2023 Cambridge IGCSE International Mathematics (0607) 模拟试题及答案详解

Multiply the terms:
\((3x - 2)(2x + 1) = 3x(2x) + 3x(1) - 2(2x) - 2(1)\)
\(= 6x^2 + 3x - 4x - 2\)
\(= 6x^2 - x - 2\)

M1 for at least 3 correct terms of the expansion \(6x^2 + 3x - 4x - 2\) seen
A1 for \(6x^2 - x - 2\)

Multiply both sides by 5:
\(2x - 3 = 5(x - 3)\)
\(2x - 3 = 5x - 15\)
Subtract \(2x\) from both sides:
\(-3 = 3x - 15\)
Add 15 to both sides:
\(12 = 3x\)
\(x = 4\)

M1 for \(2x - 3 = 5(x - 3)\) or \(2x - 3 = 5x - 15\) seen
A1 for \(x = 4\)

First expand the power:
\((4y^3)^2 = 16y^6\)
Then multiply by \(y^{-4}\):
\(16y^6 \times y^{-4} = 16y^{6 - 4} = 16y^2\)

M1 for \(16y^6\) or \(k y^2\) (where \(k \neq 1\))
A1 for \(16y^2\)

Add \(3x\) to both sides:
\(5 \ge 5x - 15\)
Add 15 to both sides:
\(20 \ge 5x\)
Divide by 5:
\(4 \ge x\) or \(x \le 4\)

M1 for \(20 \ge 5x\) or \(-5x \ge -20\) or equivalent correct step collecting terms on one side
A1 for \(x \le 4\) or \(4 \ge x\)

First find \(f(2)\):
\(f(2) = 3(2) - 1 = 5\)
Now find \(g(5)\):
\(g(5) = 2(5) + 5 = 10 + 5 = 15\)

M1 for \(f(2) = 5\) or \(g(f(x)) = 2(3x - 1) + 5\) seen
A1 for 15

Simplify each surd:
\(\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}\)
\(\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}\)
Subtract the simplified surds:
\(3\sqrt{5} - 2\sqrt{5} = \sqrt{5}\)

M1 for \(3\sqrt{5}\) or \(2\sqrt{5}\) seen
A1 for \(\sqrt{5}\)

Use the laws of logarithms:
\(\log_5\left(\frac{x}{2}\right) = 2\)
Convert to exponential form:
\(\frac{x}{2} = 5^2\)
\(\frac{x}{2} = 25\)
\(x = 50\)

M1 for \(\big_5\left(\frac{x}{2}\right) = 2\) or \(\log_5 x = \log_5 25 + \log_5 2\) or equivalent
A1 for 50

Find a common denominator, which is \(a(a+1)\):
\(\frac{3(a+1) - 2a}{a(a+1)}\)
Expand the numerator:
\(\frac{3a + 3 - 2a}{a(a+1)}\)
Simplify the numerator:
\(\frac{a + 3}{a(a+1)}\)

M1 for common denominator \(a(a+1)\) with at least one correct numerator term, e.g., \(\frac{3(a+1)-2a}{a(a+1)}\)
A1 for \(\frac{a+3}{a(a+1)}\) or \(\frac{a+3}{a^2+a}\)

To simplify, we find a common denominator, which is (x-3)(x+1). This gives: \\frac{4(x+1) - 3(x-3)}{(x-3)(x+1)} = \\frac{4x + 4 - 3x + 9}{(x-3)(x+1)} = \\frac{x + 13}{(x-3)(x+1)}.

M1 for a single fraction with denominator (x-3)(x+1) and numerator showing at least one correct expansion. A1 for \\frac{x+13}{(x-3)(x+1)} or \\frac{x+13}{x^2-2x-3} as the final answer.

Let y = \\frac{5}{x-2}. Multiplying both sides by x-2 gives y(x-2) = 5, which expands to xy - 2y = 5. Rearranging to make x the subject gives xy = 2y + 5, so x = \\frac{2y+5}{y}. Thus, f^{-1}(x) = \\frac{2x+5}{x}.

M1 for correct first steps to change the subject, e.g. y(x-2) = 5 or x(y-2) = 5. A1 for \\frac{2x+5}{x} or 2 + \\frac{5}{x} as the final answer.

1. Simplify the surds in the numerator:
$$\sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5}$$
$$\sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5}$$

2. Subtract the terms in the numerator:
$$\sqrt{180} - \sqrt{125} = 6\sqrt{5} - 5\sqrt{5} = \sqrt{5}$$

3. Write the simplified fraction:
$$\frac{\sqrt{5}}{\sqrt{5} - 2}$$

4. Rationalise the denominator by multiplying the numerator and denominator by $(\sqrt{5} + 2)$:
$$\frac{\sqrt{5}(\sqrt{5} + 2)}{(\sqrt{5} - 2)(\sqrt{5} + 2)} = \frac{5 + 2\sqrt{5}}{5 - 4} = 5 + 2\sqrt{5}$$

This gives the final simplified answer of $5 + 2\sqrt{5}$.

B1 for expressing $\sqrt{180}$ as $6\sqrt{5}$ and $\sqrt{125}$ as $5\sqrt{5}$
B1 for numerator simplified to $\sqrt{5}$
M1 for rationalising method: multiplying by $\frac{\sqrt{5} + 2}{\sqrt{5} + 2}$
A1 for final answer $5 + 2\sqrt{5}$

1. Find the inverse function $f^{-1}(x)$:
Let $y = \frac{x+3}{2}$
$$2y = x + 3 \Rightarrow x = 2y - 3$$
$$f^{-1}(x) = 2x - 3$$

2. Calculate $f^{-1}(5)$:
$$f^{-1}(5) = 2(5) - 3 = 7$$

3. Find the composite function $g(f(x))$:
$$g(f(x)) = g\left(\frac{x+3}{2}\right) = \frac{5}{\frac{x+3}{2} - 1} = \frac{5}{\frac{x+1}{2}} = \frac{10}{x+1}$$

4. Solve the equation $g(f(x)) = f^{-1}(5)$:
$$\frac{10}{x+1} = 7$$
$$10 = 7(x + 1)$$
$$10 = 7x + 7$$
$$3 = 7x \Rightarrow x = \frac{3}{7}$$

Thus, the solution is $x = \frac{3}{7}$.

