Problem-Solving Playbook
14 recipes — match the input form, follow the steps, score full marks
- 01LAMay 2024 MCQ1
Recipe 1: You are given a matrix $A$ and several candidate column vectors; the question asks which one is an eigenvector of $A$.
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May 2024 MCQ1, $A = \begin{pmatrix} 2 & 0 & 1 \\ 0 & -2 & -1 \\ 1 & 0 & 3 \end{pmatrix}$, candidates: (A) $(2,0,-1)^T$, (B) $(3,1,2)^T$, (C) $(0,1,0)^T$, (D) none. *Option C — $(0,1,0)^T$ has a zero first and third entry, so test it first:* $$A\begin{pmatrix}0\\1\\0\end{pmatrix} = \begin{pmatrix}2\cdot0+0\cdot1+1\cdot0\\0\cdot0+(-2)\cdot1+(-1)\cdot0\\1\cdot0+0\cdot1+3\cdot0\end{pmatrix} = \begin{pmatrix}0\\-2\\0\end{pmatrix} = -2\begin{pmatrix}0\\1\\0\end{pmatrix}.$$ The result is $-2$ times the original vector. By the definition of an eigenpair (§2.6 of `01_Linear_Algebra.md`): $A\mathbf{v} = \lambda\mathbf{v}$ with $\lambda = -2$. **Answer: C.** *(For completeness: option A gives $(3,1,-1)^T$; checking $3=2\lambda$ vs $-1=-\lambda$ forces $\lambda=3/2$ and $\lambda=1$ simultaneously — contradiction. Option B gives $(8,-4,9)^T$; $8=3\lambda$ and $-4=\lambda$ are contradictory.)* - 02DCMay 2024 MCQ2
Recipe 2: Given a $C^{(2)}$ function $f(x,y)$ and a centre $(a,b)$, you must identify the correct quadratic approximation $f(x,y) \approx f(a,b) + \nabla f(a,b)\cdot\mathbf{h} + \frac{1}{2}\mathbf{h}^T\nabla^2 f(a,b)\mathbf{h}$ where $\mathbf{h}=(x-a,y-b)$.
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May 2024 MCQ2, $f(x,y)=x^2+x\cos y$, centre $(1,-\pi/2)$. *Step 1:* $\cos(-\pi/2)=0$, so $f(1,-\pi/2)=1+1\cdot0=1$. *Step 2:* $f_x=2x+\cos y$; $f_y=-x\sin y$. At $(1,-\pi/2)$: $\cos(-\pi/2)=0$, $\sin(-\pi/2)=-1$, giving $\nabla f(1,-\pi/2)=[2,\,1]$. *Step 3:* $f_{xx}=2$; $f_{yy}=-x\cos y$; $f_{xy}=-\sin y$. At $(1,-\pi/2)$: $f_{xx}=2$, $f_{yy}=-1\cdot0=0$, $f_{xy}=-(-1)=1$. *Step 4:* With $h_1=x-1$, $h_2=y+\pi/2$: $$f = 1 + 2h_1 + h_2 + h_1^2 + h_1 h_2 + 0 + o(\|\mathbf{h}\|^2).$$ *Step 5:* Expand $h_1=x-1$, $h_2=y+\pi/2$ and collect. The constant terms: $1+2(-1)+1\cdot(\pi/2)+1+(-1)(\pi/2)=0$. Linear in $x$: $2+(-2)+(\pi/2)=\pi/2$. Linear in $y$: $1+(-1)=0$. Quadratic: $x^2$ from $h_1^2=(x-1)^2$; $xy$ from $h_1 h_2=(x-1)(y+\pi/2)$. Result: $$\boxed{f(x,y)=\tfrac{\pi}{2}\,x+x^2+xy+o(\|\mathbf{h}\|^2).}$$ This matches **none of A, B, C** (options A–C all omit the $\tfrac{\pi}{2}x$ term or have wrong signs). **Answer: D (none of the others).** - 03ICMay 2024 MCQ3
Recipe 3: Given an improper integral $\int_1^{+\infty}f(x)\,dx$ (or $\int_a^b$ with a singularity), you must determine whether it converges and/or how it compares to a standard integral.
