Chapter 11 — The Challenge Set

Answer: A

Explanation

Step 1 — Rationalise each fraction separately.

Multiply each one, top and bottom, by the conjugate of its own denominator. The conjugate of $\sqrt{3} - \sqrt{2}$ is $\sqrt{3} + \sqrt{2}$, and vice versa:

$$\frac{1}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{3} + \sqrt{2}}{3 - 2} = \sqrt{3} + \sqrt{2}$$

$$\frac{1}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2}$$

Both denominators become $3 - 2 = 1$, which is why this question is doable without a calculator at all.

Step 2 — Add them.

$$(\sqrt{3} + \sqrt{2}) + (\sqrt{3} - \sqrt{2}) = 2\sqrt{3}$$

The $\sqrt{2}$ terms cancel because one fraction produced $+\sqrt{2}$ and the other produced $-\sqrt{2}$.

Why the distractors are wrong:

B) $2\sqrt{2}$ subtracts the two rationalised fractions instead of adding them. The $\sqrt{3}$ terms cancel and the $\sqrt{2}$ terms survive — the exact mirror image of the right answer, which is why it looks so plausible.

C) $2\sqrt{3} + 2\sqrt{2}$ uses $\sqrt{3} + \sqrt{2}$ as the conjugate for both fractions. Each fraction has its own denominator and therefore its own conjugate; using one twice leaves the second denominator as $(\sqrt{3}+\sqrt{2})^2 = 5 + 2\sqrt{6}$, which is still irrational.

D) $\dfrac{\sqrt{3}}{3}$ comes from adding the denominators: $\dfrac{2}{2\sqrt{3}} = \dfrac{1}{\sqrt{3}}$. Fractions are never added by adding tops and bottoms.

Takeaway: Each fraction gets its own conjugate. When two conjugate surds appear as separate denominators, rationalise them one at a time and watch the middle terms cancel.

---

Answer: C

Explanation

Step 1 — See the quadratic.

$3^{2x}$ is $\left(3^x\right)^2$, because $a^{2x} = \left(a^x\right)^2$. Put $y = 3^x$:

$$y^2 - 4y + 3 = 0 \quad\Longrightarrow\quad (y - 1)(y - 3) = 0 \quad\Longrightarrow\quad y = 1 \text{ or } y = 3$$

Step 2 — Go back to $x$. This is the step the question is testing.

$y$ is not the answer. Convert each value back:

$$3^x = 1 \;\Longrightarrow\; x = 0 \qquad\qquad 3^x = 3 \;\Longrightarrow\; x = 1$$

Step 3 — Add them. $0 + 1 = 1$.

Note that both values of $y$ were usable here. Had one come out negative, it would have had to be discarded, because $3^x$ is positive for every real $x$ — always check that before converting back.

Why the distractors are wrong:

A) $0$ is one of the two solutions on its own. The question asks for the sum, and a sum of two things needs both of them.

B) $3$ is a value of $y$, not of $x$. It is the answer to "what is $3^x$?", not to "what is $x$?".

D) $4$ is $1 + 3$, the sum of the two $y$-values. That is the sum of the substitution, not of the solutions — and it equals the coefficient of $y$, which makes it feel like it must mean something.

Takeaway: When a substitution turns an equation into a quadratic, solving the quadratic is the middle of the work, not the end. Write down what the letter you introduced actually stands for, and convert every root back before answering.

---

Answer: B

Explanation

Step 1 — Break 12 into prime factors.

$$12 = 4 \times 3 = 2^2 \times 3$$

The whole question is this line. Everything after it is one law applied twice.

Step 2 — Apply the log laws.

The log of a product is the sum of the logs, and the log of a power brings the power down in front:

$$\log_a 12 = \log_a\left(2^2 \times 3\right) = \log_a 2^2 + \log_a 3 = 2\log_a 2 + \log_a 3$$

Step 3 — Substitute. $\log_a 2 = p$ and $\log_a 3 = q$, so the answer is $2p + q$.

Check it in a base you know. Take $a = 10$: $p \approx 0.301$, $q \approx 0.477$, and $2p + q \approx 1.079$. Since $10^1 = 10$ and $10^2 = 100$, $\log_{10} 12$ must be a little over $1$. ✓

Why the distractors are wrong:

A) $p + q$ is $\log_a 6$, not $\log_a 12$. It comes from splitting 12 as $2 \times 6$ and then forgetting that the 6 breaks up further, or from splitting it as $3 \times 4$ and reading $\log_a 4$ as $p$.