M1 for finding $f^{-1}(5) = 7$
M1 for finding the composite expression $g(f(x)) = \frac{10}{x+1}$
M1 for setting up the equation $\frac{10}{x+1} = 7$ and attempting to solve
A1 for $x = \frac{3}{7}$ (or equivalent fraction)

We can find this probability using the complement event: $P(\text{at least one blue}) = 1 - P(\text{both red})$.

1. Calculate the probability of selecting two red pens without replacement:
- Probability of first pen being red: $P(R_1) = \frac{5}{8}$
- Probability of second pen being red given the first was red: $P(R_2 | R_1) = \frac{4}{7}$
$$P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$$

2. Subtract this from 1 to find the probability of at least one blue pen:
$$P(\text{at least one blue}) = 1 - \frac{5}{14} = \frac{9}{14}$$

Alternatively, sum the probabilities of the cases RB, BR, and BB:
- $P(RB) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}$
- $P(BR) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}$
- $P(BB) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56}$
$$\text{Sum} = \frac{15}{56} + \frac{15}{56} + \frac{6}{56} = \frac{36}{56} = \frac{9}{14}$$

M1 for recognizing correct probability structure with decreasing denominators (8 then 7)
M1 for calculating $P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{5}{14}$ (or $\frac{20}{56}$), or individual case probabilities
M1 for correct approach to subtract from 1 or sum the cases
A1 for $\frac{9}{14}$ (or equivalent fraction)

1. Factorise each quadratic expression:
- $2x^2 - 5x - 3 = (2x + 1)(x - 3)$
- $4x^2 - 1 = (2x - 1)(2x + 1)$
- $x^2 - 9 = (x - 3)(x + 3)$
- $2x^2 + 7x + 3 = (2x + 1)(x + 3)$

2. Rewrite the division as multiplication by the reciprocal:
$$\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} \times \frac{(2x + 1)(x + 3)}{(x - 3)(x + 3)}$$

3. Cancel out common factors from the numerators and denominators:
- Cancel $(2x + 1)$ from the first fraction
- Cancel $(x - 3)$ from the numerator of the first and denominator of the second
- Cancel $(x + 3)$ from the numerator of the second and denominator of the second

This leaves:
$$\frac{2x + 1}{2x - 1}$$

Thus, the simplified expression is $\frac{2x + 1}{2x - 1}$.

M1 for factorising $2x^2 - 5x - 3$ to $(2x + 1)(x - 3)$ and $4x^2 - 1$ to $(2x - 1)(2x + 1)$
M1 for factorising $2x^2 + 7x + 3$ to $(2x + 1)(x + 3)$ and $x^2 - 9$ to $(x - 3)(x + 3)$
M1 for inverting the second fraction and showing multiplication
A1 for final simplified answer $\frac{2x + 1}{2x - 1}$ or $(2x + 1)/(2x - 1)$

1. Find the midpoint, $M$, of $AB$:
$$M = \left(\frac{-2 + 4}{2}, \frac{5 + 2}{2}\right) = \left(1, \frac{7}{2}\right) = (1, 3.5)$$

2. Find the gradient of line $L_1$:
$$m_1 = \frac{2 - 5}{4 - (-2)} = \frac{-3}{6} = -\frac{1}{2}$$

3. Find the gradient of line $L_2$, which is perpendicular to $L_1$:
$$m_2 = -\frac{1}{m_1} = 2$$

4. Find the equation of line $L_2$ passing through $M\left(1, \frac{7}{2}\right)$ with gradient $m_2 = 2$:
$$y - \frac{7}{2} = 2(x - 1)$$
$$y - 3.5 = 2x - 2$$
Multiply the entire equation by 2 to clear the fraction:
$$2y - 7 = 4x - 4$$
Rearrange into the form $ax + by = c$ with $a > 0$:
$$4x - 2y = -3$$

Thus, the equation is $4x - 2y = -3$.

M1 for finding the midpoint of $AB$ as $(1, 3.5)$ or equivalent
M1 for finding the gradient of $L_1$ as $-\frac{1}{2}$
M1 for finding the perpendicular gradient as $2$ and attempting to form the equation $y - y_1 = m_2(x - x_1)$
A1 for final equation in the required form, $4x - 2y = -3$ (or equivalent integer coefficients)

(a)
Using the formula for compound interest:
\(800 \times \left(1 + \frac{r}{100}\right)^6 = 1014.26\)
\(\left(1 + \frac{r}{100}\right)^6 = \frac{1014.26}{800}\)
\(\left(1 + \frac{r}{100}\right)^6 \approx 1.267825\)
\(1 + \frac{r}{100} = (1.267825)^{\frac{1}{6}}\)
\(1 + \frac{r}{100} \approx 1.04\)
\(\frac{r}{100} = 0.04 \implies r = 4\)

(b)
Using \(r = 4\):
\(800 \times (1.04)^n > 2000\)
\(1.04^n > \frac{2000}{800}\)
\(1.04^n > 2.5\)
Taking logarithms on both sides:
\(n \log(1.04) > \log(2.5)\)
\(n > \frac{\log(2.5)}{\log(1.04)}\)
\(n > \frac{0.39794}{0.017033}\)
\(n > 23.36\)
Since the question asks for the number of complete years, we round up to the next integer:
\(n = 24\)

(c)
Using the formula for simple interest:
\(P \times \left(1 + \frac{3.5 \times 5}{100}\right) = 1175\)
\(P \times (1 + 0.175) = 1175\)
\(1.175P = 1175\)
\(P = \frac{1175}{1.175}\)
\(P = 1000\)

(a)
M1 for \( 800 \times (1+r/100)^6 = 1014.26 \) or \( (1+r/100)^6 = 1.2678... \)
M1 for \( 1+r/100 = 1.0400... \)
A1 for \( r = 4 \) (accept 4.0)

(b)
M1 for \( 800 \times 1.04^n > 2000 \) or better
M1 for \( 1.04^n > 2.5 \)
M1 for correct use of logarithms, e.g. \( n > \log(2.5)/\log(1.04) \)
A1 for 23.36 or 23.4
A1 for 24 (must be integer, dep on previous A1)

(c)
M1 for \( P(1 + 3.5 \times 5 / 100) = 1175 \)
M1 for \( 1.175P = 1175 \)
A1 for 1000

Total number of pens = \(5 + 4 + 3 = 12\).

