How to deliberately lose some of your marks (add math 2025, paper 1)
. . . Ramai ibu bapa meluahkan rasa simpati terhadap anak-anak mereka yang menduduki kertas peperiksaan Matematik Tambahan Sijil Pelajaran Malaysia (SPM) tahun ini yang dikatakan susah dan di luar jangkaan. . . .
. . . ramai calon SPM menangis lepas jawab kertas ‘add math’ . . .
Halisya Isma Zulkurnain
Kosmo! (16 Disember 2025)
Question 1
The '08-batch was asked on December 15 to guess the value of \(p\) if the following objects are parallel.
$$ \left\{ \begin{array}{l} 2y-8x+1 = 0\\ y = 6\color{brown}{p}x + 3 \end{array} \right. $$The simplest way to tackle this problem, if one insists on doing it efficiently rather than floundering in despair, is perhaps to begin by multiplying the second equation by 2.
$$ \underbrace{y}_{\textrm{multiply by 2}} = \underbrace{6px + 3}_{\textrm{multiply by 2}} $$This is not because multiplying by 2 is inherently interesting, but because it makes everything else slightly less terrifying.
$$ \left\{ \begin{array}{l} \color{limegreen}{2} y-\color{brown}{8}x+1 = 0\\ \color{limegreen}{2}y - \color{brown}{12p}x - 6 = 0 \end{array} \right. $$Paralleism requires \(8 = 12p\) or \(p = \frac{2}{3}\).
Question 2
We have a straight line with \(X\)-intercept at \(X = -6\) and \(Y\)-intercept at \(Y = 3\). Since the slope of the given straight line is simply the ratio of the height to the base of the triangle, that is,
$${\rm gradient} = \tfrac{\Delta Y}{\Delta X} = \tfrac{3}{6} = \tfrac{1}{2}$$The equation of the straight line is then
$$Y= \tfrac{1}{2}X + 3$$Snce \(Y = \frac{y}{x}\) and \(X = x^2 y \), we must have
$$\frac{y}{x}= \tfrac{1}{2}x^2 y + 3$$ $$y\left(\tfrac{1}{x} - \tfrac{1}{2}x^2\right)= 3$$ $$y\left(1 - \tfrac{1}{2}x^3\right)= 3x$$ $$y = \frac{3x}{1 - \tfrac{1}{2}x^3} = \frac{6x}{2 - x^3}$$Question 3
My son was given a list of numbers:
$$9, 3, -3, \ldots, -105$$and was told to guess the number of terms in the sequence and the sum of all the terms in the sequence.
$$(1,9), (2,3), (3,-3), \ldots, $$Since the slope of the equation is \(\frac{3-9}{2-1} = -6\), we can write the equation of the straight line as \(y - 9 = -6(x - 1)\), since \(y = -105\) is a member on this line, we must have:
$$-105 - 9 = -6(x - 1)$$ $$19 = x - 1$$ $$x = 20$$ $$(1,9), (2,3), (3,-3), \ldots, (20, -105)$$The last term is the 20th term. The sum is easy since you only need to invoke Gauss's trick
$$S = 9 + \ldots + (-105)$$ $$S = (-105) + \ldots + 9$$ $$ \begin{align} 2S &= \underbrace{(-105 + 9) + \ldots + (9 - 105)}_{\textrm{20 identical terms of -96}}\\ S &= \tfrac{1}{2} \times 20 \times (-96) = -960 \end{align} $$Question 4(a)
We have two parallelograms with different bases \(b_1 = 3 + \sqrt{2}\) and \(b_2 = \sqrt{8}\) and the fact that the two quadrangles sum to \(9r + 3r\sqrt{2}\).