C) $2pq$ multiplies the two logarithms. The law is $\log(mn) = \log m + \log n$ — a product inside becomes a sum outside. Multiplying logs corresponds to nothing at all.

D) $p + 2q$ is $\log_a 18$, because $18 = 2 \times 3^2$. It is the right shape with the wrong number squared. Which one gets doubled is decided by the prime factorisation, so factorise first and the ambiguity disappears.

Takeaway: Factorise the number into primes before touching a log law. $\log_a(2^m 3^n) = mp + nq$, and the exponents in the factorisation are the coefficients in the answer.

---

Answer: D

Explanation

Step 1 — Find the cycle.

Last digits of powers of 7 repeat. Work out the first few and watch:

$$7^1 = 7 \qquad 7^2 = 49 \qquad 7^3 = 343 \qquad 7^4 = 2401 \qquad 7^5 = 16\,807$$

The last digits run $7, 9, 3, 1$, then $7$ again. The cycle has length 4.

Step 2 — Find where 2026 sits in the cycle.

Divide the exponent by the cycle length and keep the remainder:

$$2026 = 4 \times 506 + 2$$

A remainder of $2$ means $7^{2026}$ ends in the same digit as $7^2$, which is $9$.

The part that goes wrong is the counting, not the division. Remainder $1$ means the first entry, remainder $2$ the second, remainder $3$ the third — and remainder $0$ means the last, because a remainder of zero puts you at the end of a complete cycle, not at the start of a new one.

Why the distractors are wrong:

A) $7$ is the first entry in the cycle, which would be right if the remainder were $1$. But $2026$ is even, and $2025$ is the multiple of $4$ plus one.

B) $1$ is the fourth entry, which would be right if $4$ divided $2026$ exactly. It does not: $2026 \div 4$ leaves $2$ over. Only exponents ending in a multiple of 4 land here.

C) $3$ is the third entry, from a remainder of $3$.

Takeaway: Powers of any digit cycle their last digit, usually with period 1, 2 or 4. Write out the cycle, divide the exponent by its length, and read the remainder as a position — with remainder 0 meaning the last position, not the first.

---

Answer: A

Explanation

Step 1 — Turn "touches at one point" into algebra.

A parabola that touches the $x$-axis has exactly one root, which means the two roots have collapsed into each other. That happens precisely when the discriminant is zero:

$$\Delta = b^2 - 4ac = 0$$

Step 2 — Substitute. Here $a = 1$, $b = k$, $c = 9$:

$$k^2 - 4(1)(9) = 0 \quad\Longrightarrow\quad k^2 = 36 \quad\Longrightarrow\quad k = 6 \text{ or } k = -6$$

Step 3 — Use the condition you were given. $k > 0$, so $k = 6$.

Check: $x^2 + 6x + 9 = (x+3)^2$, which touches the axis at $x = -3$ and nowhere else. ✓ The $k > 0$ line is in the question for a reason — without it there are two answers, and a multiple-choice question cannot have two.

Why the distractors are wrong:

B) $3$ is $\sqrt{9}$ — the square root of the constant term rather than of $4ac$. It is what you get by setting $k^2 = c$ instead of $k^2 = 4ac$, and $x^2 + 3x + 9$ has discriminant $9 - 36 = -27$, so it never reaches the axis at all.

C) $9$ simply repeats the constant term.

D) $36$ is $k^2$, the value found one line before the end. The equation solves to $k^2 = 36$, and stopping there is the single most common way to lose this question.

Takeaway: One root means $\Delta = 0$; two roots means $\Delta > 0$; none means $\Delta < 0$. Solve for $k^2$, then remember to take the square root — and read the question for the condition that tells you which sign to keep.

---

Section 11.2 — Trigonometry Without a Diagram

Two questions, neither with a picture. Both are decided by an identity and by counting carefully.

---

Answer: B

Explanation

Step 1 — Square both sides.

