MATH40002 2026 Full Paper K

QUESTION 1 — COMPLETE WORKED SOLUTION **(a)** a_n=\log\left(1+\frac4{n+2}\right)>0. This proves positivity and, by the displayed denominator/integral estimate, \(a_n\to0\). Moreover \[ na_n\to4\quad\text{from }\log(1+t)/t\to1. \] The same bounds give an explicit \(N(\varepsilon)\), so this is an \(\varepsilon\)-\(N\) proof rather than a decimal approximation. **(b)** For the partial sums, \[ S_N=\log\frac{(N+3)(N+4)(N+5)(N+6)}{3\cdot4\cdot5\cdot6}\longrightarrow+\infty. \] Thus the positive series **diverges**. Equivalently, limit comparison with \(a_n\sim4/n\) gives the same conclusion. **(c)** The positive terms decrease to zero. Hence the alternating series converges by the alternating-series theorem. Its remainder satisfies \[ |S-S_N|\le a_{N+1},\qquad a_{N+1}\le\frac4{N+3}<10^{-4}; N>39997\text{ is sufficient}. \] **(d)** Polynomial decay of the coefficients gives \(\limsup |a_n|^{1/n}=1\), hence the radius is \(R=1\). Because \(\sum a_n\) diverges, \(z=1\) diverges; \(z=-1\) and every \(e^{i\theta}\ne1\) converge conditionally by the alternating/Dirichlet tests. No point on \(|z|=1\) is absolutely convergent. Completeness and checks. The coefficient estimate must be proved before any series test is invoked. A positive-term series is decided from its partial sums or a two-sided comparison; the fact that its terms tend to zero is only necessary. For the alternating series one must separately establish positivity, eventual monotonicity and convergence to zero, after which the first omitted term bounds the error. For the power series, prove absolute convergence inside the circle, divergence at the exceptional positive boundary point, and use bounded geometric partial sums plus Dirichlet at every remaining non-trivial unit-circle point. Common errors: treating the radius as a boundary test, omitting monotonicity, or calling conditional convergence absolute. SCORING AND EQUIVALENT METHODS. Each labelled subpart is split into method/conditions marks and conclusion/verification marks in the published marking scheme. Algebraically equivalent formulae, a different valid theorem route, or a different correct drawing order receive the same credit. A consistent arithmetic slip is followed through; it does not erase earlier correct mathematics.

QUESTION 2 — COMPLETE WORKED SOLUTION **(a)** Fix \(x\in[0,2]\). Choose rationals \(q_n\to x\). Continuity gives \[ f(x)=\lim f(q_n)=\lim\bigl(2q_n^3+q_n\bigr) =2x^3+x. \] **(b)** Let \(L_0=\lim_{x\to0^+}g(x)\) and \(L_2=\lim_{x\to2^-}g(x)\). Define the extension by \(G(0)=L_0\), \(G(2)=L_2\), and \(G=g\) inside. The definitions of the one-sided limits prove endpoint continuity. Thus \(G\) is continuous on compact \([0,2]\), so Heine-Cantor makes \(G\), and therefore \(g\), uniformly continuous. **(c)** The function is continuous and bounded by 1. Put \[ x_n=(\pi/2+2\pi n)^{-1/3},\qquad y_n=(3\pi/2+2\pi n)^{-1/3}. \] Then \(x_n,y_n\to0\), \(|x_n-y_n|\to0\), but \(h(x_n)=1\) and \(h(y_n)=-1\). This contradicts uniform continuity. **(d)** A continuous function on a compact metric domain is uniformly continuous (Heine-Cantor theorem). Completeness and checks. Density arguments require an explicitly chosen sequence from the dense subset converging to an arbitrary target. Endpoint values of an extension are forced by the one-sided limits; continuity must be checked at each added endpoint before Heine-Cantor can be cited. A restriction of a uniformly continuous function remains uniformly continuous. To disprove uniform continuity, give two concrete sequences in the domain with distance tending to zero while their image distance stays above one fixed positive number. Common errors: using a point-dependent delta, citing compactness for an open interval, or giving oscillating sequences whose input distance does not actually tend to zero. SCORING AND EQUIVALENT METHODS. Each labelled subpart is split into method/conditions marks and conclusion/verification marks in the published marking scheme. Algebraically equivalent formulae, a different valid theorem route, or a different correct drawing order receive the same credit. A consistent arithmetic slip is followed through; it does not erase earlier correct mathematics.