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May 2024 MCQ3, $\int_1^{+\infty}\frac{x+\ln x}{x^3}\,dx$. *Step 1–2:* For $x\ge1$, $x>0$ and $\ln x\ge0$, so the integrand is positive. Singularity at $+\infty$. *Step 3:* As $x\to+\infty$, $\ln x=o(x)$, so $x+\ln x\sim x$, hence: $$\frac{x+\ln x}{x^3}\sim\frac{x}{x^3}=\frac{1}{x^2}.$$ *Step 4:* Here $p=2>1$, so $\int_1^\infty 1/x^2\,dx$ converges (equals $[-1/x]_1^\infty=0-(-1)=1$). By asymptotic comparison, the original integral **converges to a finite positive number**. Rules out A (diverges to $+\infty$) and B (does not exist). *Step 5 — Direction:* For $x>1$, $\ln x>0$, so $(x+\ln x)/x^3>x/x^3=1/x^2$. Pointwise domination. By Theorem 1928 (monotonicity of improper integrals): $$\int_1^{+\infty}\frac{x+\ln x}{x^3}\,dx\ge\int_1^{+\infty}\frac{1}{x^2}\,dx=1.$$ **Answer: D** (finite and $\ge\int_1^\infty 1/x^2\,dx$). - 04PRMay 2024 MCQ4
Recipe 4: Given a Poisson distribution with parameter $\lambda$ on $\Omega=\mathbb{N}$, you must compute the probability of the event $\{n\ge k\}$ for some threshold $k$.
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May 2024 MCQ4, Poisson with $\lambda=2$, event $E=\{n\ge3\}$. *Step 1–2:* Use complement: $P(E)=1-P(n\le2)=1-[P(0)+P(1)+P(2)]$. *Step 3–4:* Using $P(n=i)=2^i e^{-2}/i!$: $$P(n=0)=\frac{1}{1}\cdot e^{-2}=e^{-2},\quad P(n=1)=\frac{2}{1}\cdot e^{-2}=2e^{-2},\quad P(n=2)=\frac{4}{2}\cdot e^{-2}=2e^{-2}.$$ *Step 5:* $$P(n\le2)=e^{-2}+2e^{-2}+2e^{-2}=5e^{-2}.$$ $$P(n\ge3)=1-5e^{-2}.$$ **Answer: A** ($1-5e^{-2}\approx1-0.677=0.323$). - 05PRMay 2024 MCQ5
Recipe 5: Given $V(g)$, the correlation $\rho(f,g)$, and the absolute value $|\operatorname{Cov}(\alpha f,\beta g)|$ for known scalars $\alpha,\beta$, you must find $\sigma(f)$.
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May 2024 MCQ5, $V(g)=100$, $\rho(f,g)=-0.1$, $|\operatorname{Cov}(2f,5g)|=90$. *Step 1–2:* $|\operatorname{Cov}(2f,5g)|=2\cdot5\cdot|\operatorname{Cov}(f,g)|=10|\operatorname{Cov}(f,g)|=90$, so $|\operatorname{Cov}(f,g)|=9$. *Step 3:* $\rho=-0.1<0$, so $\operatorname{Cov}(f,g)=-9$. *Step 4:* $\sigma(g)=\sqrt{100}=10$. *Step 5:* $-0.1=\frac{-9}{\sigma(f)\cdot10}$. Cross-multiply: $-\sigma(f)=-9$, so $\sigma(f)=9$. **Answer: A** ($\sigma(f)=9$). - 06MFMay 2024 MCQ6
Recipe 6: Given a **pure investment** cash flow ($a_0<0$, $a_s>0$ for $s\ge1$) and sample values $G(i_1),G(i_2),\ldots$ of the discounted cash flow function $G(i)=\sum_{s=0}^n a_s(1+i)^{-s}$, you must determine which statement about the IRR is true.
