0606/22 May June 2026: Problems and Solutions

1. By completing the square, find the minimum point of $ \displaystyle y = x^2 - 6x + 10 $.
  
  Solution
  $\displaystyle y = x^2 - 6x + 10$ $\displaystyle y = (x^2 - 6x + 3^2) - 3^2 + 10$ $\displaystyle y = (x - 3)^2 - 9 + 10$ $\displaystyle y = (x - 3)^2 + 1$ The minimum point is $ (3, 1) $.
2. Point $ B $ is $ (6,7) $.
  
  Point $ C $ is where the graph of $ y $ crosses the $ y $-axis.
  
  Find the equation of the line that crosses point $ B $ given that the line is parallel to the line $ AC $.
  
  Solution
  From (a), $ A(3, 1) $.
  
  The point $ C $ is the $ y $-intercept of $ \displaystyle y = x^2 - 6x + 10 $.
  
  When $ x = 0 $, $ y = 10 \implies C(0, 10) $.
  
  The gradient of line $ \displaystyle AC = \frac{10 - 1}{0 - 3} = \frac{9}{-3} = -3 $.
  
  Since the required line is parallel to $ AC $, its gradient is $ m = -3 $.
  
  Equation of the line passing through $ B(6,7) $: $\begin{align*} y - 7 &= -3(x - 6) \\ y - 7 &= -3x + 18 \\ y &= -3x + 25 \end{align*}$ The equation of the line is $ y = -3x + 25 $.
Show that there is no real value of $ k $ that satisfies $$$ 3x^2 - (2k+3)x + (k-2) = 0 $$$ that has equal roots.

Solution
For a quadratic equation to have equal roots, its discriminant must be zero. $\begin{align*} \text{discriminant} &= b^2 - 4ac \\ \text{discriminant} &= [-(2k+3)]^2 - 4(3)(k-2) \\ \text{discriminant} &= (4k^2 + 12k + 9) - 12(k-2) \\ \text{discriminant} &= 4k^2 + 12k + 9 - 12k + 24 \\ \text{discriminant} &= 4k^2 + 33 \end{align*}$ For any real value of $ k $, $ \displaystyle k^2 \ge 0 $, which implies $ \displaystyle 4k^2 \ge 0 $.

Therefore, $ \displaystyle 4k^2 + 33 \ge 33 > 0 $.

Since $ \displaystyle \text{discriminant} > 0 $ for all real values of $ k $, it can never be equal to $ 0 $.
Hence, there is no real value of $ k $ that gives equal roots.
1. Find $ \displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} $ if $ \displaystyle y = \sqrt{x+1} + \cos(5x+3) $.
  
  Solution
  $ \displaystyle y = (x+1)^{\frac{1}{2}} + \cos(5x+3) $ $\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{1}{2}(x+1)^{-\frac{1}{2}} \cdot (1) - \sin(5x+3) \cdot (5) \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{1}{2\sqrt{x+1}} - 5\sin(5x+3) \end{aligned}$
2. Find $ \displaystyle \int \sec^2 \frac{x}{3} \, \mathrm{d}x $.
  
  Solution
  $\int \sec^2 \left(\frac{1}{3}x\right) \, \mathrm{d}x = \frac{1}{\frac{1}{3}} \tan \left(\frac{x}{3}\right) + C = 3\tan\left(\frac{x}{3}\right) + C$
1. Sketch the graph of $ \displaystyle y = |x^2 - 5| $.
  
  Solution
  The $ x $-intercepts of $ \displaystyle y = |x^2 - 5| $ are $ (\sqrt{5},0) $ and $ (-\sqrt{5},0) $.
  
  The $ y $-intercept is at $ \displaystyle |0^2 - 5| = 5 $.
2. $ \mathrm{f}(x) $ is in the form of $ ax+b $.
  
  Find possible expressions for $ \mathrm{f}(x) $ if the two possible solutions for $ |\mathrm{f}(x)| = 4 $ are $ x = 2.5 $ and $ x = -1.5 $.
  
  Solution
  $ \displaystyle |\mathrm{f}(x)| = 4 $.
  
  $ \displaystyle |ax+b| = 4 $
  
  $ \displaystyle ax+b = 4 $ or $ \displaystyle ax+b = -4 $.
  
  Since there are exactly two solutions $ x = 2.5 $ and $ x = -1.5 $, there are two cases.
  