(a)(i)
\(P(R_1 \text{ and } B_2) = P(R_1) \times P(B_2 | R_1) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132} = \frac{5}{33}\)

(a)(ii)
\(P(B_1 \text{ and } G_2) = P(B_1) \times P(G_2 | B_1) = \frac{4}{12} \times \frac{3}{11} = \frac{12}{132} = \frac{1}{11}\)

(b)
Both same colour can be Red-Red, Blue-Blue, or Green-Green:
\(P(RR) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132}\)
\(P(BB) = \frac{4}{12} \times \frac{3}{11} = \frac{12}{132}\)
\(P(GG) = \frac{3}{12} \times \frac{2}{11} = \frac{6}{132}\)
\(P(\text{same colour}) = \frac{20 + 12 + 6}{132} = \frac{38}{132} = \frac{19}{66}\) (or approx 0.288)

(c)
Using the complement rule:
\(P(\text{at least one red}) = 1 - P(\text{no red})\)
The number of non-red pens is \(4 \text{ (blue)} + 3 \text{ (green)} = 7\).
\(P(\text{no red}) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132} = \frac{7}{22}\)
\(P(\text{at least one red}) = 1 - \frac{7}{22} = \frac{15}{22}\) (or approx 0.682)

(a)(i) M1 for \( \frac{5}{12} \times \frac{4}{11} \), A1 for \( \frac{5}{33} \) (accept \( \frac{20}{132} \) or 0.152)
(a)(ii) M1 for \( \frac{4}{12} \times \frac{3}{11} \), A1 for \( \frac{1}{11} \) (accept \( \frac{12}{132} \) or 0.0909)
(b) M1 for adding three correct product probabilities (RR, BB, GG), M1 for \( \frac{20}{132} + \frac{12}{132} + \frac{6}{132} \), A1 for \( \frac{19}{66} \) (accept \( \frac{38}{132} \) or 0.288)
(c) M1 for finding total number of non-red pens = 7, M1 for \( \frac{7}{12} \times \frac{6}{11} \) (or equivalent addition method), M1 for \( 1 - \text{their } P(\text{no red}) \), A1 for \( \frac{15}{22} \) (accept \( \frac{90}{132} \) or 0.682)

(a)
\(V_{\text{cylinder}} = \pi r^2 h = \pi r^2 (2r) = 2\pi r^3\)
\(V_{\text{hemisphere}} = \frac{1}{2} \left(\frac{4}{3}\pi r^3\right) = \frac{2}{3}\pi r^3\)
\(V_{\text{total}} = 2\pi r^3 + \frac{2}{3}\pi r^3 = \frac{8}{3}\pi r^3\)

(b)(i)
\(\frac{8}{3}\pi r^3 = 576\pi\)
\(\frac{8}{3}r^3 = 576\)
\(r^3 = 576 \times \frac{3}{8}\)
\(r^3 = 72 \times 3 = 216\)
\(r = \sqrt[3]{216} = 6\)

(b)(ii)
The total surface area consists of:
1. The flat circular bottom base of the cylinder: \(\pi r^2\)
2. The curved surface area of the cylinder: \(2\pi r h = 2\pi r (2r) = 4\pi r^2\)
3. The curved surface area of the hemisphere: \(2\pi r^2\)

\(A_{\text{total}} = \pi r^2 + 4\pi r^2 + 2\pi r^2 = 7\pi r^2\)
Substituting \(r = 6\):
\(A_{\text{total}} = 7\pi (6^2) = 7\pi (36) = 252\pi\) \(\text{cm}^2\)

(a) M1 for cylinder volume \( \pi r^2(2r) \), M1 for hemisphere volume \( \frac{2}{3}\pi r^3 \), A1 for correct sum leading to \( \frac{8}{3}\pi r^3 \) with no errors seen
(b)(i) M1 for \( \frac{8}{3}\pi r^3 = 576\pi \), M1 for \( r^3 = 216 \), A1 for \( r = 6 \)
(b)(ii) M1 for curved cylinder area \( 2\pi r(2r) = 4\pi r^2 \), M1 for curved hemisphere area \( 2\pi r^2 \), M1 for bottom base area \( \pi r^2 \), M1 for total sum \( 7\pi r^2 \) substituted with their \( r \), A1 for \( 252\pi \)

(a)
Angle \(BAC = 135^\circ - 65^\circ = 70^\circ\).
Using the Cosine Rule in triangle \(ABC\):
\(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(BAC)\)
\(BC^2 = 15^2 + 24^2 - 2(15)(24)\cos(70^\circ)\)
\(BC^2 = 225 + 576 - 720\cos(70^\circ)\)
\(BC^2 \approx 801 - 720(0.34202)\)
\(BC^2 \approx 801 - 246.25 = 554.75\)
\(BC = \sqrt{554.75} \approx 23.553\text{ km}\)
So \(BC \approx 23.6\text{ km}\) (correct to 3 s.f.).

(b)
Using the Sine Rule to find angle \(ABC\):
\(\frac{\sin(ABC)}{AC} = \frac{\sin(70^\circ)}{BC}\)
\(\sin(ABC) = \frac{24 \times \sin(70^\circ)}{23.553}\)
\(\sin(ABC) \approx \frac{24 \times 0.93969}{23.553} \approx 0.95674\)
Angle \(ABC \approx \arcsin(0.95674) \approx 73.1^\circ\).
Since the bearing of \(B\) from \(A\) is \(065^\circ\), the bearing of \(A\) from \(B\) is \(65^\circ + 180^\circ = 245^\circ\).
To find the bearing of \(C\) from \(B\), we subtract angle \(ABC\) from this back bearing because \(C\) lies to the south-east of \(A\) and \(B\) is to the north-east:
\(\text{Bearing} = 245^\circ - 73.1^\circ = 171.9^\circ\).

