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Paper 1 Answers

Time2 hours

Marks60

SetSet 3

PaperLevel 1 - Math I (Physics) Paper 1

Information

Section A: Real Numbers and Algebra
Section B: Trigonometry and Limits
Section C: Differentiation and Integration
Section D: Complex Numbers

Marking Guidance

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

Section A: Real Numbers and Algebra

Answer 1. Calibration algebra and finite sums

[15 marks]

A dimensionless calibration setting \(x\) is adjusted during a bench test. The same test also uses a small error parameter \(\epsilon\) and a sequence of linearly increasing readings.

a) Solve the inequality \(0<\frac{2x-1}{x+2}<1\).

[5 marks]

First solve the two inequalities separately. For \[ \frac{2x-1}{x+2}>0, \] the critical values are \(x=-2\) and \(x=\frac{1}{2}\). A sign chart gives \(x<-2\) or \(x>\frac{1}{2}\). Next, \[ \frac{2x-1}{x+2}<1 \] is equivalent to \[ \frac{2x-1-(x+2)}{x+2}<0, \] so \[ \frac{x-3}{x+2}<0. \] This holds for \(-2<x<3\). Intersecting the two conditions gives \[ \frac{1}{2}<x<3. \]

Answer 1. Calibration algebra and finite sums

[15 marks]

b) Expand \((1+2\epsilon)^5\) up to and including the term in \(\epsilon^3\). Use this to estimate \((1.02)^5\).

[5 marks]

Using the binomial theorem, \[ (1+2\epsilon)^5=1+5(2\epsilon)+10(2\epsilon)^2+10(2\epsilon)^3+5(2\epsilon)^4+(2\epsilon)^5. \] Keeping terms up to \(\epsilon^3\), \[ (1+2\epsilon)^5\approx 1+10\epsilon+40\epsilon^2+80\epsilon^3. \] For \(\epsilon=0.01\), \[ 1+10(0.01)+40(0.01)^2+80(0.01)^3=1.10408. \] The cubic approximation is \(1.10408\).

Answer 1. Calibration algebra and finite sums

[15 marks]

c) Find \(\sum_{r=1}^{n}(4r-1)\). Hence find \(n\) if this sum is \(171\).

[5 marks]

Use the standard sums: \[ \sum_{r=1}^{n}(4r-1)=4\sum_{r=1}^{n}r-\sum_{r=1}^{n}1. \] Therefore \[ \sum_{r=1}^{n}(4r-1)=4\cdot\frac{n(n+1)}{2}-n=2n(n+1)-n. \] So \[ \sum_{r=1}^{n}(4r-1)=2n^2+n. \] If the sum is \(171\), then \[ 2n^2+n=171. \] Hence \[ 2n^2+n-171=0=(n-9)(2n+19). \] Since \(n\) is positive, \(n=9\).

Section B: Trigonometry and Limits

Answer 2. Oscillatory signal model

[15 marks]

An oscillatory signal is modelled by \(s(t)=4\sin t+3\cos t\), where \(t\) is measured in radians. A separate small-angle calculation is used to compare two nearby signals.

a) Write \(s(t)=4\sin t+3\cos t\) in the form \(R\sin(t+\alpha)\), where \(R>0\) and \(0<\alpha<\frac{\pi}{2}\).

[5 marks]

Write \[ 4\sin t+3\cos t=5\sin(t+\alpha). \] Expanding the right-hand side gives \[ 5\sin(t+\alpha)=5\sin t\cos\alpha+5\cos t\sin\alpha. \] Comparing coefficients, \[ 5\cos\alpha=4,\qquad 5\sin\alpha=3. \] Thus \[ \cos\alpha=\frac{4}{5},\qquad \sin\alpha=\frac{3}{5},\qquad \alpha=\tan^{-1}\left(\frac{3}{4}\right). \] Therefore \[ s(t)=5\sin\left(t+\tan^{-1}\left(\frac{3}{4}\right)\right). \]

Answer 2. Oscillatory signal model

[15 marks]

b) Find the maximum value of \(s(t)\), and find the first \(t\) in \([0,2\pi)\) at which it occurs.

[5 marks]

From part (a), \[ s(t)=5\sin(t+\alpha),\qquad \alpha=\tan^{-1}\left(\frac{3}{4}\right). \] The maximum value of the sine function is \(1\), so the maximum signal value is \(5\). This first occurs when \[ t+\alpha=\frac{\pi}{2}. \] Therefore \[ t=\frac{\pi}{2}-\tan^{-1}\left(\frac{3}{4}\right). \]

Answer 2. Oscillatory signal model

[15 marks]

c) Evaluate \(\lim_{u\to0}\frac{\sin(5u)-\sin(2u)}{u}\).