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May 2024 MCQ6, cash flow with $a_0<0$, $a_1,a_2,a_3>0$; $G(0.03)=5$, $G(0.06)=-3$, $G(0.09)=-7$. *Step 1:* Pure investment confirmed ($a_0<0$, $a_s>0$) — $G$ strictly decreasing. *Step 2:* Sign change between $3\%$ and $6\%$: $G(0.03)=+5>0>-3=G(0.06)$. So $i^*\in(3\%,6\%)$. *Step 3:* Rules out A ($i^*<3\%$ would require $G(0.03)\le0$) and C ($i^*\in(6\%,9\%)$ would require a sign change there, but $G(0.06)=-3$ and $G(0.09)=-7$ are both negative). *Step 4:* $\lim_{i\to\infty}G(i)=a_0$ and $G(i)>a_0$ for all finite $i$. In particular $G(0.09)=-7>a_0$, so $a_0<-7$. **Answer: B** ($a_0<-7$). - 07LAMay 2024 Q7
Recipe 7: Given a $3\times3$ symmetric matrix $A$ (or a quadratic form from which you extract $A$), you must classify it as positive definite, positive semi-definite, negative definite, negative semi-definite, or indefinite.
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May 2024 Q7, $A=\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}$, quadratic form $q(x_1,x_2,x_3)=\mathbf{x}^TA\mathbf{x}$. *Step 1:* All $a_{ij}=1$, so by Proposition 1201: $q=(x_1+x_2+x_3)^2=x_1^2+x_2^2+x_3^2+2x_1x_2+2x_1x_3+2x_2x_3$. *Step 2:* $D_1=1$, $D_2=\det\begin{pmatrix}1&1\\1&1\end{pmatrix}=0$, $D_3=\det A=0$ (rank-1 matrix). *Step 3:* $D_2=D_3=0$ — Sylvester is inconclusive; switch to eigenvalues. *Step 4:* Characteristic polynomial: $\det(A-\lambda I)=-\lambda^2(\lambda-3)=0$. Eigenvalues: $\lambda=0$ (multiplicity 2), $\lambda=3$ (multiplicity 1). All eigenvalues $\ge0$, and $\lambda=0$ occurs, so $A$ is **positive semi-definite but not positive definite**. *Witness for "not PD"*: $q(1,-1,0)=(1-1+0)^2=0$ with $(1,-1,0)\ne\mathbf{0}$, confirming $q$ vanishes away from the origin. - 08DCMay 2024 Q8
Recipe 8: Given a $C^{(2)}$ function $f(x,y)$ on an open domain, you must (a) find all stationary points, (b) classify each via the Hessian, and (c) determine whether $f$ is convex/concave on the whole domain.
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May 2024 Q8, $f(x,y)=e^x(x-1)y+y^2$. *Part (a):* $f_x=xye^x$; $f_y=e^x(x-1)+2y$. From $xye^x=0$ (and $e^x>0$ always): $xy=0$, so $x=0$ or $y=0$. - Case $x=0$: $f_y=e^0(0-1)+2y=-1+2y=0\Rightarrow y=\tfrac{1}{2}$. Stationary point $P=(0,\tfrac{1}{2})$. - Case $y=0$: $f_y=e^x(x-1)=0\Rightarrow x-1=0\Rightarrow x=1$. Stationary point $Q=(1,0)$. *Part (b):* $f_{xx}=ye^x(x+1)$, $f_{yy}=2$, $f_{xy}=xe^x$. So $\nabla^2 f=\begin{pmatrix}ye^x(x+1)&xe^x\\xe^x&2\end{pmatrix}$. At $P=(0,\tfrac{1}{2})$: $\nabla^2 f=\begin{pmatrix}\tfrac{1}{2}&0\\0&2\end{pmatrix}$. $D_1=\tfrac{1}{2}>0$, $D_2=1>0$ $\Rightarrow$ **positive definite** $\Rightarrow$ $P$ is a **strict local minimum**. At $Q=(1,0)$: $\nabla^2 f=\begin{pmatrix}0&e\\e&2\end{pmatrix}$. $D_2=0\cdot2-e^2=-e^2<0$ $\Rightarrow$ **indefinite** $\Rightarrow$ $Q$ is a **saddle point**. *Part (c):* At $Q=(1,0)$, the Hessian is indefinite — hence not PSD. Therefore $f$ is **not convex on $\mathbb{R}^2$**. - 09ICMay 2024 Q9
Recipe 9: Part (a) asks to state and/or prove the First Fundamental Theorem of Calculus (Newton–Leibniz); part (b) asks to find an antiderivative of $f(x)=x^n\ln x$ (or a similar product) and evaluate a definite integral.
