MATH40004 2026 Exam Standard Full Paper B

Put \(h=x-1\). **(a)** If \(f\) is \(n\)-times differentiable near \(a\), a local Taylor form is \[ f(x)=\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a)^k+o((x-a)^n). \] Under the usual extra \(C^{n+1}\) hypothesis one may instead give the Lagrange or integral remainder. **(b)** Differentiation at \(x=1\) gives \[ \arctan(1+h)=\frac\pi4+\frac h2-\frac{h^2}4+ \frac{h^3}{12}+O(h^4). \] Before substituting another point one must check that it lies inside the interval/radius on which the Taylor series converges (the nearest complex singularities are at \(\pm i\)). **(c)** \[ \lim_{x\to0^+}g(x)=\frac\pi4,\qquad \lim_{x\to1}g(x)=\frac12,\qquad \lim_{x\to\infty}g(x)=0. \] Moreover \[ g(1+h)=\frac12-\frac h4+\frac{h^2}{12}+O(h^3). \] The numerator is analytic and has a simple zero at \(x=1\), so the quotient has a removable singularity and its extension is \(C^\infty\) (indeed real analytic) at \(1\). **(d)** Near \(0\), \(g/(1+x)\to\pi/4\), so there is no singularity. At \(1\) the quotient has the removable finite value \(1/4\). As \(x\to\infty\), \[ g(x)\sim\frac{\pi}{4x},\qquad \frac{g(x)}{1+x}\sim\frac{\pi}{4x^2}. \] Comparison with \(x^{-2}\) proves ordinary absolute convergence of the integral after splitting, for example, at \(1\) and at any large finite point. No principal-value interpretation is needed.

Let \(h(0)=h_0\). **(a)** Since the stored volume is \(V=Sh\), \[ S\dot h=-c\sqrt h-kh. \] Thus \(k\) has dimensions area/time (so that \(kh\) is volume/time). With \(u=\sqrt h\), \(dh=2u\,du\), hence \[ dt=-\frac{2S\,du}{c+ku}. \] The emptying time is \[ T=2S\int_0^{\sqrt{h_0}}\frac{du}{c+ku} =\frac{2S}{k}\log\left(1+\frac{k\sqrt{h_0}}c\right). \] **(b)** As \(k\to0\), \(S\dot h=-c\sqrt h\) and \[ T_0=\frac{2S\sqrt{h_0}}c. \] Because \(\log(1+x)<x\) for \(x>0\), \(T<T_0\). The porous-wall loss adds outflow, so the tank empties sooner. **(c)** With the standard \(2\pi\)-periodic convention, \[ a_0=2,\quad a_n=0,\quad b_n=\frac{2}{\pi n}(1-(-1)^n). \] Therefore \[ f(x)\sim1+\frac4\pi\left(\sin x+\frac{\sin3x}{3} +\frac{\sin5x}{5}+\cdots\right). \] At each jump the series equals the midpoint, \(1\). At \(x=\pi/2\), \[ 2=1+\frac4\pi\left(1-\frac13+\frac15-\cdots\right), \] so \[ \frac\pi4=1-\frac13+\frac15-\frac17+\cdots. \]

**(a)** The characteristic polynomial is \(\lambda^2+1\), so \(\lambda=\pm i\). For \(\lambda=i\), one may take \(v=(2+i,1)^T\); the conjugate is an eigenvector for \(-i\). Equivalently \(A^2=-I\), hence \[ X_h(t)=(I\cos t+A\sin t)C,\qquad C\in\mathbb R^2. \] All nonzero solutions are bounded periodic closed orbits. The origin is a neutrally stable centre; there is no convergence to or escape from the origin as \(t\to\infty\). **(b)** Neither 1 nor 0 is an eigenvalue, so there is no resonance. For the \(e^t(1,0)^T\) term set \(X_{p1}=pe^t\); \((I-A)p=(1,0)^T\) gives \(p=(3/2,1/2)^T\). For the constant forcing set \(X_{p2}=q\); \(Aq=(0,-3)^T\) gives \(q=(-15,-6)^T\). Thus \[ X(t)=(I\cos t+A\sin t)C+ e^t\binom{3/2}{1/2}+\binom{-15}{-6}. \] **(c)** Here \(M=2x^2y\), \(N=x^3+3xy^2\), and \(M_y=2x^2\ne N_x=3x^2+3y^2\), so the form is not exact. On either half-plane \(x>0\) or \(x<0\), multiply by \(x^my^n\). Equality of the cross-partials forces \[ m=-1,\qquad n=0. \] The resulting form is \[ 2xy\,dx+(x^2+3y^2)\,dy=0, \] which is exact. Integrating its first component gives \(\Phi=x^2y+\phi(y)\); matching \(\Phi_y=x^2+3y^2\) yields \(\phi(y)=y^3\). Hence \[ x^2y+y^3=C. \]

**(a)** Write \[ F_r(y)=\frac{y(2r+(r-1)y)}{2+y}. \] The singular state is \(y=-2\). The equilibria are \(y_0=0\) for all \(r\), and, when \(r\ne1\), \[ y_*(r)=\frac{2r}{1-r}. \] The linearisation derivatives are \[ F_r'(0)=r,\qquad F_r'(y_*)=r(r-1). \] Thus \(y=0\) is stable for \(r<0\) and unstable for \(r>0\). The second branch is stable for \(0<r<1\) and unstable for \(r<0\) or \(r>1\). At \(r=0\) the two finite branches cross and exchange stability: a transcritical bifurcation. At \(r=1\) the second branch escapes through infinity rather than meeting a finite equilibrium. The line \(y=-2\) must be drawn as a singular barrier, and the phase-line arrows on each side follow from the sign of \(F_r(y)\). **(b)** Take \[ \widehat u(\xi)=\int_{\mathbb R}u(x)e^{-i\xi x}\,dx,\qquad u(x)=\frac1{2\pi}\int_{\mathbb R}\widehat u(\xi)e^{i\xi x}\,d\xi. \] Then \[ \mathcal F(u'')=-\xi^2\widehat u,\qquad \mathcal F(xu)=i\widehat u', \] so \[ -\xi^2\widehat u-i\widehat u'=0,\qquad \widehat u(\xi)=C e^{i\xi^3/3}. \] Interpreting the inverse as the standard oscillatory integral and pairing positive and negative frequencies gives \[ u(x)=\frac{C}{\pi}\int_0^\infty \cos\left(\frac{t^3}{3}+xt\right)\,dt=C\,\operatorname{Ai}(x). \] This is the real decaying Airy branch (up to scale). **(c)** At \((0,1,0)\), \(F=0\) and \[ F_x=ze^{xz}+y=1,\quad F_y=x=0,\quad F_z=xe^{xz}-1=-1\ne0. \] The implicit-function theorem applies and \[ z_x=-F_x/F_z=1,\qquad z_y=-F_y/F_z=0. \] Hence, near \((0,1)\), \[ z(x,y)=x+o\!\left(\sqrt{x^2+(y-1)^2}\right). \]