  Case 1: $ 2.5a + b = 4 \quad \text{--- (1)} $
  $ -1.5a + b = -4 \quad \text{--- (2)} $
  
  Subtracting (2) from (1): $ 4a = 8 \implies a = 2 $.
  Substitute into (1): $ 2.5(2) + b = 4 \implies 5 + b = 4 \implies b = -1 $.
  So, $ \mathrm{f}(x) = 2x - 1 $.
  
  Case 2: $ 2.5a + b = -4 \quad \text{--- (3)} $
  $ -1.5a + b = 4 \quad \text{--- (4)} $
  
  Subtracting (4) from (3): $ 4a = -8 \implies a = -2 $.
  Substitute into (3): $ 2.5(-2) + b = -4 \implies -5 + b = -4 \implies b = 1 $.
  So, $ \mathrm{f}(x) = -2x + 1 $.
  Possible expressions are $ \mathrm{f}(x) = 2x - 1 $ and $ \mathrm{f}(x) = -2x + 1 $.
1. Find the number of ways to form a $ 6 $-digit number that has $ 6 $ different digits, is divisible by $ 2 $ and does not start with $ 0 $.
  
  Solution
2. A committee of 10 people are selected from 9 students and 7 teachers. Find the number of different committees that could be formed containing at least 4 students and at least 4 teachers.
  
  Solution
  To form a committee of 10 people with at least 4 students and at least 4 teachers from a group of 9 students and 7 teachers, the possible combinations are:
  
  (1) 4 students and 6 teachers $ \displaystyle = {}^{9}C_{4} \times {}^{7}C_{6} = 882 $
  
  (2) 5 students and 5 teachers $ \displaystyle = {}^{9}C_{5} \times {}^{7}C_{5} = 2646 $
  
  (3) 6 students and 4 teachers $ \displaystyle = {}^{9}C_{6} \times {}^{7}C_{4} = 2940 $
  
  $\therefore\quad\textbf{Total number of committees} = 882 + 2646 + 2940 = 6468$
$ \displaystyle (a+x)^4(1+ax)^5 $ has $ \displaystyle 16 + bx + cx^2 $ as its first three terms. Find the possible values of $ a, b, c $.

Solution
$(a+x)^4(1+ax)^5 = (a^4 + 4a^3x + 6a^2x^2 + \dots)(1 + 5ax + 10a^2x^2 + \dots)$ $ \displaystyle \text{Constant term} = a^4 $

$ \displaystyle \therefore\quad a^4 = 16 \implies a = \pm 2 $

$ \displaystyle \text{Coefficient of } x = 5a^5 + 4a^3 $
$ \displaystyle b = 5a^5 + 4a^3 $

$ \displaystyle \text{Coefficient of } x^2 = 10a^6 + 20a^4 + 6a^2 $
$ \displaystyle \therefore\quad c = 10a^6 + 20a^4 + 6a^2 $

If $ a = 2 $: $ \displaystyle b = 5(2)^5 + 4(2)^3 = 5(32) + 4(8) = 160 + 32 = 192 $
$ \displaystyle c = 10(2)^6 + 20(2)^4 + 6(2)^2 = 10(64) + 20(16) + 6(4) = 640 + 320 + 24 = 984 $

If $ a = -2 $: $ \displaystyle b = 5(-2)^5 + 4(-2)^3 = 5(-32) + 4(-8) = -160 - 32 = -192 $
$ \displaystyle c = 10(-2)^6 + 20(-2)^4 + 6(-2)^2 = 10(64) + 20(16) + 6(4) = 984 $
Possible values are $ (a=2, b=192, c=984) $ and $ (a=-2, b=-192, c=984) $.
$ \displaystyle \mathrm{f}(x) = 1 + 2^x $, $ \displaystyle x \ge -1 $. $ \displaystyle \mathrm{g}(x) = x(x+k) $, $ \displaystyle x \ge -1 $
1. Find the range of $ \mathrm{f}(x) $.
  
  Solution
  Since $ \displaystyle x \ge -1 $, $ \displaystyle 2^x \ge 2^{-1} \implies 2^x \ge 0.5 $.
  Therefore, $ \displaystyle \mathrm{f}(x) = 1 + 2^x \ge 1 + 0.5 = 1.5 $.
  The range is $ \mathrm{f}(x) \ge 1.5 $.
2. Find the domain of $ \mathrm{f}^{-1}(x) $.
  
