SPM KSSM Diagnostic Drill
Form 4 Chapter 4: Indices, Surds & Logarithms
Total Marks: 40 Marks • Time Allowed: 50 Minutes
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Date
A Bahagian A: SPM Paper 1 Format [16 Marks]
Answer all questions
Soalan 1
[4 Marks]
Simplify the algebraic expression: $$\frac{2^{n+2} \times 4^{n-1}}{8^n}$$ Hence, find the value of $n$ if $\frac{2^{n+2} \times 4^{n-1}}{8^n} = \frac{1}{2^n}$.
Express all terms in common base 2:
$4^{n-1} = (2^2)^{n-1} = 2^{2n - 2}$ and $8^n = (2^3)^n = 2^{3n}$. [P1]
$\frac{2^{n+2} \times 2^{2n-2}}{2^{3n}} = \frac{2^{(n+2) + (2n-2)}}{2^{3n}} = \frac{2^{3n}}{2^{3n}} = 2^{3n - 3n} = 2^0 = 1$. [K1]
Given $1 = \frac{1}{2^n} \implies 2^n = 1 \implies 2^n = 2^0 \implies n = 0$.
Simplified value $= 1$; Value of $n = 0$.
[N1, N1]
Soalan 2
[4 Marks]
Rationalize the denominator of $\frac{4\sqrt{3} - 2}{\sqrt{3} + 2}$ and express the answer in the form $a + b\sqrt{3}$, where $a$ and $b$ are integers.
Conjugate of $(\sqrt{3} + 2)$ is $(\sqrt{3} - 2)$:
$\frac{4\sqrt{3} - 2}{\sqrt{3} + 2} \times \frac{\sqrt{3} - 2}{\sqrt{3} - 2}$ [K1]
Numerator: $(4\sqrt{3} - 2)(\sqrt{3} - 2) = 4(3) - 8\sqrt{3} - 2\sqrt{3} + 4 = 12 + 4 - 10\sqrt{3} = 16 - 10\sqrt{3}$. [K1]
Denominator: $(\sqrt{3})^2 - (2)^2 = 3 - 4 = -1$.
$\frac{16 - 10\sqrt{3}}{-1} = -16 + 10\sqrt{3}$.
Answer: $-16 + 10\sqrt{3} \quad (a = -16, b = 10)$
[N1, N1]
Soalan 3
[4 Marks]
Solve the logarithmic equation: $$\log_3(x + 2) + \log_3(x - 4) = 3$$
Combine logs using product law: $\log_3[(x + 2)(x - 4)] = 3$ [P1]
Convert to index form: $(x + 2)(x - 4) = 3^3 = 27$ [K1]
$x^2 - 2x - 8 = 27 \implies x^2 - 2x - 35 = 0$
$(x - 7)(x + 5) = 0 \implies x = 7$ or $x = -5$.
Domain check: $\log_3(x - 4)$ requires $x > 4$. Hence $x = -5$ is extraneous and rejected.
Valid solution: $x = 7$
[K1 check, N1 7]
Soalan 4
[4 Marks]
Given that $\log_2 3 = p$ and $\log_2 5 = q$, express each of the following in terms of $p$ and $q$:
(a) $\log_2 45$. [2 marks]
(b) $\log_4 15$. [2 marks]
(a) $45 = 9 \times 5 = 3^2 \times 5$
$\log_2 45 = \log_2(3^2 \times 5) = 2\log_2 3 + \log_2 5 = 2p + q$
[K1] $2\log_2 3 + \log_2 5$$\log_2 45 = \log_2(3^2 \times 5) = 2\log_2 3 + \log_2 5 = 2p + q$
[N1] $2p + q$
(b) Change base to 2:
$\log_4 15 = \frac{\log_2 15}{\log_2 4} = \frac{\log_2(3 \times 5)}{\log_2(2^2)} = \frac{\log_2 3 + \log_2 5}{2} = \frac{p + q}{2}$
[K1] change base$\log_4 15 = \frac{\log_2 15}{\log_2 4} = \frac{\log_2(3 \times 5)}{\log_2(2^2)} = \frac{\log_2 3 + \log_2 5}{2} = \frac{p + q}{2}$
[N1] $\frac{p+q}{2}$
B Bahagian B: SPM Paper 2 Format [24 Marks]
Detailed solutions required
Soalan 5
[10 Marks]
(a) Solve the exponential equation:
$$3^{2x+1} - 10(3^x) + 3 = 0$$
[5 marks]
(b) Solve the simultaneous equations:
$$\log_2 x - \log_2 y = 2$$
$$x + 2y = 12$$
[5 marks]
(a) Quadratic substitution:
Notice $3^{2x+1} = 3^1 \cdot (3^x)^2 = 3(3^x)^2$. Let $u = 3^x$: [P1]
$3u^2 - 10u + 3 = 0$ [K1]
$(3u - 1)(u - 3) = 0$
Case 1: $u = 3 \implies 3^x = 3^1 \implies x = 1$ [N1]
Case 2: $u = \frac{1}{3} \implies 3^x = 3^{-1} \implies x = -1$ [N1]
$\therefore x = 1$ or $x = -1$.