$$\underbrace{p \times (b_1 + b_2)}_{\textrm{sum of parallelograms}} = 9r + 3r\sqrt{2}$$ $$p ( 3 + \sqrt{2} + \sqrt{8}) = 3p(1 +\sqrt{2}) = 3r(3 + \sqrt{2})$$ $$ p\left(\tfrac{1}{\sqrt{2}} +1 \right) = r\left(\tfrac{3}{\sqrt{2}} + 1 \right)$$ $$ \begin{align} \tfrac{p}{r} &= \tfrac{1 + 3\sin 45^\circ }{1 + \sin 45^\circ} = \tfrac{3(1 + \sin 45^\circ) -2 }{1 + \sin 45^\circ} \\ & = 3 - \tfrac{2 }{\sin 90^\circ + \sin 45^\circ} = 3 - \tfrac{1}{\sin(\frac{135}{2})\cos(\frac{45}{2})}\\ &= 3 - \sec^2(\tfrac{45}{2}) \end{align} $$Question 4(b)
Given that \(\log_p 3 - \log_p x = \tfrac{1}{3}\log_p (2-y)\), express \(x\) in terms of \(y\).
$$\log_p \tfrac{3}{x} = \log_p (2-y)^{1/3}$$ $$ \tfrac{3}{x} = (2-y)^{1/3}$$ $$ \tfrac{x}{3} = (2-y)^{-1/3}$$ $$ x = 3(2-y)^{-1/3}$$Question 5
Since the axis of symmetry of \(f(x) = k(x - p)^2 + q\) is associated with the extremum point, we must have
$$x = p$$And its value is independent of the choice of \(k\), so is the value of its minimum point, which remains locked in \(f_{\rm min}(x) = q\)
Question 6
A triangle is to be formed with the given vectors, \(\mathbf{x} = \tbinom{1}{1} \) and \(\mathbf{y} = \tbinom{2}{-1} \). The side formed by \(2\mathbf{x} - \mathbf{y}\) is actually a vertical line spanning three units. Thus \(|2\mathbf{x} - \mathbf{y}| = 3\).
Question 7
State the complementary angle, in radian, of \(\alpha\) in terms of \(\pi\) and \(\alpha\), and in the given x-y quadrants, sketch a line to represent \(-225^\circ\).
Question 8(a)
Given that \(f(x) = \frac{3}{(1-2x)^8}\) and \(f'(x) = \frac{a}{(1-2x)^b}\). Guess the value of \(a\) and \(b\). The value \(b\) is apparently 9. The value of \(a\) is \(3\times -8 \times -2 = 48\).
Question 8(b)
The scenario described in this problem is, in fact, an accurate reconstruction of real life. It closely resembles the moment when your boss begins the morning by saying something illogical, immediately treats it as a profound insight, and then prescribes a workflow for solving a business problem. In practice, the workflow is irrelevant. The exercise exists purely to determine whether you will nod at the correct moments and display an acceptable level of political correctness, rather than attempt to understand anything at all. Accordingly, the employee nodded smilingly and pretended to agree that the area of a sector of a circle can, quite sensibly, be computed analytically using a quadratic arc, \(y = x^2 - 4x + 10p\), (\(x\) is the radius, \(p\) is a constant) something she will make a point of not teaching her children.
When the given expression is differentiated, we arrive at \(y'(x) = 2x - 4\), which dutifully informs us that the local extremum occurs at \(x = 2\). Substituting this back in produces \(y_{\rm max} = 10p-12\), which, coincidentally, is exactly the number the boss was hoping to see, thereby concluding the performance. Unfortunately, reality then intrudes. Since the area may also be written as \(y = x^2 \left(\tfrac{1}{2} \theta\right)\), where \(\theta\) is the angle subtended by the arc, it becomes blindingly obvious that this cannot possibly be reconciled with the expression \(x^2 - 4x + 10p\)given that the boss has insisted—quite firmly—that \(p\) is a constant. At this point, the mathematics has done its duty. Any remaining inconsistency is clearly a managerial choice.
