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Paper 1 Answers

Time2 hours

Marks60

SetSet 2

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. Real-number and algebraic structure

[15 marks]

This question uses exact arithmetic, inequalities, binomial coefficients, and summation notation.

a) Prove that \(\sqrt{3}+\sqrt{7}\) is irrational.

[3 marks]

Assume for contradiction that \(\sqrt{3}+\sqrt{7}=r\), where \(r\in\mathbb{Q}\). Then \[ r-\sqrt{3}=\sqrt{7}. \] Squaring both sides gives \[ r^2-2r\sqrt{3}+3=7. \] Hence \[ 2r\sqrt{3}=r^2-4. \] If \(r=0\), then \(\sqrt{3}+\sqrt{7}=0\), impossible. If \(r\ne0\), then \[ \sqrt{3}=\frac{r^2-4}{2r}, \] which is rational. This contradicts the irrationality of \(\sqrt{3}\). Therefore \(\sqrt{3}+\sqrt{7}\) is irrational.

Answer 1. Real-number and algebraic structure

[15 marks]

b) Solve the inequality \(\frac{2x+1}{x-3}\ge1\).

[4 marks]

Move all terms to one side: \[ \frac{2x+1}{x-3}-1\ge0. \] Combine into one fraction: \[ \frac{2x+1-(x-3)}{x-3}\ge0, \] so \[ \frac{x+4}{x-3}\ge0. \] The critical values are \(x=-4\) and \(x=3\). A sign chart gives positive or zero on \(( -\infty,-4]\) and \((3,\infty)\). The point \(x=3\) is excluded because it is not in the domain. Therefore \[ x\in(-\infty,-4]\cup(3,\infty). \]

Answer 1. Real-number and algebraic structure

[15 marks]

c) Find the coefficient of \(x^5\) in the expansion of \((2-x)^7\).

[4 marks]

The general term in \((2-x)^7\) is \[ \binom{7}{k}2^{7-k}(-x)^k. \] The term in \(x^5\) occurs when \(k=5\). Its coefficient is \[ \binom{7}{5}2^2(-1)^5=21\cdot4\cdot(-1)=-84. \] So the coefficient of \(x^5\) is \(-84\).

Answer 1. Real-number and algebraic structure

[15 marks]

d) Evaluate \(\sum_{r=1}^{10}(2r^2-r)\).

[4 marks]

Separate the summation: \[ \sum_{r=1}^{10}(2r^2-r)=2\sum_{r=1}^{10}r^2-\sum_{r=1}^{10}r. \] Use \[ \sum_{r=1}^{10}r=\frac{10\cdot11}{2}=55 \] and \[ \sum_{r=1}^{10}r^2=\frac{10\cdot11\cdot21}{6}=385. \] Therefore \[ \sum_{r=1}^{10}(2r^2-r)=2(385)-55=715. \]

Section B: Trigonometry and Limits

Answer 2. Trigonometric identities and limiting values

[15 marks]

Use radians unless degrees are explicitly stated. Exact trigonometric values are preferred.

a) Show that \(\frac{\sin(2x)}{1+\cos(2x)}=\tan x\) wherever both sides are defined.

[4 marks]

Use \(\sin(2x)=2\sin x\cos x\) and \(1+\cos(2x)=2\cos^2x\). Then \[ \frac{\sin(2x)}{1+\cos(2x)}=\frac{2\sin x\cos x}{2\cos^2x}. \] Where the expression is defined, cancel \(2\cos x\): \[ \frac{\sin(2x)}{1+\cos(2x)}=\frac{\sin x}{\cos x}=\tan x. \]

Answer 2. Trigonometric identities and limiting values

[15 marks]

b) Find the exact value of \(\sin\left(\arctan\left(\frac{5}{12}\right)\right)\).