Notice $3^{2x+1} = 3^1 \cdot (3^x)^2 = 3(3^x)^2$. Let $u = 3^x$: [P1]
$3u^2 - 10u + 3 = 0$ [K1]
$(3u - 1)(u - 3) = 0$
Case 1: $u = 3 \implies 3^x = 3^1 \implies x = 1$ [N1]
Case 2: $u = \frac{1}{3} \implies 3^x = 3^{-1} \implies x = -1$ [N1]
$\therefore x = 1$ or $x = -1$.
(b) Simultaneous Logarithms:
From first equation: $\log_2\left(\frac{x}{y}\right) = 2 \implies \frac{x}{y} = 2^2 = 4 \implies x = 4y$ [P1, K1]
Substitute $x = 4y$ into $x + 2y = 12$:
$4y + 2y = 12 \implies 6y = 12 \implies y = 2$ [K1, N1]
$x = 4(2) = 8$ [N1]
Check domain: $x = 8 > 0, y = 2 > 0$ (both valid). Solution: $(x, y) = (8, 2)$.
From first equation: $\log_2\left(\frac{x}{y}\right) = 2 \implies \frac{x}{y} = 2^2 = 4 \implies x = 4y$ [P1, K1]
Substitute $x = 4y$ into $x + 2y = 12$:
$4y + 2y = 12 \implies 6y = 12 \implies y = 2$ [K1, N1]
$x = 4(2) = 8$ [N1]
Check domain: $x = 8 > 0, y = 2 > 0$ (both valid). Solution: $(x, y) = (8, 2)$.
Soalan 6 • SPM Paper 2 & KBAT
[14 Marks]
Archeometry & Biomedical Modeling
Marine Archeology Radiocarbon Dating & Vaccine Culture Kinetics
Part I (Ancient Wooden Shipwreck Radiocarbon Dating):
Marine archaeologists recover timber beams from a sunken galleon in the Straits of Malacca. The remaining activity of Carbon-14 in living organisms is modeled by:
$$N(t) = N_0 e^{-\lambda t}$$
where $N_0$ is the initial Carbon-14 activity, $t$ is the elapsed time in years, and the decay constant is $\lambda = 1.21 \times 10^{-4}\text{ year}^{-1}$.