(c)
Area of triangle \(ABC = \frac{1}{2} \times AB \times AC \times \sin(BAC)\)
\(\text{Area} = \frac{1}{2} \times 15 \times 24 \times \sin(70^\circ)\)
\(\text{Area} = 180 \times 0.93969 \approx 169.14\text{ km}^2\)
So \(\text{Area} \approx 169\text{ km}^2\) (correct to 3 s.f.).

(a) M1 for finding angle \( BAC = 70^\circ \), M1 for substituting into correct Cosine Rule formula, M1 for evaluating \( BC^2 \approx 554.75 \), A1 for 23.6 (accept 23.55 - 23.6)
(b) M1 for setting up correct Sine Rule equation, M1 for evaluating angle \( ABC \approx 73.1^\circ \) (or 73.09), M1 for back bearing \( 245^\circ \) or diagram showing North angle relation, A1 for 171.9 (accept 171.9 - 172.0)
(c) M1 for setting up correct area formula \( \frac{1}{2} \times 15 \times 24 \times \sin(70^\circ) \), M1 for calculating \( 180 \times \sin(70^\circ) \), A1 for 169 (accept 169 - 169.2)

(a)
Gradient \(m\) of \(L_1 = \frac{1 - 5}{6 - (-2)} = \frac{-4}{8} = -0.5\).
Using the point-slope formula with \(Q(6, 1)\):
\(y - 1 = -0.5(x - 6)\)
\(y - 1 = -0.5x + 3\)
\(y = -0.5x + 4\)

(b)
Midpoint of \(PQ = \left(\frac{-2 + 6}{2}, \frac{5 + 1}{2}\right) = \left(\frac{4}{2}, \frac{6}{2}\right) = (2, 3)\).

(c)
Since \(L_2\) is perpendicular to \(L_1\), its gradient \(m_2\) is the negative reciprocal of \(-0.5\):
\(m_2 = -\frac{1}{-0.5} = 2\).
\(L_2\) passes through the midpoint \((2, 3)\):
\(y - 3 = 2(x - 2)\)
\(y - 3 = 2x - 4\)
\(y = 2x - 1\)
Rearranging into the form \(ax + by = d\):
\(2x - y = 1\)

(d)
\(L_2\) crosses the \(x\)-axis when \(y = 0\):
\(2x - 0 = 1\)
\(2x = 1 \implies x = 0.5\).
So, the coordinates are \((0.5, 0)\).

(a) M1 for gradient of \( L_1 = -4/8 \) or \( -0.5 \), M1 for substituting point and gradient into linear equation, A1 for \( y = -0.5x + 4 \) (or equivalent)
(b) M1 for correct midpoint formula used, A1 for \( (2, 3) \)
(c) M1 for perpendicular gradient \( m_2 = 2 \), M1 for using \( y - 3 = 2(x - 2) \), M1 for expanding and grouping terms, A1 for \( 2x - y = 1 \) (or equivalent integer equation, e.g. \( -2x + y = -1 \))
(d) M1 for setting \( y = 0 \) in their \( L_2 \) equation, A1 for \( (0.5, 0) \) (accept \( (1/2, 0) \))

(a)
\(\text{f}(7) = \frac{3}{7-1} = \frac{3}{6} = 0.5\)

(b)
Let \(y = 2x + 5\)
\(2x = y - 5\)
\(x = \frac{y-5}{2}\)
So, \(\text{g}^{-1}(x) = \frac{x-5}{2}\)

(c)
\(\text{f}(\text{g}(x)) = \text{f}(2x + 5) = \frac{3}{(2x+5)-1} = \frac{3}{2x+4}\)

(d)
\(\text{f}(x) = \text{g}(x)\)
\(\frac{3}{x-1} = 2x + 5\)
\(3 = (2x+5)(x-1)\)
\(3 = 2x^2 - 2x + 5x - 5\)
\(3 = 2x^2 + 3x - 5\)
\(2x^2 + 3x - 8 = 0\)

Using the quadratic formula:
\(x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-8)}}{2(2)}\)
\(x = \frac{-3 \pm \sqrt{9 + 64}}{4}\)
\(x = \frac{-3 \pm \sqrt{73}}{4}\)

Evaluating the roots:
\(x_1 = \frac{-3 + 8.544}{4} \approx 1.386\)
\(x_2 = \frac{-3 - 8.544}{4} \approx -2.886\)

So, \(x \approx 1.39\) and \(x \approx -2.89\) (correct to 3 s.f.).

(a) A1 for 0.5 (or \( \frac{1}{2} \))
(b) M1 for correct algebraic process of changing subject, A1 for \( \frac{x-5}{2} \) (or equivalent)
(c) M1 for correct substitution of \( g(x) \) into \( f(x) \), M1 for expanding denominator \( (2x+5)-1 \), A1 for \( \frac{3}{2x+4} \) (accept \( \frac{3}{2(x+2)} \))
(d) M1 for setting up \( \frac{3}{x-1} = 2x + 5 \), M1 for multiplying by \( (x-1) \) correctly, M1 for reducing to \( 2x^2 + 3x - 8 = 0 \), M1 for using the quadratic formula correctly with their coefficients, A1 for 1.39 and -2.89

(a)
Factorise the numerator:
\(2x^2 - 5x - 3 = (2x + 1)(x - 3)\)

Factorise the denominator:
\(x^2 - 9 = (x - 3)(x + 3)\)

Simplify the fraction by cancelling the common factor \((x-3)\):
\(\frac{(2x+1)(x-3)}{(x-3)(x+3)} = \frac{2x+1}{x+3}\)

(b)
Find a common denominator for the left-hand side:
\(\frac{4(x-1) + 3(x+2)}{(x+2)(x-1)} = 2\)