Question 8(c)
A cone has a volume of \(16\pi\) cm³ with height of 3 cm. If the height of the cone is fixed, using differentiation, find the small change in the volume, in cm³, of the cone when its radius increases by 0.5%. We know the value of \(h\) is \(\frac{16}{3}\) since \(\frac{1}{3}\pi h (3)^2 = 16\pi\).
$$V_{\rm cone}' = \tfrac{2}{3}h \pi r \approx \frac{\Delta V_{\rm cone}}{\Delta r}$$ $$ \Delta V_{\rm cone} \approx \tfrac{32}{3} \pi \Delta r$$Increase in cone volume can then be estimated by
$$\tfrac{32}{3} \pi \Delta r = \tfrac{32}{3} \pi \times \tfrac{1/2}{100} \times 3 = \tfrac{16}{100}\pi $$which is approximately one percent.
Question 9
In an audition for a reality TV show, each contestant is evaluated based on three types of talents which are singing, dancing and acting. Table 9 shows a probability distribution for discrete random variable of \(X\),
| \(X = r\) | 0 | 1 | 2 | 3 |
| \(P(X = r)\) | \(\tfrac{17}{50}\) | \(\tfrac{22}{50}\) | \(\tfrac{10}{50}\) | \(k\) |
such that \(X\) represents the number of talents that the contestants passed in the audition from the previous season. Determine the value of \(k\) and sketch and label the distribution. The contestant who pass with at least two types of talent will qualify to the next stage, estimate the number of contestants who will qualify if the total number of contestants in this season is 1850.
Question 10(a)
Determine teh number of different ways to arrange 7 letters from the word SEMPURNA in a row without repetition.
$$^8\!P_7 = 8! = 40320$$
Question 10(b)
A four-digit number is to be formed by using the digits 1, 2, 3, 4, 5, and 6, such that the repetition of 1, 2, 3, 4, 5 is not allowed. Find the number of different ways the four-digit numbers can be formed. The easiest way to solve this counting problem is to set up the following table.
| 1st digit | 2nd digit | 3rd digit | 4th digit |
|---|---|---|---|
| 5 | 4 | 3 | 2 |
| ⑥ | 5 | 4 | 3 |
| 5 | ⑥ | 4 | 3 |
| 5 | 4 | ⑥ | 3 |
| 5 | 4 | 3 | ⑥ |
| ⑥ | ⑥ | 5 | 4 |
| ⑥ | 5 | ⑥ | 4 |
| ⑥ | 5 | 4 | ⑥ |
| 5 | ⑥ | ⑥ | 4 |
| 5 | ⑥ | 4 | ⑥ |
| 5 | 4 | ⑥ | ⑥ |
| 5 | ⑥ | ⑥ | ⑥ |
| ⑥ | 5 | ⑥ | ⑥ |
| ⑥ | 5 | 5 | ⑥ |
| ⑥ | ⑥ | ⑥ | 5 |
| ⑥ | ⑥ | ⑥ | ⑥ |
The total number of ways is therefore given by
$$\tbinom{4}{0} (^5\!P_4) + \tbinom{4}{1} (^5\!P_3) + \tbinom{4}{2} (^5\!P_2) + \ldots$$ $$\ldots + \tbinom{4}{3} (^5\!P_1) + \tbinom{4}{4}{^5\!P_0} = 501$$Question 11
There are 25 students present at the general meeting of STEM Society. Find the number of different ways to choose 4 students to lead the committee. After the leaders are chosen, the remaining number of boys to the number of girls is in the ratio 4:3. A boy called Zam and a girl called Min are among them. Find the number of different ways 5 committee members can be chosen, such that the number of boys is more than the number of girls and must include either Zam or Min.
Question 12
Diagram 12 shows a geometrical mural drawn by members of the Art Club. The mural consists of a rectangle QRST and a semicircle PQBAT with centre P. SR = 1.84 m and \(\angle SPR = 72^\circ\). The club has allocated RM50 to decorate the mural. Sectors TPA and OPB will be painted green. A tube of paint costing RM6 will paint an area of 0.25 m2. Paint is only sold in whole tubes. The perimeter of the blue colored region will be taped with yellow ribbon. A roll of ribbon measuring 2m cost RM10. Ribbon is only sold in whole rolls. Using \(\pi = 3.142\), determine \(\angle APB\) in radians. Hence, determined whether the allocation is sufficient or not.Question 13
Given \(f(x) = kx^2 + 3kx + 10k^2\) and its maximum value is \(\frac{49}{4}\), work out the value of \(k\).
