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Paper 1 Answers

Time2 hours

Marks60

SetSet 4

PaperLevel 1 - Math I (Physics) Paper 1

Information

Section A: Functions and Inverses
Section B: Limits and Derivatives
Section C: Integration
Section D: Complex Numbers and De Moivre's Theorem

Marking Guidance

These worked answers show the expected method and final result. Equivalent correct reasoning should receive credit.

Section A: Functions and Inverses

Answer 1. Logarithmic response functions

[15 marks]

A logarithmic response model is \(R(q)=2+\ln(q+1)\), where \(q>-1\). A read-back formula is \(A(y)=e^{y-2}-1\).

a) Find the range of \(R\), and find an expression for \(R^{-1}(y)\).

[5 marks]

Since \(q>-1\), we have \(q+1>0\), so \(\ln(q+1)\) can take every real value. Therefore the range of \(R\) is all real numbers. To find the inverse, write \[ y=2+\ln(q+1). \] Then \[ y-2=\ln(q+1). \] Exponentiating gives \[ e^{y-2}=q+1. \] Hence \[ R^{-1}(y)=e^{y-2}-1. \]

Answer 1. Logarithmic response functions

[15 marks]

b) Simplify \(A(R(q))\), and state what this means for the read-back formula.

[5 marks]

Substitute \(R(q)=2+\ln(q+1)\) into \(A(y)=e^{y-2}-1\): \[ A(R(q))=e^{2+\ln(q+1)-2}-1. \] This simplifies to \[ A(R(q))=e^{\ln(q+1)}-1. \] Since \(q>-1\), \(q+1>0\), so \[ A(R(q))=q+1-1=q. \] Thus the read-back formula recovers the original setting.

Answer 1. Logarithmic response functions

[15 marks]

c) Find the setting \(q\) for which \(R(q)=\ln6\).

[5 marks]

Set the model equal to the given reading: \[ 2+\ln(q+1)=\ln6. \] Then \[ \ln(q+1)=\ln6-2. \] Since \(2=\ln(e^2)\), this is \[ \ln(q+1)=\ln\left(\frac{6}{e^2}\right). \] Therefore \[ q+1=\frac{6}{e^2},\qquad q=6e^{-2}-1. \]

Section B: Limits and Derivatives

Answer 2. Continuity, first principles, and small angles

[15 marks]

A removable discontinuity is used to test a numerical routine. A separate polynomial model is differentiated from first principles.

a) Let \(f(x)=\frac{x^2-4}{x-2}\) for \(x\ne2\), and let \(f(2)=k\). Find \(k\) so that \(f\) is continuous at \(x=2\).

[5 marks]

For \(x\ne2\), factor the numerator: \[ \frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2. \] Thus \[ \lim_{x\to2}\frac{x^2-4}{x-2}=\lim_{x\to2}(x+2)=4. \] For the function to be continuous at \(x=2\), the defined value must equal this limit. Hence \[ k=4. \]

Answer 2. Continuity, first principles, and small angles

[15 marks]

b) Using first principles, differentiate \(g(x)=x^2-4x\).

[5 marks]

Using first principles, \[ g'(x)=\lim_{h\to0}\frac{g(x+h)-g(x)}{h}. \] For \(g(x)=x^2-4x\), \[ g(x+h)=(x+h)^2-4(x+h). \] Then \[ g(x+h)-g(x)=x^2+2xh+h^2-4x-4h-(x^2-4x). \] This simplifies to \[ g(x+h)-g(x)=2xh+h^2-4h. \] Divide by \(h\): \[ \frac{g(x+h)-g(x)}{h}=2x+h-4. \] Taking \(h\to0\) gives \[ g'(x)=2x-4. \]

Answer 2. Continuity, first principles, and small angles

[15 marks]

c) Evaluate \(\lim_{x\to0}\frac{\cos(3x)-1}{x^2}\).

[5 marks]

Use the identity \(\cos a-1=-2\sin^2\left(\frac{a}{2}\right)\). Then \[ \frac{\cos(3x)-1}{x^2}=\frac{-2\sin^2\left(\frac{3x}{2}\right)}{x^2}. \] Rewrite this as \[ -2\left(\frac{\sin\left(\frac{3x}{2}\right)}{x}\right)^2. \] Now \[ \frac{\sin\left(\frac{3x}{2}\right)}{x}=\frac{3}{2}\frac{\sin\left(\frac{3x}{2}\right)}{\frac{3x}{2}}\to\frac{3}{2}. \] Therefore \[ \lim_{x\to0}\frac{\cos(3x)-1}{x^2}=-2\left(\frac{3}{2}\right)^2=-\frac{9}{2}. \]

Section C: Integration

Answer 3. Definite integrals from rate models

[15 marks]

Accumulated quantities are computed from three rate models over short intervals: a polynomial rate, a trigonometric rate, and an exponential rate.

a) Evaluate \(\int_0^1(6u^2-4u+5)\,du\).