  Solution
  The domain of the inverse function is equal to the range of the original function.
  The domain is $ x \ge 1.5 $.
3. Find an expression of $ \mathrm{f}^{-1}(x) $.
  
  Solution
  $\begin{aligned} &\text { Let } \mathrm{f}^{-1}(x)=y \text { then }\\ &\begin{aligned} & \mathrm{f}(y)=x \\ & 1+2^y=x \\ & 2^y=x-1 \\ & y=\log _2(x-1) \\ & \mathrm{f}^{-1}(x)=\log _2(x-1) \end{aligned} \end{aligned}$
4. Sketch the graphs of $ \mathrm{f}(x) $ and $ \mathrm{f}^{-1}(x) $.
  
  Solution
  The curves of $ \displaystyle y=\mathrm{f}(x) $ and $ \displaystyle y=\mathrm{f}^{-1}(x) $ are symmetrical about the line $ \displaystyle y=x $.
5. State why $ \mathrm{gf}(x) $ exists.
  
  Solution
  For the composite function $ \mathrm{gf}(x) $ to exist, the range of $ \mathrm{f}(x) $ must be a subset of the domain of $ \mathrm{g}(x) $.
  
  The range of $ \mathrm{f}(x) :$ $ \mathrm{f}\ge 1.5 $.
  
  The domain of $ \mathrm{g}(x) :$ $ \mathrm{g}\ge -1 $.
  Hence, $ \mathrm{gf}(x) $ exists.
6. Find an expression for $ \mathrm{gf}(x) $.
  
  Solution
  $\begin{aligned} \mathrm{gf}(x) &= \mathrm{g}(\mathrm{f}(x)) \\ \mathrm{gf}(x) &= \mathrm{g}(1+2^x) \\ \mathrm{gf}(x) &= (1+2^x)(1+2^x+k) \end{aligned}$
7. Find $ k $ if $ \mathrm{gf}(3) = 126 $.
  
  Solution
  $\begin{aligned} \mathrm{gf}(3) &= (1+2^3)(1+2^3+k) \\ 126 &= (1+8)(1+8+k) \\ 126 &= 9(9+k) \\ 14 &= 9+k \\ k &= 5 \end{aligned}$ The value is $ k = 5 $.
A particle's displacement function is described as
$$ \mathrm{s}(t) = e^{4t} - 10e^{2t} + 9 $$
1. Find the value of $ t $ which the particle has stationary velocity.
  
  Solution
  Velocity $ \displaystyle \mathrm{v}(t) = \mathrm{s}'(t) = 4e^{4t} - 20e^{2t} $.
  
  Stationary velocity means the acceleration is zero ($ \mathrm{v}'(t) = 0 $).
  
  Acceleration $ \displaystyle \mathrm{a}(t) = \mathrm{v}'(t) = 16e^{4t} - 40e^{2t} $.
  
  Setting $ \mathrm{a}(t) = 0 $: $\begin{align*} 16e^{4t} - 40e^{2t} &= 0 \\ 8e^{2t}(2e^{2t} - 5) &= 0 \end{align*}$ Since $ e^{2t} > 0 $, we have: $\begin{align*} 2e^{2t} - 5 &= 0 \implies e^{2t} = 2.5 \\ 2t &= \ln 2.5 \\ t &= \frac{1}{2} \ln 2.5 \end{align*}$
2. Find the acceleration when $ t = \ln 2 $.
  
  Solution
  When $ t = \ln 2 $, we find $ \displaystyle e^{2t} = e^{2\ln 2} = e^{\ln 4} = 4 $.
  Also, $ \displaystyle e^{4t} = (e^{2t})^2 = 4^2 = 16 $.
  Substitute into the acceleration equation: $\begin{align*} \mathrm{a}(\ln 2) &= 16(16) - 40(4) \\ \mathrm{a}(\ln 2) &= 256 - 160 = 96 \end{align*}$ The acceleration is $ 96 $.
In this question, distance is in kilometers and time is in hours.

Point $ P $ has position vector $ 2\mathbf{i} + 5\mathbf{j} $. At 13:00, Boat $ A $ travels from point $ P $ with velocity vector $ 8\mathbf{i} + 8\mathbf{j} $.