$$f(x) = \underbrace{k(x^2 + 3x + \left(\tfrac{3}{2}\right)^2)}_{\textrm{this bit can be ignored}} \underbrace{-\tfrac{9k}{4} + 10k^2}_{\substack{\textrm{this bit to be equated}\\ \textrm{with the given value}}}$$ $$-\tfrac{9k}{4} + 10k^2 = \tfrac{49}{4}$$ $$ 40k^2 -9k - 49 = 0$$The value of \(k\) is either \(k = -1\) or \(k = \frac{49}{40}\), which becomes painfully obvious once the aforementioned equation is factorized $$(40k-49)(k + 1) = 0$$ assuming, of course, one remembers how to factorize anything at all. However, one must discard the positive root, because the examiner very helpfully informed us that \(f(x)\) has a local maximum, as if we might have thought otherwise.
The graph of \(f(x) = -(x+\frac{3}{2})^2 + \frac{49}{4}\) should be symmetrical about \(x = -\frac{3}{2}\). We should explicitly show \(f(-6)=-8\) and \(f(3)=-8\) and the two x-intercepts \( f \begin{pmatrix}2\\-5\end{pmatrix} = \mathbf{0}\) and the y-intercept \(f(0) = 10\).
Question 14
We were presented with the innocent-looking function \(f(x) = x + 3\), which seems perfectly harmless at first glance. But, in a move that can only be described as theatrical, the examiner then demanded that the student find \(f^{n}(x)\), thus transforming a simple exercise into a subtle descent into mild panic. One examinee, however, did not panic. Instead, she began calmly by writing down the first few iterations, as though this were all part of a sensible plan and not the beginning of something that might spiral out of control at any moment.
$$\begin{align} f^2(x) = f(f(x)) &= f(x + 3)\\ &= (x + 3) + 3\\ &= x + 3(\color{brown}{2}) \end{align} $$ $$\begin{align} f^3(x) = f^2(f(x)) &= f^2(x + 3)\\ &= (x + 3) + 3(1+1)\\ &= x + 3(\color{brown}{3}) \end{align} $$ $$\begin{align} f^4(x) = f^3(f(x)) &= f^3(x + 3)\\ &= (x + 3) + 3(2+1)\\ &= x + 3(\color{brown}{4}) \end{align} $$Thus she was able to convince at least herself that \(f^n(x) = x + 3n\), hence \(f^{10}(x) = x + 30\). As for the solution to problem of determining the identity of \(h(x)\) given \(h f^{10}(x) = (x-30)(x + 30)\), she merely wrote down the following:
$$h(x + 30) = (x - 30)(x + 30)$$ $$h(\chi) = (\chi - 60) \chi$$Hence \(h(x) = x(x - 60)\).
Question 15
A right-angled triangle is given, with base \(b\), height \(a\) and hypotenuse \(c\) and the student is asked to show that \(\sin^2 \theta + \cos^2 \theta = 1\). With Pythagoras's theorem as our starting point, we can write \(c^2 = a^2 + b^2\). We then divide the equation by \(c^2\) to form \(\tfrac{c^2}{c^2} = \tfrac{a^2}{c^2} +\tfrac{b^2}{c^2}\) or
$$1= (\tfrac{a}{c})^2 + (\tfrac{b}{c})^2$$Since \(\sin \theta = \frac{a}{c}\) and \(\cos \theta = \frac{b}{c}\), we must have
$$1= (\sin \theta)^2 + (\cos \theta)^2$$The student is then instructed to demonstrate the following awful, grisly, grotesque, and hideous-looking identity.
$$\frac{\sin^2 \theta + \cos \theta + 1}{2 - \cos \theta} = \frac{\sec \theta + 1}{\sec \theta}$$One can sneekily introduce the magical zero-term (\(0 = \cos^2 \theta - \cos^2\theta\)) in the numerator, so that the sinful term in the numerator can be annihilated.