[5 marks]

Integrate term by term: \[ \int_0^1(6u^2-4u+5)\,du=\left[2u^3-2u^2+5u\right]_0^1. \] At \(u=1\), this is \[ 2-2+5=5. \] At \(u=0\), it is \(0\). Therefore \[ \int_0^1(6u^2-4u+5)\,du=5. \]

Answer 3. Definite integrals from rate models

[15 marks]

b) Use a substitution to evaluate \(\int_0^{\pi/4}\frac{\sin\theta}{\cos^3\theta}\,d\theta\).

[5 marks]

Let \(y=\cos\theta\). Then \(dy=-\sin\theta\,d\theta\). When \(\theta=0\), \(y=1\). When \(\theta=\frac{\pi}{4}\), \(y=\frac{\sqrt{2}}{2}\). Thus \[ \int_0^{\pi/4}\frac{\sin\theta}{\cos^3\theta}\,d\theta=\int_1^{\sqrt{2}/2}-y^{-3}\,dy. \] Reverse the limits: \[ \int_{\sqrt{2}/2}^{1}y^{-3}\,dy=\left[-\frac{1}{2y^2}\right]_{\sqrt{2}/2}^{1}. \] Therefore \[ -\frac{1}{2}-\left(-1\right)=\frac{1}{2}. \]

Answer 3. Definite integrals from rate models

[15 marks]

c) Use integration by parts to evaluate \(\int_0^1 u e^{2u}\,du\).

[5 marks]

Use integration by parts with \(a=u\) and \(db=e^{2u}\,du\). Then \(da=du\) and \(b=\frac{1}{2}e^{2u}\). Hence \[ \int_0^1 u e^{2u}\,du=\left[\frac{u}{2}e^{2u}\right]_0^1-\int_0^1\frac{1}{2}e^{2u}\,du. \] The boundary term is \(\frac{e^2}{2}\). The remaining integral is \[ \int_0^1\frac{1}{2}e^{2u}\,du=\left[\frac{1}{4}e^{2u}\right]_0^1=\frac{e^2-1}{4}. \] Therefore \[ \int_0^1 u e^{2u}\,du=\frac{e^2}{2}-\frac{e^2-1}{4}=\frac{e^2+1}{4}. \]

Section D: Complex Numbers and De Moivre's Theorem

Answer 4. Complex powers and roots

[15 marks]

Let \(z=\cos\theta+i\sin\theta\). De Moivre's theorem connects powers of \(z\) with trigonometric identities and roots of complex numbers.

a) Use De Moivre's theorem to show that \(z^3+z^{-3}=2\cos3\theta\).

[5 marks]

By De Moivre's theorem, \[ z^3=(\cos\theta+i\sin\theta)^3=\cos3\theta+i\sin3\theta. \] Also \[ z^{-3}=\cos(-3\theta)+i\sin(-3\theta)=\cos3\theta-i\sin3\theta. \] Adding gives \[ z^3+z^{-3}=2\cos3\theta. \]

Answer 4. Complex powers and roots

[15 marks]

b) Deduce that \(\cos3\theta=4\cos^3\theta-3\cos\theta\).

[5 marks]

Expand the cube: \[ (\cos\theta+i\sin\theta)^3=\cos^3\theta+3i\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta. \] The real part is \[ \cos^3\theta-3\cos\theta\sin^2\theta. \] By De Moivre's theorem, this real part equals \(\cos3\theta\). Use \(\sin^2\theta=1-\cos^2\theta\): \[ \cos3\theta=\cos^3\theta-3\cos\theta(1-\cos^2\theta). \] Therefore \[ \cos3\theta=4\cos^3\theta-3\cos\theta. \]

Answer 4. Complex powers and roots

[15 marks]

c) Find all fourth roots of \(-16\), giving your answers in exponential form.

[5 marks]

Write \[ -16=16e^{i(\pi+2m\pi)},\qquad m\in\mathbb{Z}. \] The fourth roots have modulus \(16^{1/4}=2\). Their arguments are \[ \frac{\pi+2m\pi}{4}=\frac{\pi}{4}+\frac{m\pi}{2}. \] Taking four distinct values, \[ w=2e^{i(\pi/4+k\pi/2)},\qquad k=0,1,2,3. \] These are the four fourth roots of \(-16\).