Expand the numerators and combine:
\(\frac{4x - 4 + 3x + 6}{x^2 + x - 2} = 2\)
\(\frac{7x + 2}{x^2 + x - 2} = 2\)

Multiply both sides by the denominator:
\(7x + 2 = 2(x^2 + x - 2)\)
\(7x + 2 = 2x^2 + 2x - 4\)

Rearrange into standard quadratic form:
\(2x^2 - 5x - 6 = 0\)

Use the quadratic formula:
\(x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-6)}}{2(2)}\)
\(x = \frac{5 \pm \sqrt{25 + 48}}{4}\)
\(x = \frac{5 \pm \sqrt{73}}{4}\)

Evaluate the roots (correct to 3 s.f.):
\(x_1 = \frac{5 + 8.544}{4} \approx 3.386\)
\(x_2 = \frac{5 - 8.544}{4} \approx -0.886\)

So, \(x \approx 3.39\) and \(x \approx -0.886\).

(a) M2 for factorising numerator \( (2x+1)(x-3) \) (M1 for partial factorisation, e.g. \( (2x+a)(x+b) \) with \( ab=-3 \) or \( 2b+a=-5 \)), M1 for factorising denominator \( (x-3)(x+3) \), A1 for \( \frac{2x+1}{x+3} \)
(b) M1 for finding a common denominator, M1 for correct expansion of numerators \( 4x-4+3x+6 \), M1 for \( 7x+2 = 2(x^2+x-2) \), M1 for standard quadratic equation \( 2x^2-5x-6=0 \), M1 for correct substitution into quadratic formula, A1 for 3.39, A1 for -0.886 (Accept 3.386 and -0.886)

(a)
To sketch \(y = 4 - 2^{-x}\):
- At \(x = -2\), \(y = 4 - 2^2 = 0\).
- At \(x = 0\), \(y = 4 - 1 = 3\).
- At \(x = 3\), \(y = 4 - 2^{-3} = 3.875\).
- The curve starts at \((-2, 0)\), passes through \((0, 3)\) and rises towards the asymptote \(y = 4\).

(b)
As \(x \to \infty\), \(2^{-x} \to 0\), so \(y \to 4\).
Equation of horizontal asymptote: \(y = 4\).

(c)
To sketch \(y = x^2 - 3x\):
- It is a parabola opening upwards with roots at \(x = 0\) and \(x = 3\).
- The vertex is at \(x = 1.5\), with \(y = (1.5)^2 - 3(1.5) = -2.25\).

(d)
Using a graphic display calculator to find the intersection of \(y = 4 - 2^{-x}\) and \(y = x^2 - 3x\):
- Find the left intersection: \(x \approx -0.660\)
- Find the right intersection: \(x \approx 3.993\)

Correct to 3 significant figures, the \(x\)-coordinates of the points of intersection are \(-0.66\) and \(3.99\).

(a) B1 for curve showing growth with a positive y-intercept, B1 for starting at \( (-2, 0) \), B1 for asymptotic behavior as \( x \to 3 \)
(b) B1 for \( y = 4 \)
(c) B1 for upwards opening parabola, B1 for passing through roots \( (0, 0) \) and \( (3, 0) \), B1 for vertex in the fourth quadrant (approx at \( (1.5, -2.25) \))
(d) M1 for attempting to find intersection on GDC, A1 for \( -0.66 \) (accept \( -0.660 \)), A1 for \( 3.99 \) (accept \( 3.993 \)), A1 for showing both coordinates on graph or correctly listing them

(a) Expected population after 12 years:
\(P = 450 \times (1 + 0.035)^{12}\)
\(P = 450 \times (1.035)^{12} \approx 679.98\)
Rounded to 3 significant figures, the population is 680.

(b) We set up the inequality:
\(450 \times (1.035)^t > 900\)
\(1.035^t > 2\)
Taking logarithms on both sides:
\(t \log(1.035) > \log(2)\)
\(t > \frac{\log(2)}{\log(1.035)}\)
\(t > 20.15\) years.
Since we need the number of complete years, \(t = 21\).

(c) Using the formula for decay over 5 years:
\(800 \times \left(1 - \frac{x}{100}\right)^5 = 550\)
\(\left(1 - \frac{x}{100}\right)^5 = \frac{550}{800} = 0.6875\)
\(1 - \frac{x}{100} = (0.6875)^{\frac{1}{5}} \approx 0.927457\)
\(\frac{x}{100} = 1 - 0.927457 = 0.072543\)
\(x = 7.2543\)
Rounded to 3 significant figures, \(x = 7.25\).

(a) [3 marks]
M1 for \(450 \times (1.035)^{12}\)
A1 for 679.98...
A1 for 680

(b) [4 marks]
M1 for \(450 \times 1.035^t = 900\) or \(1.035^t = 2\)
M1 for logarithmic method or showing at least two trials for \(t = 20\) and \(t = 21\)
A1 for \(20.1...\) or \(20.15\) seen
A1 for 21

(c) [4 marks]
M1 for \(800 \times \left(1 - \frac{x}{100}\right)^5 = 550\) (or with \(r^5\))
M1 for \(1 - \frac{x}{100} = \sqrt[5]{0.6875}\) oe or \(0.927...\) seen
A1 for \(0.0725...\) or \(7.254\%\)
A1 for 7.25

(a) Total number of pens = \(5 + 4 + 3 = 12\).
\(P(\text{Same colour}) = P(\text{Red, Red}) + P(\text{Blue, Blue}) + P(\text{Green, Green})\)
\(P(\text{Red, Red}) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132}\)
\(P(\text{Blue, Blue}) = \frac{4}{12} \times \frac{3}{11} = \frac{12}{132}\)
\(P(\text{Green, Green}) = \frac{3}{12} \times \frac{2}{11} = \frac{6}{132}\)
\(P(\text{Same colour}) = \frac{20 + 12 + 6}{132} = \frac{38}{132} = \frac{19}{66} \approx 0.288\)