(a) Show that the half-life of Carbon-14 is approximately $5\,730\text{ years}$. [3 marks]
(b) Laboratory mass spectrometry indicates that the timber sample retains only $76.5\%$ of its original Carbon-14 activity. Determine the estimated age of the shipwreck to the nearest year. [4 marks]
(a) Show that the half-life of Carbon-14 is approximately $5\,730\text{ years}$. [3 marks]
(b) Laboratory mass spectrometry indicates that the timber sample retains only $76.5\%$ of its original Carbon-14 activity. Determine the estimated age of the shipwreck to the nearest year. [4 marks]
Part II (Biomedical Bacterial Doubling Kinetics):
A laboratory cultivates bacteria for antibiotic research. The population $P(t)$ after $t$ hours grows exponentially according to:
$$P(t) = 800 \times (2.5)^{0.4 t}$$
(c) State the initial population of bacteria. [1 mark]
(d) Find the population of bacteria after 5 hours. [2 marks]
(e) Calculate the time, in hours and minutes, for the bacterial colony to surpass $100\,000$. [4 marks]
(c) State the initial population of bacteria. [1 mark]
(d) Find the population of bacteria after 5 hours. [2 marks]
(e) Calculate the time, in hours and minutes, for the bacterial colony to surpass $100\,000$. [4 marks]
(a) Half-life Derivation:
When $N(t) = 0.5 N_0 \implies e^{-\lambda t} = 0.5 \implies -\lambda t = \ln(0.5) = -\ln 2$. [K1]
$t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693147}{1.21 \times 10^{-4}} \approx 5\,728.5 \approx 5\,730\text{ years}$ (Shown). [K1] $\frac{\ln 2}{\lambda}$
[N1] $5\,730$
When $N(t) = 0.5 N_0 \implies e^{-\lambda t} = 0.5 \implies -\lambda t = \ln(0.5) = -\ln 2$. [K1]
$t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693147}{1.21 \times 10^{-4}} \approx 5\,728.5 \approx 5\,730\text{ years}$ (Shown). [K1] $\frac{\ln 2}{\lambda}$
[N1] $5\,730$
(b) Shipwreck Age:
$N(t) = 0.765 N_0 \implies e^{-\lambda t} = 0.765$ [K1]
$-\lambda t = \ln(0.765) \approx -0.26788$ [K1]
$t = \frac{-0.26788}{-1.21 \times 10^{-4}} = \frac{0.26788}{0.000121} \approx 2\,213.9\text{ years}$.
Estimated age $= 2\,214\text{ years}$ (around 188 BCE). [K1 division]
[N1] $2\,214\text{ years}$
$N(t) = 0.765 N_0 \implies e^{-\lambda t} = 0.765$ [K1]
$-\lambda t = \ln(0.765) \approx -0.26788$ [K1]
$t = \frac{-0.26788}{-1.21 \times 10^{-4}} = \frac{0.26788}{0.000121} \approx 2\,213.9\text{ years}$.
Estimated age $= 2\,214\text{ years}$ (around 188 BCE). [K1 division]
[N1] $2\,214\text{ years}$
(c) Initial Population:
$P(0) = 800 \times (2.5)^0 = 800$ bacteria. [N1] 800
$P(0) = 800 \times (2.5)^0 = 800$ bacteria. [N1] 800
(d) Population after 5 hours:
$P(5) = 800 \times (2.5)^{0.4 \times 5} = 800 \times (2.5)^2 = 800 \times 6.25 = 5\,000$ bacteria. [K1] $800 \times 2.5^2$
[N1] 5,000
$P(5) = 800 \times (2.5)^{0.4 \times 5} = 800 \times (2.5)^2 = 800 \times 6.25 = 5\,000$ bacteria. [K1] $800 \times 2.5^2$
[N1] 5,000
(e) Time to surpass 100,000:
$800 \times (2.5)^{0.4 t} > 100\,000 \implies (2.5)^{0.4 t} > \frac{100\,000}{800} = 125$ [K1]
Take $\log_{10}$ on both sides: $0.4 t \lg(2.5) > \lg(125)$ [K1]
$0.4 t (0.39794) > 2.09691 \implies 0.15918 t > 2.09691$
$t > \frac{2.09691}{0.15918} \approx 13.173\text{ hours}$ [K1]
$0.173 \times 60 \approx 10.4\text{ minutes}$.
$\therefore$ Time required is 13 hours and 11 minutes. [N1] 13h 11m
$800 \times (2.5)^{0.4 t} > 100\,000 \implies (2.5)^{0.4 t} > \frac{100\,000}{800} = 125$ [K1]
Take $\log_{10}$ on both sides: $0.4 t \lg(2.5) > \lg(125)$ [K1]
$0.4 t (0.39794) > 2.09691 \implies 0.15918 t > 2.09691$
$t > \frac{2.09691}{0.15918} \approx 13.173\text{ hours}$ [K1]
$0.173 \times 60 \approx 10.4\text{ minutes}$.
$\therefore$ Time required is 13 hours and 11 minutes. [N1] 13h 11m