There is no way to get a product out of a sum directly, but squaring produces one:

$$(\sin\theta + \cos\theta)^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta$$

Step 2 — Use the identity. $\sin^2\theta + \cos^2\theta = 1$, so the right-hand side collapses to

$$1 + 2\sin\theta\cos\theta$$

Step 3 — Set the two sides equal and solve.

$$\left(\tfrac{1}{2}\right)^2 = 1 + 2\sin\theta\cos\theta \quad\Longrightarrow\quad \tfrac{1}{4} - 1 = 2\sin\theta\cos\theta$$

$$-\tfrac{3}{4} = 2\sin\theta\cos\theta \quad\Longrightarrow\quad \sin\theta\cos\theta = -\tfrac{3}{8}$$

Should the answer be negative? Yes, and it is worth seeing why. In the first quadrant $\sin\theta$ and $\cos\theta$ are both positive, and their sum there is always at least $1$. A sum of only $\tfrac{1}{2}$ therefore forces one of them to be negative, and their product must be negative too.

Why the distractors are wrong:

A) $\dfrac{1}{4}$ is $\left(\tfrac{1}{2}\right)^2$ — the square of what you were given, with the identity never applied. It is the first thing you write down and the last thing that is an answer.

C) $\dfrac{3}{8}$ is the right size with the sign lost, from writing $1 - \tfrac{1}{4}$ instead of $\tfrac{1}{4} - 1$.

D) $-\dfrac{3}{4}$ is $2\sin\theta\cos\theta$, one halving short. The identity leaves a 2 in front, and it has to be divided out.

Takeaway: $(\sin\theta \pm \cos\theta)^2 = 1 \pm 2\sin\theta\cos\theta$. Squaring a sum of sine and cosine is the standard route to their product — and the answer is always $\dfrac{(\text{sum})^2 - 1}{2}$, halved and signed.

---

Answer: D

Explanation

Step 1 — Factorise it as a quadratic in $\cos x$.

$$2\cos^2 x - \cos x - 1 = 0 \quad\Longrightarrow\quad (2\cos x + 1)(\cos x - 1) = 0$$

so $\cos x = -\dfrac{1}{2}$ or $\cos x = 1$.

Step 2 — Count the angles for each value separately. This is where the question is won or lost, because the two values do not give the same number of angles.

$\cos x = -\dfrac{1}{2}$: cosine is negative in the second and third quadrants. The reference angle is $60°$, giving $x = 180° - 60° = 120°$ and $x = 180° + 60° = 240°$. That is two angles.

$\cos x = 1$: cosine reaches $1$ only at $x = 0°$ within the range, since $360°$ is excluded. That is one angle.

Step 3 — Add. $2 + 1 = 3$ values: $0°$, $120°$ and $240°$.

Why the distractors are wrong:

A) $1$ solves only $\cos x = 1$ and drops the other factor. A product is zero when either factor is zero, and both have to be followed through.

B) $2$ solves only $\cos x = -\tfrac{1}{2}$ and drops $\cos x = 1$. It is the more tempting half to keep, because it is the one that needs the quadrant work.

C) $4$ assumes each value of $\cos x$ gives two angles. Usually it does — but $\cos x = 1$, $\cos x = -1$, $\sin x = 1$ and $\sin x = -1$ are the maximum and minimum, and each of those is reached only once per revolution.

Takeaway: Factorise, then count each root's angles on its own. Two per revolution is the normal case; a value of exactly $+1$ or $-1$ gives one, and a value outside $[-1, 1]$ gives none.

---

Section 11.3 — Geometry and Money

---

Answer: C

Explanation

Step 1 — Read the circle.

$x^2 + y^2 = 20$ is centred at the origin with radius $r = \sqrt{20} = 2\sqrt{5}$. Note that $20$ is $r^2$, not $r$.

Step 2 — Say what "tangent" means.

A line is a tangent exactly when its perpendicular distance from the centre equals the radius. Closer and it cuts the circle twice; further and it misses.

Step 3 — Use the distance formula.

Write the line in the form $ax + by + c_0 = 0$: from $y = 2x + c$ we get $2x - y + c = 0$. The distance from the origin is

$$\frac{|2(0) - 1(0) + c|}{\sqrt{2^2 + (-1)^2}} = \frac{|c|}{\sqrt{5}}$$

Step 4 — Set it equal to the radius.

$$\frac{|c|}{\sqrt{5}} = 2\sqrt{5} \quad\Longrightarrow\quad |c| = 2\sqrt{5} \times \sqrt{5} = 2 \times 5 = 10$$

Taking the positive value, $c = 10$. (The other tangent with gradient 2 is $y = 2x - 10$, on the far side of the circle.)