(b) \(P(\text{At least one Blue}) = 1 - P(\text{No Blue selected})\)
Total non-blue pens = \(5 + 3 = 8\).
\(P(\text{No Blue}) = \frac{8}{12} \times \frac{7}{11} = \frac{56}{132}\)
\(P(\text{At least one Blue}) = 1 - \frac{56}{132} = \frac{76}{132} = \frac{19}{33} \approx 0.576\)

(c) The probability of selecting one Red, one Blue, and one Green in any order:
Number of permutations of 3 different colours = \(3! = 6\).
\(P(\text{All different}) = 6 \times \left(\frac{5}{12} \times \frac{4}{11} \times \frac{3}{10}\right)\)
\(P(\text{All different}) = 6 \times \frac{60}{1320} = \frac{360}{1320} = \frac{3}{11} \approx 0.273\)

(a) [4 marks]
M1 for any one correct probability: \(\frac{5}{12} \times \frac{4}{11}\) or \(\frac{4}{12} \times \frac{3}{11}\) or \(\frac{3}{12} \times \frac{2}{11}\)
M1 for adding three correct product pairs
A1 for \(\frac{38}{132}\) oe
A1 for \(\frac{19}{66}\) or 0.288

(b) [4 marks]
M1 for \(P(\text{No Blue}) = \frac{8}{12} \times \frac{7}{11}\) oe (or listing all positive outcomes)
M1 for subtracting their \(P(\text{No Blue})\) from 1, or adding the correct combinations
A1 for \(\frac{76}{132}\) oe
A1 for \(\frac{19}{33}\) or 0.576

(c) [3 marks]
M1 for \(\frac{5}{12} \times \frac{4}{11} \times \frac{3}{10}\) oe
M1 for multiplying their product by 6
A1 for \(\frac{3}{11}\) oe or 0.273

(a) Using the Cosine Rule:
\(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(B)\)
\(AC^2 = 120^2 + 150^2 - 2 \times 120 \times 150 \times \cos(72^\circ)\)
\(AC^2 = 14400 + 22500 - 36000 \times 0.309017\)
\(AC^2 = 36900 - 11124.6 = 25775.4\)
\(AC = \sqrt{25775.4} \approx 160.55\) m.
Rounded to 3 significant figures, \(AC = 161\) m.

(b) Using the Sine Rule:
\(\frac{\sin(\angle BAC)}{BC} = \frac{\sin(B)}{AC}\)
\(\frac{\sin(\angle BAC)}{150} = \frac{\sin(72^\circ)}{160.55}\)
\(\sin(\angle BAC) = \frac{150 \times \sin(72^\circ)}{160.55} \approx \frac{150 \times 0.9510565}{160.55} \approx 0.88856\)
\(\angle BAC = \arcsin(0.88856) \approx 62.69^\circ\).
Rounded to 1 decimal place, \(\angle BAC = 62.7^\circ\).

(c) Let \(d\) be the perpendicular distance from B to AC.
Using triangle ABD where \(d\) is perpendicular to AC:
\(\sin(\angle BAC) = \frac{d}{AB}\)
\(d = 120 \times \sin(62.69^\circ) = 120 \times 0.88856 \approx 106.6\) m.
Rounded to 3 significant figures, the length of the path is 107 m.
(Alternative method using area of triangle: \(\text{Area} = \frac{1}{2} \times 120 \times 150 \times \sin(72^\circ) = \frac{1}{2} \times 160.55 \times d\))

(a) [3 marks]
M1 for \(120^2 + 150^2 - 2 \times 120 \times 150 \times \cos(72)\)
A1 for \(25775.4\) or \(\sqrt{25775.4}\) seen
A1 for 161 (accept 160.5 to 160.6)

(b) [4 marks]
M1 for \(\frac{\sin(BAC)}{150} = \frac{\sin(72)}{\text{their } AC}\)
M1 for isolating \(\sin(BAC) = \frac{150 \sin(72)}{\text{their } AC}\)
A1 for \(\sin(BAC) = 0.888...\) oe
A1 for 62.7 (accept 62.69)

(c) [4 marks]
M1 for \(\text{Area} = \frac{1}{2} \times 120 \times 150 \times \sin(72)\) (or using right-angled triangle trig: \(\sin(BAC) = \frac{d}{120}\))
M1 for calculating Area = 8560 oe or setting up \(d = 120 \sin(\text{their } BAC)\)
M1 for \(d = \frac{2 \times \text{Area}}{\text{their } AC}\) or carrying out calculation for \(d\)
A1 for 107 (accept 106.6)

(a) The sequence of dots is: 1, 7, 19, 37, ...
The differences between consecutive terms are:
7 - 1 = 6
19 - 7 = 12
37 - 19 = 18
These differences form an arithmetic sequence with a common difference of 6.
The next difference is 18 + 6 = 24.
So, the number of dots for a hexagon of side 5 is 37 + 24 = 61.

(b) Since the second difference is constant (6), the general formula is quadratic: $D_n = a n^2 + b n + c$.
The coefficient $a = 6 / 2 = 3$.
So, $D_n = 3n^2 + bn + c$.
Using $n=1$, $D_1 = 3(1)^2 + b(1) + c = 1 \implies b+c = -2$.
Using $n=2$, $D_2 = 3(2)^2 + b(2) + c = 7 \implies 12 + 2b + c = 7 \implies 2b+c = -5$.
Subtracting the first equation from the second gives:
$b = -3$.
Then $c = -2 - (-3) = 1$.
Thus, the expression is $3n^2 - 3n + 1$.

(c) Set $D_n = 169$:
$3n^2 - 3n + 1 = 169 \implies 3n^2 - 3n - 168 = 0 \implies n^2 - n - 56 = 0$.
Factoring the quadratic equation:
$(n - 8)(n + 7) = 0$.
Since $n$ must be positive, the side length is $n = 8$.

Total Marks: 7.5

(a) [2 marks]
- B1: for identifying the pattern in first differences (6, 12, 18, 24).
- B1: for the correct next term, 61.