Point $ Q $ has position vector $ 20\mathbf{i} + 8\mathbf{j} $. At 15:00, Boat $ B $ travels from point $ Q $ at bearing of $ 300^\circ $ with constant speed $ 5\sqrt{3} $.
1. Find position vector of boat $ A $ after $ 2 $ hours.
  
  Solution
  $\begin{aligned} &\mathbf{r}_A = \mathbf{r}_P + 2 \mathbf{v}_A \\ &\mathbf{r}_A = (2\mathbf{i} + 5\mathbf{j}) + 2(8\mathbf{i} + 8\mathbf{j}) \\ &\mathbf{r}_A = (2\mathbf{i} + 5\mathbf{j}) + (16\mathbf{i} + 16\mathbf{j}) \\ &\mathbf{r}_A = 18\mathbf{i} + 21\mathbf{j} \end{aligned}$
2. Find position vector of boat $ A $ after $ t $ hours.
  
  Solution
  $\begin{aligned} \mathbf{r}_A &= \mathbf{r}_P + t \mathbf{v}_A \\ \mathbf{r}_A &= (2\mathbf{i} + 5\mathbf{j}) + t(8\mathbf{i} + 8\mathbf{j}) \\ \mathbf{r}_A &= (2 + 8t)\mathbf{i} + (5 + 8t)\mathbf{j} \end{aligned}$
3. Find position vector of boat $ B $ at 17:00.
  
  Solution
  Velocity vector of boat $ B $ with bearing $ 300^\circ $ ($ 60^\circ $ West of North): $\begin{align*} \mathbf{v}_B &= 5\sqrt{3} (-\sin 60^\circ \mathbf{i} + \cos 60^\circ \mathbf{j}) \\ \mathbf{v}_B &= 5\sqrt{3} \left(-\frac{\sqrt{3}}{2}\mathbf{i} + \frac{1}{2}\mathbf{j}\right) \\ \mathbf{v}_B &= -7.5\mathbf{i} + 2.5\sqrt{3}\mathbf{j} \end{align*}$ Boat $ B $ starts at 15:00. At 17:00, time elapsed $ t_B = 2 $ hours. $\begin{align*} \mathbf{r}_B &= \mathbf{r}_Q + 2\mathbf{v}_B \\ \mathbf{r}_B &= (20\mathbf{i} + 8\mathbf{j}) + 2(-7.5\mathbf{i} + 2.5\sqrt{3}\mathbf{j}) \\ \mathbf{r}_B &= (20\mathbf{i} + 8\mathbf{j}) + (-15\mathbf{i} + 5\sqrt{3}\mathbf{j}) \\ \mathbf{r}_B &= 5\mathbf{i} + (8 + 5\sqrt{3})\mathbf{j} \end{align*}$
4. Find the distance between two boats at 17:00.
  
  Solution
  At 17:00, Boat $ A $ has traveled for $ 4 $ hours (since 13:00).
  
  $ \displaystyle \mathbf{r}_A = (2 + 8(4))\mathbf{i} + (5 + 8(4))\mathbf{j} = 34\mathbf{i} + 37\mathbf{j} $.
  
  Position vector relative to each other ($ \mathbf{AB} $): $\begin{align*} \mathbf{AB} &= \mathbf{r}_B - \mathbf{r}_A \\ \mathbf{AB} &= (5 - 34)\mathbf{i} + (8 + 5\sqrt{3} - 37)\mathbf{j} \\ \mathbf{AB} &= -29\mathbf{i} + (5\sqrt{3} - 29)\mathbf{j} \end{align*}$ Distance $ d = |\mathbf{AB}| $: $\begin{align*} d^2 &= (-29)^2 + (5\sqrt{3} - 29)^2 \\ d^2 &= 841 + (75 - 290\sqrt{3} + 841) \\ d^2 &= 1757 - 290\sqrt{3} \\ d &= \sqrt{1757 - 290\sqrt{3}} \approx 35.42 \text{ km} \end{align*}$
Solve $ \displaystyle \tan(3x+1) = 5\sin(3x+1), \quad -\frac{\pi}{3} \leqslant x \leqslant \frac{\pi}{3} $

Solution
let $ u = 3x+1 $ and hence $ 1-\pi \leqslant u \leqslant 1+\pi $ or $ -122.7^\circ \leqslant u \leqslant 237.3^\circ $ (for reference)