$$ \begin{align} \tfrac{\sin^2 \theta + \cos \theta + 1}{2 - \cos \theta} &= \tfrac{\sin^2 \theta + \color{brown}{(\cos^2 \theta - \cos^2\theta)} + \cos \theta + 1}{2 - \cos \theta}\\ &= \tfrac{\color{brown}{(\sin^2 \theta + \cos^2 \theta)} - \cos^2\theta + \cos \theta + 1}{2 - \cos \theta}\\ &= \tfrac{2 - \cos^2\theta + \cos \theta}{2 - \cos \theta}\\ &= \tfrac{2 + \cos \theta - \cos^2\theta }{2 - \cos \theta}\\ &= \tfrac{(2 - \cos\theta)(1 + \cos\theta)}{2 - \cos \theta}\\ &= 1 + \cos\theta \\ &= \tfrac{\left(1 + \cos\theta\right) \times \frac{1}{\cos\theta}}{\frac{1}{\cos\theta}} \\ &= \tfrac{\sec \theta + 1}{\sec \theta} \end{align} $$- It is not, strictly speaking, compulsory to reduce every equation to the comforting shape \(y = mx + c\), despite this being the only form some students appear to trust. They will usually rewrite the first equation as
$$y = 4x - \tfrac{1}{2}$$
because it looks familiar and therefore feels correct. They then compare the mysterious \(m\)-bits of the equations and announce, with relief, that \( 4 = 6p\), as though the numbers had always intended this outcome. A more convenient approach is to notice the central idea, which is that equations of the form
$$
\begin{array}
\mathbf{a} \cdot \mathbf{x} = b_1\\
\mathbf{a} \cdot \mathbf{x} = b_2\\
\vdots\\
\mathbf{a} \cdot \mathbf{x} = b_m
\end{array}
$$
all describe parallel objects in \(n\)-dimensional space. This avoids the need to rearrange anything at all, which is helpful, because rearranging equations does not actually make them understand themselves any better.
- I apologize for my earlier sarcastic ramblings. For this question, you may need to summon the most fearsome of mathematical beasts: \(y = mx + c\), as if it were a dragon in a spreadsheet.
- The examiner, in a display of smugness that could only have been perfected over years of marking, warned the student not to list out all the terms of the sequence. Presumably, he wants the student to obediently deploy the sacred formulas
$$
\begin{array}{l}
T_n = a + (n-1)d\\
S_n = \tfrac{n[2a + (n-1)d]}{2}
\end{array}
$$
as if summoning them by rote somehow counts as thinking. However, if the student instead reframes the whole ordeal as a coordinate geometry problem, the examiner—so smug, so confident in his spreadsheet of sequences—will almost certainly experience mild existential fury.
- If you are the sort of person who clings desperately to the old ways, like a historian hoarding quills, then you may proceed with the following manipulations. Use them as if each step were carved into stone tablets by mathematicians long dead.
$$ \tfrac{p}{r} = \tfrac{3(3 + \sqrt{2})}{3(1 + \sqrt{2})} = \tfrac{ 3 + \sqrt{2} }{1 + \sqrt{2}} \tfrac{1 - \sqrt{2}}{1 - \sqrt{2}}$$
$$ \tfrac{p}{r} = \tfrac{ (3 + \sqrt{2})(1 - \sqrt{2})}{1-2} = \tfrac{ 3 -3\sqrt{2} + \sqrt{2} - 2 }{-1} $$
$$ p = -r(1 -2\sqrt{2}) = -r + 2r\sqrt{2} $$
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- The algebraic solution to the problem is simply
$$2\mathbf{x} - \mathbf{y} = 2 \tbinom{1}{1} - \tbinom{2}{-1} = \tbinom{0}{3} $$
- The so-called angles complémentaires is a French invention that somehow migrated into English like a tourist who doesn’t read the map. Euclid? Never heard of it. In his Proposition 13, he didn’t bother with cute labels. No, he went full geometric pragmatist:
ἐὰν εὐθεῖα ἐπʼ εὐθεῖαν σταθεῖσα γωνίας ποιῇ, ἤτοι δύο ὀρθὰς ἢ δυσὶν ὀρθαῖς ἴσας ποιήσει.