(b) [3.5 marks]
- M1: for attempting to find first and second differences.
- M1: for setting up quadratic equations to find coefficients $b$ and $c$.
- A1.5: for the correct final simplified expression $3n^2 - 3n + 1$ (award partial credit of 1 mark if $a=3$ is found but mistakes are made finding $b$ and $c$).

(c) [2 marks]
- M1: for setting their expression equal to 169 and attempting to solve the quadratic equation.
- A1: for $n = 8$.

(a) In a $4 \times 6$ grid, the count of squares of each size is:
- $1 \times 1$ squares: $4 \times 6 = 24$
- $2 \times 2$ squares: $3 \times 5 = 15$
- $3 \times 3$ squares: $2 \times 4 = 8$
- $4 \times 4$ squares: $1 \times 3 = 3$
Total squares = $24 + 15 + 8 + 3 = 50$.

(b) We are given $S_n = \frac{1}{3} n^3 + b n^2 + c n$.
For $n = 1$, $S_1 = 3$:
$\frac{1}{3}(1)^3 + b(1)^2 + c(1) = 3 \implies \frac{1}{3} + b + c = 3 \implies b + c = \frac{8}{3}$.

For $n = 2$, $S_2 = 11$:
$\frac{1}{3}(2)^3 + b(2)^2 + c(2) = 11 \implies \frac{8}{3} + 4b + 2c = 11 \implies 4b + 2c = \frac{25}{3} \implies 2b + c = \frac{25}{6}$.

Subtracting the first equation from the second:
$(2b + c) - (b + c) = \frac{25}{6} - \frac{16}{6} \implies b = \frac{9}{6} = 1.5$ (or $\frac{3}{2}$).

Then substitute $b$ back to find $c$:
$c = \frac{8}{3} - \frac{3}{2} = \frac{16 - 9}{6} = \frac{7}{6}$ (or $1.17$ to 3 s.f.).

(c) For $n = 3$, using the formula with $b = 1.5$ and $c = \frac{7}{6}$:
$S_3 = \frac{1}{3}(3)^3 + 1.5(3)^2 + \frac{7}{6}(3) = 9 + 13.5 + 3.5 = 26$.
This matches the given value of 26.

Total Marks: 7.5

(a) [2.5 marks]
- M1: for demonstrating a systematic count of different sizes of squares ($1\times1, 2\times2, 3\times3, 4\times4$).
- A1.5: for the correct total of 50.

(b) [3.5 marks]
- M1: for substituting $n = 1$ to get an equation in terms of $b$ and $c$.
- M1: for substituting $n = 2$ to get a second equation in terms of $b$ and $c$.
- A1: for $b = 1.5$ (or $3/2$).
- A0.5: for $c = 7/6$ (or $1.17$ to 3 s.f.).

(c) [1.5 marks]
- M1: for substituting $n = 3$, $b = 1.5$, and $c = 7/6$ into the formula.
- A0.5: for showing that the sum evaluates exactly to 26.

(a) At $t = 0$, $T = 92$:
$92 = 22 + A b^0 \implies 92 = 22 + A \implies A = 70$.

(b) At $t = 10$, $T = 57$:
$57 = 22 + 70 b^{10} \implies 35 = 70 b^{10} \implies b^{10} = 0.5$.
Taking the 10th root:
$b = (0.5)^{1/10} \approx 0.93303 \approx 0.9330$.

(c) At $t = 25$:
$T = 22 + 70 (0.93303)^{25}$
$T = 22 + 70 \times 0.17678 = 22 + 12.37 = 34.37^\circ\text{C} \approx 34.4^\circ\text{C}$.

Total Marks: 7.5

(a) [2 marks]
- M1: for substituting $t=0$ and $T=92$ into the model.
- A1: for $A = 70$.

(b) [3.5 marks]
- M1: for substituting $t=10, T=57$ and their $A$ into the model.
- M1: for isolating $b^{10} = 0.5$ (or equivalent).
- A1.5: for $b = 0.9330$ correct to 4 decimal places.

(c) [2 marks]
- M1: for substituting $t=25$ and their values of $A$ and $b$ into the formula.
- A1: for $34.4$ (accept $34.3$ or $34.4$ depending on rounding of intermediate values).

(a) The constant $r$ represents the average level of the water, which is the vertical shift:
$r = \frac{\text{High tide} + \text{Low tide}}{2} = \frac{4.8 + 1.2}{2} = 3.0$ metres.

(b) The constant $p$ is the amplitude of the cosine wave:
$p = \frac{\text{High tide} - \text{Low tide}}{2} = \frac{4.8 - 1.2}{2} = 1.8$.

The low tide occurs at $t = 6$, which corresponds to the first minimum of the cosine function (occurring when the angle is $180^\circ$):
$q \times 6 = 180 \implies q = 30$.

(c) Substituting $p = 1.8$, $q = 30$, $r = 3.0$ and $t = 4$:
$h = 1.8 \cos(30 \times 4^\circ) + 3.0 = 1.8 \cos(120^\circ) + 3.0$
Since $\cos(120^\circ) = -0.5$:
$h = 1.8(-0.5) + 3.0 = -0.9 + 3.0 = 2.1$ metres.

Total Marks: 7.5

(a) [2 marks]
- M1: for using the average of high and low tides.
- A1: for $r = 3.0$.
- Note: award full 2 marks for stating 3.0 without working.

(b) [3.5 marks]
- M1: for showing amplitude calculation: $(4.8 - 1.2)/2 = 1.8$.
- M1: for setting up equation for $q$, e.g., $6q = 180$ or $\cos(6q) = -1$.
- A1.5: for $q = 30$.

(c) [2 marks]
- M1: for substituting $t=4$ into their completed model.
- A1: for $2.1$ metres.