[3 marks]

Let \(\theta=\arctan\left(\frac{5}{12}\right)\). Then \(\tan\theta=\frac{5}{12}\). Use a right triangle with opposite side \(5\), adjacent side \(12\), and hypotenuse \(13\). Hence \[ \sin\theta=\frac{5}{13}. \] Therefore \[ \sin\left(\arctan\left(\frac{5}{12}\right)\right)=\frac{5}{13}. \]

Answer 2. Trigonometric identities and limiting values

[15 marks]

c) Solve \(\arctan x+\arctan(2x)=\frac{\pi}{4}\).

[4 marks]

Let \(A=\arctan x\) and \(B=\arctan(2x)\). Taking tangent of both sides gives \[ \tan(A+B)=1. \] Using the addition formula, \[ \frac{x+2x}{1-2x^2}=1. \] Thus \[ 3x=1-2x^2, \] so \[ 2x^2+3x-1=0. \] Solving, \[ x=\frac{-3\pm\sqrt{17}}{4}. \] The negative root makes both arctangent terms negative, so their sum cannot be \(\frac{\pi}{4}\). Therefore \[ x=\frac{-3+\sqrt{17}}{4}. \]

Answer 2. Trigonometric identities and limiting values

[15 marks]

d) Evaluate \(\lim_{h\to0}\frac{1-\cos(4h)}{h^2}\).

[4 marks]

Use \(1-\cos u=2\sin^2\left(\frac{u}{2}\right)\). With \(u=4h\), \[ 1-\cos(4h)=2\sin^2(2h). \] Hence \[ \lim_{h\to0}\frac{1-\cos(4h)}{h^2}=2\lim_{h\to0}\left(\frac{\sin(2h)}{h}\right)^2. \] Now \[ \frac{\sin(2h)}{h}=2\frac{\sin(2h)}{2h}, \] and \(\lim_{h\to0}\frac{\sin(2h)}{2h}=1\). Therefore the limit is \[ 2(2)^2=8. \]

Section C: Differentiation and Integration

Answer 3. Reconstructing volume from a rate

[15 marks]

A storage vessel contains \(30\) litres of liquid at \(t=0\). For \(0\le t\le5\), the signed rate of change of volume is \(R(t)=2t^2-8t+6\), measured in litres per minute.

a) Determine when the volume is increasing and when it is decreasing.

[3 marks]

Factor the rate: \[ R(t)=2t^2-8t+6=2(t^2-4t+3)=2(t-1)(t-3). \] So \(R(t)=0\) at \(t=1\) and \(t=3\). Since the quadratic opens upwards, \(R(t)>0\) on \(0\le t<1\) and \(3<t\le5\), while \(R(t)<0\) on \(1<t<3\). Therefore the volume increases on \(0\le t<1\) and \(3<t\le5\), and decreases on \(1<t<3\).

Answer 3. Reconstructing volume from a rate

[15 marks]

b) Find an expression for the volume \(V(t)\).

[4 marks]

Since \(R(t)=\frac{dV}{dt}\), integrate: \[ V(t)=\int(2t^2-8t+6)\,dt=\frac{2}{3}t^3-4t^2+6t+C. \] Use \(V(0)=30\), giving \(C=30\). Therefore \[ V(t)=\frac{2}{3}t^3-4t^2+6t+30. \]

Answer 3. Reconstructing volume from a rate

[15 marks]

c) Find the minimum and maximum volumes during \(0\le t\le5\).

[4 marks]

Stationary values occur at \(t=1\) and \(t=3\). Check these and the endpoints: \[ V(0)=30, \] \[ V(1)=\frac{2}{3}-4+6+30=\frac{98}{3}, \] \[ V(3)=18-36+18+30=30, \] \[ V(5)=\frac{250}{3}-100+30+30=\frac{130}{3}. \] Thus the minimum volume is \(30\) litres, at \(t=0\) and \(t=3\). The maximum volume is \[ \frac{130}{3}\text{ litres}\approx43.3\text{ litres}, \] at \(t=5\).