Why the distractors are wrong:

A) $4\sqrt{5}$ divides by $2$ instead of by $\sqrt{5}$ — using only the coefficient of $x$ under the root and forgetting the $-1$ from the $y$ term. The denominator is $\sqrt{a^2 + b^2}$ and both coefficients belong in it.

B) $2\sqrt{5}$ is the radius. It is the number you need in the middle of the work, not at the end.

D) $20$ treats the right-hand side of the circle's equation as the radius. In $x^2 + y^2 = r^2$ the number on the right is the radius squared.

Takeaway: Tangent means "distance from the centre equals the radius". Put the line in the form $ax + by + c_0 = 0$, use $\dfrac{|ax_0 + by_0 + c_0|}{\sqrt{a^2+b^2}}$, and remember that $x^2 + y^2 = k$ has radius $\sqrt{k}$.

---

Answer: A

Explanation

Step 1 — Investment P, simple interest.

Simple interest is calculated on the original amount every year, so it is the same amount each year:

$$A = P(1 + in) = 10\,000\,(1 + 0.10 \times 2) = 10\,000 \times 1.2 = \text{R}12\,000$$

Step 2 — Investment Q, compound interest.

$$A = P(1 + i)^n = 10\,000\,(1.1)^2 = 10\,000 \times 1.21 = \text{R}12\,100$$

$1.1^2 = 1.21$ is worth knowing by heart; it saves the multiplication.

Step 3 — Subtract. $12\,100 - 12\,000 = \text{R}100$.

Where that R100 comes from. In year 2, compound interest pays 10% on the R1 000 of interest earned in year 1, and $10\%$ of $1\,000$ is exactly R100. Over two years that is the whole difference between the two methods — which is why compounding looks unimpressive over short periods and overwhelming over long ones.

Why the distractors are wrong:

B) R1 000 is one year's interest on the original amount, under either method. It is not a difference between them.

C) R2 100 is the total interest Q earns. The question asks how much more Q earns, so P's R2 000 has to come off.

D) R200 is 10% of the R2 000 of simple interest. The extra interest is earned on the first year's R1 000 only — the second year's interest has not been sitting in the account long enough to earn anything.

Takeaway: Simple interest is $P(1+in)$; compound is $P(1+i)^n$. When a question asks how much more, compute both totals in full and subtract — and expect the gap to be small over two years and large over twenty.

---

Answer: D

Explanation

Step 1 — Turn "at least one" into its complement.

"At least one red" covers red-then-blue, blue-then-red and red-then-red: three cases. Its opposite is a single case — no red at all, meaning both are blue. Always take the complement when "at least one" appears.

$$P(\text{at least one red}) = 1 - P(\text{no red})$$

Step 2 — Work out $P(\text{both blue})$, without replacement.

There are $3 + 5 = 8$ balls. The first draw is blue with probability $\tfrac{5}{8}$. That ball is not put back, so for the second draw there are 4 blue balls left out of 7:

$$P(\text{both blue}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$$

Both numbers drop — the numerator because a blue ball left, and the denominator because a ball left.

Step 3 — Subtract.

$$1 - \frac{5}{14} = \frac{9}{14}$$

Why the distractors are wrong:

A) $\dfrac{5}{14}$ is $P(\text{no red})$ — the complement, computed correctly and then not subtracted from 1.

B) $\dfrac{3}{8}$ is the probability that the first ball is red. It ignores the second draw entirely, and it is smaller than the right answer because it counts none of the ways the red one comes second.

C) $\dfrac{39}{64}$ is the answer with replacement: $1 - \left(\tfrac{5}{8}\right)^2$. Notice it is smaller than $\tfrac{9}{14}$. Removing a blue ball makes the next one likelier to be red, so drawing without replacement improves your chances here.

Takeaway: "At least one" means compute the complement. Without replacement, reduce the top and the bottom for the second draw, and write the two fractions side by side before multiplying so the drop is visible.

---

Answer: B

Explanation

Step 1 — Know which formula "reducing balance" means.