(a) Rate of change from \(t=0\) to \(t=10\) is \(\frac{59-85}{10} = -2.6\).
Rate of change from \(t=20\) to \(t=40\) is \(\frac{28-43}{20} = -0.75\).
Since these rates of change are different, a linear model is not suitable because a linear model has a constant rate of change.
(b)(i) At \(t=0\), \(C=85 \implies 85 = p \cdot q^0 + 20 \implies 85 = p + 20 \implies p = 65\).
At \(t=20\), \(C=43 \implies 43 = 65 \cdot q^{20} + 20 \implies 23 = 65 \cdot q^{20} \implies q^{20} = \frac{23}{65}\).
\(q = (23/65)^{1/20} \approx 0.94954\).
To 3 decimal places, \(q = 0.950\).
(b)(ii) \(C = 65 \cdot (0.950)^t + 20\).
(c)(i) When \(t=12\), \(C = 65 \cdot (0.950)^{12} + 20 \approx 65 \cdot 0.54036 + 20 = 35.12 + 20 = 55.12 \approx 55.1^\circ\text{C}\).
(c)(ii) \(30 = 65 \cdot (0.950)^t + 20 \implies 10 = 65 \cdot (0.950)^t \implies 0.950^t = \frac{10}{65} \approx 0.1538\).
\(t \ln(0.950) = \ln(0.1538) \implies t = \frac{\ln(0.1538)}{\ln(0.950)} \approx \frac{-1.8718}{-0.051293} \approx 36.49\) minutes.
To 1 decimal place, \(t = 36.5\) minutes.

(a) M1 for attempting to calculate at least one rate of change. A1 for correct calculation of both rates of change and stating they are different, thus concluding the linear model is unsuitable.
(b)(i) B1 for showing \(p=65\) using \(t=0, C=85\). M1 for substituting \(t=20, C=43\) and solving for \(q^{20}\). A1 for obtaining \(q \approx 0.950\).
(b)(ii) B1 for \(C = 65 \cdot (0.950)^t + 20\) (or using their \(q\)).
(c)(i) M1 for substituting \(t=12\) into their model. A1 for \(55.1\) (accept \(55.1^\circ\text{C}\) or awrt \(55.1\)).
(c)(ii) M1 for setting their model equation equal to 30 and isolating the exponential term. A1 for correct use of logarithms to solve. A1 for \(36.5\) minutes (accept awrt \(36.5\)).

(a) When \(t = 0\), \(h = 2 \implies a(0)^2 + b(0) + c = 2 \implies c = 2.0\).
(b)(i) Substitute \(t = 1, h = 11.5\) and \(c = 2\):
\(a + b + 2 = 11.5 \implies a + b = 9.5\) (Eq 1).
Substitute \(t = 2, h = 16.0\) and \(c = 2\):
\(4a + 2b + 2 = 16.0 \implies 4a + 2b = 14.0 \implies 2a + b = 7.0\) (Eq 2).
Subtract Eq 1 from Eq 2:
\((2a + b) - (a + b) = 7.0 - 9.5 \implies a = -2.5\).
Then \(b = 9.5 - (-2.5) = 12\).
(b)(ii) \(h = -2.5t^2 + 12t + 2\).
(c)(i) Vertex \(t = -\frac{b}{2a} = -\frac{12}{2(-2.5)} = 2.4\) seconds.
\(h(2.4) = -2.5(2.4)^2 + 12(2.4) + 2 = 16.4\) metres.
So coordinates of the vertex are \((2.4, 16.4)\) and the maximum height is \(16.4\) metres.
(c)(ii) Set \(h = 0 \implies -2.5t^2 + 12t + 2 = 0 \implies 2.5t^2 - 12t - 2 = 0\).
Using quadratic formula:
\(t = \frac{12 \pm \sqrt{12^2 - 4(2.5)(-2)}}{2(2.5)} = \frac{12 \pm \sqrt{164}}{5}\).
Since \(t > 0\), \(t = \frac{12 + \sqrt{164}}{5} \approx 4.9612\).
So \(t = 4.96\) seconds.

(a) B1 for \(c = 2\) or \(c = 2.0\).
(b)(i) M1 for forming two simultaneous equations in \(a\) and \(b\). M1 for a valid method to solve the equations. A1 for obtaining \(a = -2.5\) and \(b = 12\) with clear steps.
(b)(ii) B1 for \(h = -2.5t^2 + 12t + 2\).
(c)(i) M1 for finding \(t\) of the vertex. M1 for substituting their \(t\) to find the maximum height. A1 for vertex coordinates \((2.4, 16.4)\) and maximum height \(16.4\) metres.
(c)(ii) M1 for setting \(h=0\) and attempting to use quadratic formula or solver. A1 for \(4.96\) seconds (accept awrt \(4.96\)).

(a) At \(t = 0\), \(N = 5 \implies A \cdot b^0 = 5 \implies A = 5\).
At \(t = 2\), \(N = 10 \implies 5 \cdot b^2 = 10 \implies b^2 = 2 \implies b = \sqrt{2}\) (since \(b > 0\)).
(b) \(N = 5 \cdot (\sqrt{2})^t\) or \(N = 5 \cdot 2^{t/2}\).
(c)(i) When \(t=5\), \(N = 5 \cdot 2^{5/2} = 5 \cdot \sqrt{32} \approx 28.284\) thousand.
To 1 decimal place, \(28.3\) thousand.
(c)(ii) \(500 = 5 \cdot 2^{t/2} \implies 2^{t/2} = 100\).
Taking logs: \(\log_{10}(2^{t/2}) = \log_{10}(100) \implies \frac{t}{2} \log_{10}(2) = 2 \implies t = \frac{4}{\log_{10}(2)}\).
\(t \approx \frac{4}{0.30103} \approx 13.2877\) hours.
To 2 decimal places, \(t = 13.29\) hours.

(a) B1 for \(A = 5\) showing working. M1 for substituting \(t=2, N=10\) and solving \(5b^2 = 10\). A1 for \(b = \sqrt{2}\).
(b) B1 for \(N = 5 \cdot (\sqrt{2})^t\) or \(N = 5 \cdot 2^{t/2}\).
(c)(i) M1 for substituting \(t=5\) into their model. A1 for \(28.3\) thousand.
(c)(ii) M1 for setting \(N = 500\) and isolating the exponential term. M1 for taking logarithms of both sides. M1 for correct rearrangement. A1 for \(13.29\) hours.