Answer 3. Reconstructing volume from a rate

[15 marks]

d) Find the average volume over the interval \(0\le t\le5\).

[4 marks]

The average volume is \[ \frac{1}{5}\int_0^5 V(t)\,dt. \] Using \(V(t)=\frac{2}{3}t^3-4t^2+6t+30\), \[ \int_0^5 V(t)\,dt=\left[\frac{t^4}{6}-\frac{4t^3}{3}+3t^2+30t\right]_0^5. \] Substitute \(t=5\): \[ \frac{625}{6}-\frac{500}{3}+75+150=\frac{975}{6}=\frac{325}{2}. \] Therefore the average volume is \[ \frac{1}{5}\cdot\frac{325}{2}=\frac{65}{2}=32.5. \] The average volume is \(32.5\) litres.

Section D: Complex Numbers

Answer 4. Complex arithmetic and De Moivre's theorem

[15 marks]

Write complex answers in exact form. Use \(i^2=-1\).

a) Evaluate \(\frac{(2-i)(3+4i)}{1+i}\) in the form \(a+bi\).

[3 marks]

First multiply the numerator: \[ (2-i)(3+4i)=6+8i-3i-4i^2. \] Since \(i^2=-1\), this is \[ 10+5i. \] Now divide by \(1+i\) by multiplying by the conjugate: \[ \frac{10+5i}{1+i}\cdot\frac{1-i}{1-i}=\frac{(10+5i)(1-i)}{2}. \] The numerator is \[ 10-10i+5i-5i^2=15-5i. \] Therefore \[ \frac{(2-i)(3+4i)}{1+i}=\frac{15}{2}-\frac{5}{2}i. \]

Answer 4. Complex arithmetic and De Moivre's theorem

[15 marks]

b) Use De Moivre's theorem to evaluate \((1+i)^6\).

[4 marks]

Write \(1+i\) in polar form. Its modulus is \(\sqrt{2}\) and its argument is \(\frac{\pi}{4}\), so \[ 1+i=\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right). \] By De Moivre's theorem, \[ (1+i)^6=(\sqrt{2})^6\left(\cos\frac{6\pi}{4}+i\sin\frac{6\pi}{4}\right). \] Now \((\sqrt{2})^6=8\) and \(\frac{6\pi}{4}=\frac{3\pi}{2}\). Therefore \[ (1+i)^6=8(0-i)=-8i. \]

Answer 4. Complex arithmetic and De Moivre's theorem

[15 marks]

c) Find all complex solutions of \(z^3=27\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)\).

[4 marks]

The right-hand side already has modulus \(27\) and argument \(\frac{\pi}{6}\). The cube roots have modulus \[ \sqrt[3]{27}=3. \] Their arguments are \[ \frac{\frac{\pi}{6}+2k\pi}{3}=\frac{\pi}{18}+\frac{2k\pi}{3},\qquad k=0,1,2. \] Thus the roots are \[ 3\left(\cos\frac{\pi}{18}+i\sin\frac{\pi}{18}\right), \] \[ 3\left(\cos\frac{13\pi}{18}+i\sin\frac{13\pi}{18}\right), \] and \[ 3\left(\cos\frac{25\pi}{18}+i\sin\frac{25\pi}{18}\right). \]

Answer 4. Complex arithmetic and De Moivre's theorem

[15 marks]

d) Describe the moduli and arguments of the roots from part (c), and state the geometric shape they form in the complex plane.

[4 marks]

Each root has modulus \(3\), so all roots lie on the circle \[ |z|=3. \] The arguments are \[ \frac{\pi}{18},\qquad \frac{13\pi}{18},\qquad \frac{25\pi}{18}. \] The difference between consecutive arguments is \[ \frac{12\pi}{18}=\frac{2\pi}{3}. \] The roots are therefore equally spaced around the circle, forming an equilateral triangle in the complex plane.