[5 marks]

Separate the two sine terms: \[ \lim_{u\to0}\frac{\sin(5u)-\sin(2u)}{u}=\lim_{u\to0}\frac{\sin(5u)}{u}-\lim_{u\to0}\frac{\sin(2u)}{u}. \] For the first term, \[ \frac{\sin(5u)}{u}=5\frac{\sin(5u)}{5u}\to5. \] For the second term, \[ \frac{\sin(2u)}{u}=2\frac{\sin(2u)}{2u}\to2. \] Therefore the limit is \[ 5-2=3. \]

Section C: Differentiation and Integration

Answer 3. Pulse rates and accumulated work

[15 marks]

A displacement model for a short pulse is \(x(t)=t^2e^{-t}\), where \(t\ge0\). A related accumulated work calculation uses a rational force model.

a) Differentiate \(x(t)=t^2e^{-t}\), and find the stationary time for \(t>0\).

[5 marks]

Differentiate using the product rule: \[ x'(t)=2t e^{-t}+t^2(-e^{-t}). \] Hence \[ x'(t)=e^{-t}(2t-t^2)=t(2-t)e^{-t}. \] For \(t>0\), the stationary point occurs when \[ t(2-t)e^{-t}=0. \] Since \(e^{-t}\ne0\) and \(t>0\), the stationary time is \[ t=2. \]

Answer 3. Pulse rates and accumulated work

[15 marks]

b) Evaluate \(\int_0^2 t e^{-t}\,dt\).

[5 marks]

Use integration by parts with \(u=t\) and \(dv=e^{-t}\,dt\). Then \(du=dt\) and \(v=-e^{-t}\). Thus \[ \int_0^2 t e^{-t}\,dt=\left[-t e^{-t}\right]_0^2+\int_0^2 e^{-t}\,dt. \] Now \[ \left[-t e^{-t}\right]_0^2=-2e^{-2} \] and \[ \int_0^2 e^{-t}\,dt=\left[-e^{-t}\right]_0^2=1-e^{-2}. \] Therefore \[ \int_0^2 t e^{-t}\,dt=1-3e^{-2}. \]

Answer 3. Pulse rates and accumulated work

[15 marks]

c) Resolve \(\frac{3x+5}{x^2+3x+2}\) into partial fractions, and evaluate \(\int_0^1\frac{3x+5}{x^2+3x+2}\,dx\).

[5 marks]

Factor the denominator: \[ x^2+3x+2=(x+1)(x+2). \] Let \[ \frac{3x+5}{x^2+3x+2}=\frac{A}{x+1}+\frac{B}{x+2}. \] Then \[ 3x+5=A(x+2)+B(x+1). \] Comparing coefficients gives \[ A+B=3,\qquad 2A+B=5. \] Thus \(A=2\) and \(B=1\). Hence \[ \int_0^1\frac{3x+5}{x^2+3x+2}\,dx=\int_0^1\left(\frac{2}{x+1}+\frac{1}{x+2}\right)dx. \] So \[ \left[2\ln(x+1)+\ln(x+2)\right]_0^1=2\ln2+\ln3-\ln2=\ln6. \]

Section D: Complex Numbers

Answer 4. Phasors and roots

[15 marks]

A phasor has complex amplitude \(z=\sqrt{3}+i\). Complex roots are used to describe equally spaced phase choices.

a) Write \(z=\sqrt{3}+i\) in exponential form, and hence find \(z^6\).

[5 marks]

The modulus is \[ |z|=\sqrt{(\sqrt{3})^2+1^2}=2. \] The argument is \[ \arg z=\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}, \] since \(z\) lies in the first quadrant. Therefore \[ z=2e^{i\pi/6}. \] By De Moivre's theorem, \[ z^6=2^6e^{i\pi}=-64. \]

Answer 4. Phasors and roots

[15 marks]

b) Find the three cube roots of \(z\), giving your answers in exponential form.

[5 marks]

Since \[ z=2e^{i\pi/6}, \] the cube roots have modulus \(2^{1/3}\). Their arguments are \[ \frac{\frac{\pi}{6}+2k\pi}{3}=\frac{\pi}{18}+\frac{2k\pi}{3},\qquad k=0,1,2. \] Therefore \[ w=2^{1/3}e^{i(\pi/18+2k\pi/3)},\qquad k=0,1,2. \]

Answer 4. Phasors and roots

[15 marks]

c) Let \(A=2e^{i\pi/3}+3e^{-i\pi/6}\). Find the real and imaginary parts of \(A\).

[5 marks]

Use Euler form term by term: \[ 2e^{i\pi/3}=2\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)=1+i\sqrt{3}. \] Also, \[ 3e^{-i\pi/6}=3\left(\frac{\sqrt{3}}{2}-i\frac{1}{2}\right)=\frac{3\sqrt{3}}{2}-\frac{3i}{2}. \] Adding gives \[ A=1+\frac{3\sqrt{3}}{2}+i\left(\sqrt{3}-\frac{3}{2}\right). \] The real part is \(1+\frac{3\sqrt{3}}{2}\) and the imaginary part is \(\sqrt{3}-\frac{3}{2}\).