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May 2024 Q9, $f(x)=x^2\ln x$; find antiderivative $P$, then compute $\int_1^e x^2\ln x\,dx$. *Step 1 (LIATE):* Factors are $\ln x$ (Logarithm) and $x^2$ (Algebraic). Logarithm has higher priority $\Rightarrow$ differentiate $\ln x$, integrate $x^2$. *Step 2:* $g=\ln x$, $g'=1/x$; $f'=x^2$, $f=x^3/3$. *Step 3:* $\int x^2\ln x\,dx = \frac{x^3}{3}\ln x - \int\frac{x^3}{3}\cdot\frac{1}{x}\,dx = \frac{x^3}{3}\ln x - \frac{1}{3}\int x^2\,dx$. *Step 4:* $\frac{1}{3}\int x^2\,dx = \frac{x^3}{9}$. Result: $P(x)=\frac{x^3}{3}\ln x-\frac{x^3}{9}+C$. Check: $P'(x)=x^2\ln x+\frac{x^3}{3}\cdot\frac{1}{x}-\frac{3x^2}{9}=x^2\ln x+\frac{x^2}{3}-\frac{x^2}{3}=x^2\ln x$. Correct. *FTC application (Theorem 1894):* $P(e)=\frac{e^3}{3}\cdot1-\frac{e^3}{9}=\frac{3e^3-e^3}{9}=\frac{2e^3}{9}$. $P(1)=\frac{1}{3}\cdot0-\frac{1}{9}=-\frac{1}{9}$. $$\int_1^e x^2\ln x\,dx=P(e)-P(1)=\frac{2e^3}{9}+\frac{1}{9}=\boxed{\frac{2e^3+1}{9}}.$$ - 10PRMay 2024 Q10
Recipe 10: The question asks you to (a) define a concept, (b) state a property or theorem, and (c) prove it. This is the "state-and-prove" flagship question type.
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May 2024 Q10 — the two-case Dirac proof above. May 2024 Q11 — the three-step LOP proof above. - 11MFMay 2024 Q12
Recipe 11: Given a bond with cash flows $(a_1,\ldots,a_n)$ at dates $(t_1,\ldots,t_n)$ and a yield-to-maturity $y$, you must (a) write the price formula, (b) define and write the Macaulay duration, and (c) estimate the new price after a yield shift $\Delta y$.
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May 2024 Q12, cash flows $a_1=60$, $a_2=60$, $a_3=1060$ at $t=1,2,3$; $y=0.06$. *Part (a):* $$P_0=\frac{60}{1.06}+\frac{60}{1.06^2}+\frac{1060}{1.06^3}\approx1000.$$ At $y=6\%$ with coupon rate $=6\%$, the bond prices at par $P_0=1000$ exactly (YTM equals coupon rate). *Part (b):* $$D(0.06)=\frac{1}{P_0}\left(1\cdot\frac{60}{1.06}+2\cdot\frac{60}{1.06^2}+3\cdot\frac{1060}{1.06^3}\right)\approx2.833\text{ years}.$$ *Part (c):* Yield shift $\Delta y=+0.001$ (positive $0.1\%$). Formula: $$P_{\text{new}}\approx P_0-\frac{D(0.06)}{1.06}\cdot(0.001)\cdot P_0\approx1000-\frac{2.833}{1.06}\times0.001\times1000\approx1000-2.673\approx997.33.$$ - 12DCFinal Exercises Q3
Recipe 12: A level curve $g(x,y)=c$ implicitly defines $y=f(x)$ near a point $(x_0,y_0)$, and you must find $f'(x_0)$.