$ \therefore \tan u = 5\sin u $

$\begin{align*} \frac{\sin u}{\cos u} - 5\sin u &= 0 \\[1ex] \sin u \left( \frac{1}{\cos u} - 5 \right) &= 0 \\[1ex] \sin u = 0 \quad &\text{or} \quad \frac{1}{\cos u} = 5 \\[1ex] \sin u = 0 \quad &\text{or} \quad \cos u = \frac{1}{5} \end{align*}$ $ \displaystyle \therefore\quad u = 0 \text{ or } \pi \quad \text{or} \quad u = \cos^{-1}\left(\frac{1}{5}\right) \text{ or } u = -\cos^{-1}\left(\frac{1}{5}\right) $

$ \displaystyle \therefore \quad u = 0 \text{ or } \pi \quad \text{or} \quad u = 1.369 \text{ or } u = -1.369 $

(i) $ \displaystyle u = 0 \implies 3x+1 = 0 \implies x = -\frac{1}{3} $

(ii) $ \displaystyle u = \pi \implies 3x+1 = \pi \implies x = \frac{\pi-1}{3} $

(iii) $ \displaystyle u = 1.369 \implies 3x+1 = 1.369 \implies x = 0.123 $

(iv) $ \displaystyle u = -1.369 \implies 3x+1 = -1.369 \implies x = -0.790 $
Solutions to this question by accurate drawing will not be accepted.

Circle $ C_1 $ has center $ (3,5) $ and point $ (1,4) $ is on the circle. Circle $ C_2 $ has center $ (12,8) $ and line with equation $ y = 19 - 3x $ is the common chord of two circles. Find the radius of $ C_2 $.

Solution

$ C_1 $: centre: $ (3,5) $

radius $ \displaystyle (r_1) = \sqrt{(3-1)^2+(5-4)^2} = \sqrt{5} $

$ C_1 $: $ \displaystyle (x-3)^2+(y-5)^2 = 5 $

$ C_2 $: centre: $ (12,8) $

Line: $ \displaystyle y = 19-3x $ is the common chord of $ C_1 $ and $ C_2 $

At the point of intersections of $ C_1 $ and the line

$ \displaystyle (x-3)^2+(19-3x-5)^2 = 5 $

$ \displaystyle x^2-6x+9+196-84x+9x^2 = 5 $

$ \displaystyle 10x^2-90x+200 = 0 $

$ \displaystyle x^2-9x+20 = 0 $

$ \displaystyle (x-4)(x-5) = 0 $

$ \displaystyle x=4 \quad \text{or} \quad x=5 $

When $ x=4 $, $ y=7 $

$ \therefore \quad (4,7) $ is one of the point of intersection of $ C_1 $ and $ C_2 $, that is it lies on $ C_2 $.

$ \therefore\quad \text{radius of } C_2 = r_2 = \sqrt{(12-4)^2+(8-7)^2} = \sqrt{65} $
Find $ V $, the volume of a sphere if $ \displaystyle \frac{\mathrm{d}V}{\mathrm{d}t} = 9\frac{\mathrm{d}r}{\mathrm{d}t} $.

Solution
The formula for the volume of a sphere is: $V = \frac{4}{3}\pi r^3$ Differentiate with respect to $ t $: $\frac{\mathrm{d}V}{\mathrm{d}t} = 4\pi r^2 \frac{\mathrm{d}r}{\mathrm{d}t}$ Given that $ \displaystyle \frac{\mathrm{d}V}{\mathrm{d}t} = 9\frac{\mathrm{d}r}{\mathrm{d}t} $, substitute this into the equation: $\begin{align*} 9\frac{\mathrm{d}r}{\mathrm{d}t} &= 4\pi r^2 \frac{\mathrm{d}r}{\mathrm{d}t} \\ 9 &= 4\pi r^2 \\ r^2 &= \frac{9}{4\pi} \\ r &= \frac{3}{2\sqrt{\pi}} \end{align*}$ Substitute $ r $ back into the volume formula: $\begin{align*} V &= \frac{4}{3}\pi \left( \frac{3}{2\sqrt{\pi}} \right)^3 \\ V &= \frac{4}{3}\pi \left( \frac{27}{8\pi\sqrt{\pi}} \right) \\ V &= \frac{9}{2\sqrt{\pi}} \end{align*}$ The volume of the sphere is $ \displaystyle \frac{9}{2\sqrt{\pi}} $.

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0606/22 May June 2026: Problems and Solutions

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