Euclid's words, when translated to English, are: If a straight line set up on a straight line make angles, it will make either two right angles or angles equal to two right angles. Why use clear, simple words when you can dress it up in Greek-to-Latin-to-English bureaucratic prose and then let centuries of textbooks invent angles complémentaires? Euclid is just sitting there, lines and right-angles behaving perfectly fine, while later mathematicians are inventing terminology such as complementary, supplementary, and explementary angles.
- When computing power is plentiful, this calculus trick should be firmly returned to the museum of unnecessary cleverness, right between the astrolabe and doing long division for fun. It is, after all, far more work than simply calculating the change directly, like a normal person with a calculator and self-respect:
$$\tfrac{\pi h}{3} (3(1 + \tfrac{0.5}{100}))^2 - \tfrac{\pi h}{3} (3)^2 = \tfrac{1203\pi h}{4000}$$
This reveals—after all that drama—a volume increase of 1.0025%, which is exactly the sort of number your picky boss demands, presumably because rounding is how civilization collapses.
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- One student, however, decided that the proper way to solve the problem was obviously to employ the trick he had memorized from a vaguely similar problem, because memory is clearly more trustworthy than logic. With great confidence, he wrote:
$$\frac{^{(5+4)}\!P_4}{4!} = 126$$
A perfect example of the ancient art of applying a trick to the wrong place, without pausing to notice that it makes about as much sense as using a spoon to hammer a nail. A simple check immediately reveals the absurdity: 126 cannot possibly be correct since it is less than 360, because the maximum number of four-digit numbers from the given set is \(6^4=1296\), and when repetition is disallowed it drops to \(^6\!P_{4}=360\). But why let facts get in the way of a good trick? The count required by examiner can also be concisely written as follows
$$\sum_{k=0}^{D} \tbinom{D}{k}\, {}^N\!P_{D-k} $$
where \(D\) is the number of digits in the number to be formed, \(N\) is the number of digits that can only be employed once. In general, the sum can be written as
$$\left({}^N\!P_{D} \right) {}_1\!F_1(-D;N-D+1;-1)$$
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- Solving this by completing the square is not always a sight for sore eyes. In this particular instance, it is positively long-winded, tediously complicated, and likely to inspire small, avoidable errors. In short, it is the mathematical equivalent of watching someone try to untangle a necklace while wearing boxing gloves. $$-\tfrac{9k}{4} + 10k^2 = \tfrac{49}{4}$$ $$ k^2 -\tfrac{9k}{40} + \left(\tfrac{9}{80}\right)^2 = \tfrac{49}{40} + \left(\tfrac{9}{80}\right)^2$$ $$ \begin{align} \left(k- \tfrac{9}{80}\right)^2 &= \tfrac{49}{40} + \left(\tfrac{9}{80}\right)^2\\ &= \tfrac{49}{40} + \tfrac{9/2}{40}\tfrac{9/2}{40}\\ &= \tfrac{1}{40}\left(49 + \tfrac{81}{160} \right)\\ &= \tfrac{49\times160 + 81}{40\times 160} \\ k-\tfrac{9}{80}&= \pm \sqrt{ \tfrac{89^2}{10^2 \times 2^2\times 4^2} } = \pm \tfrac{89}{80}\\ k &= -1, \tfrac{49}{40} \end{align} $$
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- The teacher who set this question appears to have stopped halfway through setting it. He (I think the teacher must be a man) did not specify where the students were meant to begin, presumably because he assumed they would all instinctively know to start with Pythagoras’ theorem, as one does. This is awkward, because if Pythagoras’ theorem has not yet been granted by the universe, then it cannot simply be summoned like a trusted old friend. In that case, some sort of elementary proof must be performed first, ideally without alarming anyone. The simplest proof of Pythagoras’ theorem involves a diagram. This is fortunate, because diagrams are traditionally how mathematicians indicate that something obvious is about to become extremely difficult.
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