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Final Exercises Q3, $g(x,y)=2ye^x+xe^y-2$, level $c=-2$, find $f'(0)$. *Step 1:* Set $x_0=0$ and solve: $g(0,y_0)=2y_0\cdot1+0\cdot e^{y_0}-2=2y_0-2=-2$, so $y_0=0$. Check: $g(0,0)=0+0-2=-2=c$. ✓ *Step 2:* $g_y(x,y)=2e^x+xe^y$. At $(0,0)$: $g_y(0,0)=2\cdot1+0\cdot1=2\ne0$. Dini's theorem applies. *Step 3:* $g_x(x,y)=2ye^x+e^y$. At $(0,0)$: $g_x(0,0)=2\cdot0\cdot1+e^0=1$. *Step 4:* $$f'(0)=-\frac{g_x(0,0)}{g_y(0,0)}=-\frac{1}{2}.$$ $$\boxed{f'(0)=-\tfrac{1}{2}.}$$ - 13ICFinal Exercises Q4
Recipe 13: An integral $\int_{-\infty}^{+\infty}f(x)\,dx$ does not converge in the classical sense (the two half-integrals are $+\infty$ and $-\infty$), and you must evaluate or classify the **Cauchy principal value**.
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Final Exercises Q4, $f(x)=2x-3$. *Step 1 — Classical integral:* $$\int_0^{+\infty}(2x-3)\,dx=\lim_{x\to+\infty}[x^2-3x]_0^x=+\infty.$$ $$\int_{-\infty}^0(2x-3)\,dx=\lim_{x\to-\infty}[x^2-3x]_x^0=\lim_{x\to-\infty}(-x^2+3x)=-\infty.$$ Sum $+\infty+(-\infty)$ is indeterminate — classical integral **does not exist**. Statement (a) "$=-\infty$" is **FALSE**. *Steps 2–3 — Principal value:* $$\int_{-K}^{K}(2x-3)\,dx=[x^2-3x]_{-K}^{K}=(K^2-3K)-(K^2+3K)=-6K.$$ $$\operatorname{PV}\int_{-\infty}^{+\infty}(2x-3)\,dx=\lim_{K\to+\infty}(-6K)=-\infty.$$ *Step 4 — Note on odd part:* The part $2x$ is odd, so $\int_{-K}^K2x\,dx=0$ always. The constant $-3$ gives $\int_{-K}^K(-3)\,dx=-6K\to-\infty$, which dominates. Statement (b) "PV $=0$" is **FALSE**. Statement (c) "PV $=-\infty$" is **TRUE**. - 14MFFinal Exercises Q5
Recipe 14: The instantaneous rate of capital growth is given as a function $f'(t)$, and you must compute the accumulated amount $M(t)=C\cdot\bigl(1+\int_0^t f'(s)\,ds\bigr)$ at a specific time.
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Final Exercises Q5, $f'(t)=(1.02)^t\ln(1.02)$, $C=1$ and $C=100$, find $M(2)$. *Step 1:* Recognise $(1.02)^t\ln(1.02)=\frac{d}{dt}(1.02)^t$, so $f(t)=(1.02)^t$. *Step 2:* By Theorem 1894 (FTC I): $$\int_0^2(1.02)^s\ln(1.02)\,ds=[(1.02)^s]_0^2=(1.02)^2-1=1.0404-1=0.0404.$$ *Step 3:* $$M(2)=C\cdot\bigl(1+(1.02)^2-1\bigr)=C\cdot(1.02)^2.$$ *Step 4:* For $C=1$: $M(2)=(1.02)^2=\boxed{1.0404}$. For $C=100$: $M(2)=100\cdot1.0404=\boxed{104.04}$. *Step 5:* This is compound interest at $i=2\%$ for 2 periods: $C(1.02)^2$. The integral form "reconstructs" the accumulation from the instantaneous interest rate $f'(t)$ via $\int_0^t f'(s)\,ds=f(t)-f(0)$.