head in tossing a coin is p ∈ (0, 1). The coin is independently tossed 25 times and heads appear 10 times. The Bayes estimate of p with respect to the prior Beta(5, 5) and squared error loss is
- A.3/7✓
- B.3/5
- C.1/2
- D.2/5
Solution
Posterior is Beta(5 + 10, 5 + 15) = Beta(15, 20) with mean 15/35 = 3/7.
For n ≥ 2, let X1,…,Xn be a random sample with density f(x|θ)= θxθ−1 on 0<x<1,θ>0 unknown. Which of the following is the UMVUE for 1/θ?
- A.−(1/n)∑lnXi✓
- B.−n/∑lnXi
- C.−(n−1)/∑lnXi
- D.−(2/n)∑lnXi
Solution
−lnXi ~ Exp(θ) with mean 1/θ, and ∑lnXi is complete sufficient; the sample mean of −lnXi is unbiased for 1/θ.
In a Latin square design, the degrees of freedom for the error sum of squares is 42. Then the degrees of freedom for the sum of squares due to treatments is
- A.6
- B.7✓
- C.8
- D.9
Solution
For a p × p Latin square the error d.f. is (p − 1)(p − 2) = 42 ⇒ p = 8, so treatments have p − 1 = 7 d.f.
Part C
One or more correct options. 4.75 marks, no negative marking, and credit only for exactly the right set.
Let x be a real number. Which of the following statements are true?
- A.There exists an integer n ≥ 1 such that n2sin(1/n)≥x.✓
- B.There exists an integer n ≥ 1 such that n cos(1/n) ≥ x.✓
- C.There exists an integer n ≥ 1 such that n e−n ≥ x.
- D.There exists an integer n ≥ 2 such that n (log n)−1 ≥ x.✓
Solution
n2sin(1/n) ~ n, n cos(1/n) ~ n and n/log n all tend to +∞, so each eventually exceeds any x. n e−n ≤ 1/e is bounded, so it fails for x > 1/e.
Let f:[0,∞)→R be the periodic function of period 1 given by f(x) = 1 − |2x − 1| for x ∈ [0, 1], and define g:[0,∞)→R by g(x)=f(x2). Which of the following statements are true?
- A.f is continuous on [0,∞).✓
- B.f is uniformly continuous on [0,∞).✓
- C.g is continuous on [0,∞).✓
- D.g is uniformly continuous on [0,∞).
Solution
f is a continuous tent function with f(0) = f(1) = 0, so its periodic extension is continuous; continuous periodic ⇒ uniformly continuous. g = f∘x2 is continuous but its oscillations speed up as x→∞(the tents get compressed), so it is not uniformly continuous.
For a real number λ, consider the improper integrals Iλ=∫01 dx/(1−x)λ and Kλ=∫1∞ dx/xλ. Which of the following statements are true?
- A.There exists λ such that Iλ converges, but Kλ does not converge.✓
- B.There exists λ such that Kλ converges, but Iλ does not converge.✓
- C.There exists λ such that Iλ,Kλ both converge.
- D.There exists λ such that neither Iλ nor Kλ converges.✓
Solution
Iλ converges iff λ<1;Kλ converges iff λ>1. So they never converge together, and at λ=1 neither does.
For real numbers a, b, c, d, e, f, consider the function F:R2→R2 given by F(x, y) = (ax + by + c, dx + ey + f). Which of the following statements are true?
- A.F is continuous.✓
- B.F is uniformly continuous.✓
- C.F is differentiable.✓
- D.F has partial derivatives of all orders.✓
Solution
An affine map is Lipschitz (hence uniformly continuous), differentiable with constant derivative, and all higher partials are zero.
Let A be an n × n real symmetric matrix. Which of the following statements are necessarily true?
- A.A is diagonalizable.✓
- B.If Aᵏ = I for some positive integer k, then A2=I.✓
- C.If Aᵏ = 0 for some positive integer k, then A2=0.✓
- D.All eigenvalues of A are real.✓
Solution
Spectral theorem: A = QDQᵀ with real D. Aᵏ = I forces each real eigenvalue to satisfy λk=1, so λ=±1 and A2=I;Ak=0 forces all λ=0 so A = 0.
Let n∈Z be such that n ≡ 1 (mod 7) and n ≡ 4 (mod 15). Which of the following statements are true?
- A.n ≡ 1 (mod 3).✓
- B.n ≡ 1 (mod 35).
- C.n ≡ 1 (mod 21).✓
- D.n ≡ 1 (mod 5).
Solution
n ≡ 4 (mod 15) gives n ≡ 1 (mod 3) and n ≡ 4 (mod 5). With n ≡ 1 (mod 7), CRT gives n ≡ 1 (mod 21); n ≢ 1 (mod 5) kills the mod-35 claim.
Let f(X)=X3−2∈Q[X] and let K⊂C be the splitting field of f(X) over Q. Let ω= e2πi/3. Which of the following statements are true?
- A.The Galois group of K over Q is the symmetric group S3.✓
- B.The Galois group of K over Q(ω) is the symmetric group S3.
- C.The Galois group of K over Q is Z/3Z.
- D.The Galois group of K over Q(ω) is Z/3Z.✓
Solution
K=Q(32,ω) has degree 6 over Q with non-abelian Galois group S3; over Q(ω) the degree is 3, so the group is Z/3.
For an integer k, consider the contour integral Ik=∫∣z∣=1 eᶻ/zᵏ dz. Which of the following statements are true?
- A.Ik=0 for every integer k.
- B.Ik=0 if k ≥ 1.✓
- C.|Ik| ≤ |Ik₊1| for every integer k.
- D.limk→∞ |Ik| =∞.
Solution
Ik=2πi⋅(coefficient of zk−1 in eᶻ)=2πi/(k−1)! for k ≥ 1 and 0 for k ≤ 0. These values decrease to 0, so (3) and (4) fail.
Consider the initial value problem x2y′′−2x2y′+(4x−2)y=0,y(0)=0. Suppose y=φ(x) is a polynomial solution satisfying φ(1)=1. Which of the following statements are true?
- A.φ(4)=16✓
- B.φ(2)=2
- C.φ(5)=25✓
- D.φ(3)=3
Solution
Try y=x2:2x2−4x3+4x3−2x2=0. So φ(x)=x2, giving φ(4)=16 and φ(5)=25.
Suppose U ~ Uniform(0,1) and X=tan(π(U− ½)). Then which of the following statements are true?
- A.E(X4)=3
- B.P(X ∈ {1, 2, 5}) = 1/2
- C.E(eˣ) does not exist✓
- D.P(X ≤ 0) = 1/2✓
Solution
X is standard Cauchy: symmetric about 0, continuous (so any finite set has probability 0), and with no finite moments — E(eˣ) diverges too.
Consider a population of 3 units having values 2, 4 and 6. A simple random sample (without replacement) of 2 units is drawn. Let M be the sample mean. Which of the following statements are true?
- A.E(M) = 4✓
- B.E(M2)=17
- C.E(M3)=72✓
- D.Var(M) = 1
Solution
M takes the values 3, 4, 5 with probability 1/3 each: E(M)=4,E(M2)=50/3,E(M3)=72,Var(M)=2/3.
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.