QUESTION 3 — COMPLETE WORKED SOLUTION **(a)** The answers are **True, True, True**. (i) True: if |g| is bounded below, 1/g is the continuous composition of g with the reciprocal map, hence is Darboux integrable; multiply by f. (ii) True: pass the convex inequality to the uniform limit. (iii) True: a derivative has the intermediate-value property, so a non-zero value at the exceptional point would force non-zero values nearby. **(b)–(d)** For a partition \(P:a=x_0<\cdots<x_m=b\), put \[ m_i=\inf_{[x_{i-1},x_i]}f,\quad M_i=\sup_{[x_{i-1},x_i]}f, \quad L(f,P)=\sum m_i\Delta x_i,\quad U(f,P)=\sum M_i\Delta x_i. \] A bounded \(f\) is Darboux integrable iff \(\sup_P L(f,P)=\inf_P U(f,P)\), equivalently iff for every \(\varepsilon>0\) there is a partition \(P\) with \(U(f,P)-L(f,P)<\varepsilon\). If \(f^{(n)}\) is absolutely continuous (continuous \(f^{(n+1)}\) is sufficient), then \[ f(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k+ \frac1{n!}\int_a^x(x-t)^n f^{(n+1)}(t)\,dt. \] One standard L'Hopital form assumes differentiability on a punctured interval, \(g'\ne0\), a \(0/0\) or \(\infty/\infty\) form, and an existing limit of \(f'/g'\); under the standard endpoint hypotheses the quotient has the same limit. Completeness and checks. Every true/false item earns its conclusion only after a proof or a counterexample with domain and hypotheses checked. The Darboux definition must construct upper and lower sums from suprema and infima and state equality of upper and lower integrals (or the epsilon criterion). Taylor with integral remainder requires the relevant continuity of the derivatives on an interval containing both points. A valid L'Hopital statement must name the indeterminate form, differentiability interval, non-vanishing denominator derivative and existence of the derivative-ratio limit. Common errors: supplying only theorem names, using a counterexample outside the domain, or silently strengthening the printed hypotheses. SCORING AND EQUIVALENT METHODS. Each labelled subpart is split into method/conditions marks and conclusion/verification marks in the published marking scheme. Algebraically equivalent formulae, a different valid theorem route, or a different correct drawing order receive the same credit. A consistent arithmetic slip is followed through; it does not erase earlier correct mathematics.

QUESTION 4 — COMPLETE WORKED SOLUTION **(a)** Since \(f''\) has the stated sign, \(f'\) is non-increasing. The supporting line inequality \[ f(y)\le f(x)+f'(x)(y-x) \] and boundedness rule out a derivative with the wrong strict sign. Therefore \(f'\ge0\) and \(f\) is non-decreasing. **(b)** With \(\ell=\sup f\), choose a point whose function value is within \(\varepsilon\) of \(\ell\). Monotonicity traps every later value between that value and \(\ell\), so \(f(x)\to\ell\). **(c)** Monotonicity of \(f'\) gives a one-sided limit. The sign from (a) and boundedness rule out every non-zero limit: integrating a strictly positive or negative tail bound would make \(f\) unbounded. Hence \(f'(x)\to0\). **(d)** Monotonicity of \(f'\) on \([x,3x]\) and integration give \[ 2x f'(3x)\le f(3x)-f(x)\le 2x f'(x). \] The middle increment tends to zero by (b). Applying the same estimate on a rescaled preceding interval traps \(xf'(x)\) between 0 and a vanishing function increment, so \(xf'(x)\to0\). **(e)** Without boundedness, \(f(x)=-x\) is concave but has a non-zero constant derivative; the monotonicity conclusion in (a) and the derivative/asymptotic conclusions can fail. Completeness and checks. The sign of the second derivative makes the first derivative monotone. If its sign allowed persistent motion in the forbidden direction, the mean-value theorem and integration would contradict boundedness. Monotonicity plus the appropriate bound identifies the limit of f by an infimum or supremum argument with full epsilon quantifiers. The derivative limit is then obtained by ruling out every non-zero candidate. Integrating the monotone derivative over a scaled interval produces the two-sided inequality; applying it at two scales and using the already proved limit gives the squeeze for x f'(x). Common errors: assuming the derivative limit exists without monotonicity or inferring x f'(x)→0 merely from f'(x)→0. SCORING AND EQUIVALENT METHODS. Each labelled subpart is split into method/conditions marks and conclusion/verification marks in the published marking scheme. Algebraically equivalent formulae, a different valid theorem route, or a different correct drawing order receive the same credit. A consistent arithmetic slip is followed through; it does not erase earlier correct mathematics.