Reducing balance takes the percentage off whatever the machine is worth now, so each year removes a smaller amount than the one before:

$$A = P(1 - i)^n$$

(Straight-line depreciation, $A = P(1 - in)$, takes the same amount off every year. The words in the question decide which one you use, and they are the only thing that does.)

Step 2 — Substitute.

$$A = 120\,000\,(1 - 0.25)^2 = 120\,000 \times (0.75)^2 = 120\,000 \times 0.5625$$

Step 3 — Do it without a calculator. $0.75^2 = \tfrac{9}{16}$, so

$$120\,000 \times \frac{9}{16} = 7\,500 \times 9 = \text{R}67\,500$$

Or year by year: $120\,000 \to 90\,000 \to 67\,500$. The first year removes R30 000 and the second removes only R22 500, which is reducing balance in one line.

Why the distractors are wrong:

A) R60 000 is straight-line depreciation: $25\% \times 2 = 50\%$ off the original. It is always lower than reducing balance, because reducing balance charges the second year's 25% against a smaller amount.

C) R90 000 is the value after one year. Correct arithmetic, one year short.

D) R112 500 squares the rate instead of the factor: $120\,000\,(1 - 0.25^2)$. Squaring $0.25$ gives $0.0625$, so this removes barely 6% over two years. The exponent belongs on $(1 - i)$ as a whole, not on $i$.

Takeaway: Reducing balance is $P(1-i)^n$ and straight line is $P(1-in)$. The power sits on the whole bracket. If you are unsure, step through the years one at a time — two lines of arithmetic settle it.

---

Section 11.4 — Sequences, Calculus and Spread

---

Answer: A

Explanation

Step 1 — Say what $S_n$ and $T_n$ each mean.

$S_{10}$ is the total of the first ten terms. $T_{10}$ is the tenth term on its own. They are not the same thing, and the whole question is keeping them apart.

Step 2 — Use the relationship between them.

Everything in $S_{10}$ except the tenth term is already in $S_9$, so:

$$T_{10} = S_{10} - S_9$$

Step 3 — Compute both sums.

$$S_{10} = 3(10)^2 - 2(10) = 300 - 20 = 280$$ $$S_9 = 3(9)^2 - 2(9) = 243 - 18 = 225$$ $$T_{10} = 280 - 225 = 55$$

A shortcut worth having. Doing the subtraction in general gives $T_n = S_n - S_{n-1} = 6n - 5$, which is arithmetic with a constant difference of 6. Then $T_{10} = 60 - 5 = 55$ ✓, and you can check any other term instantly.

Why the distractors are wrong:

B) $280$ is $S_{10}$ — the sum of the first ten terms, which is what you get by substituting $n = 10$ into the formula as given and stopping. It is the answer to a question that was not asked.

C) $49$ is $T_9 = S_9 - S_8$, one term early. This comes from subtracting the wrong pair — writing $S_9 - S_8$ while thinking "ninth and tenth".

D) $61$ is $T_{11} = S_{11} - S_{10}$, one term late, from using $S_{11}$ as the upper sum.

Takeaway: $T_n = S_n - S_{n-1}$. When a question hands you a formula for $S_n$ and asks for a term, subtract the previous sum — and if you will need more than one term, derive the general $T_n$ once and use it.

---

Answer: C

Explanation

Step 1 — Find $r$ from the sum to infinity.

$$S_\infty = \frac{a}{1 - r} \quad\Longrightarrow\quad \frac{8}{1 - r} = 12$$

Cross-multiply carefully — this is where the question catches people:

$$8 = 12(1 - r) \quad\Longrightarrow\quad 1 - r = \frac{8}{12} = \frac{2}{3} \quad\Longrightarrow\quad r = \frac{1}{3}$$

Note that $|r| = \tfrac{1}{3} < 1$, so the series does converge and the formula was allowed. Always confirm that.

Step 2 — Use the term formula.

$$T_n = ar^{n-1} \quad\Longrightarrow\quad T_4 = 8\left(\tfrac{1}{3}\right)^3 = \frac{8}{27}$$

The exponent is $n - 1$, not $n$: the first term has $r$ to the power zero.

Why the distractors are wrong:

A) $\dfrac{8}{9}$ is $8r^2$, which is $T_3$. It comes from writing $T_4 = ar^{4-2}$ or simply counting the multiplications one short.

B) $-1$ comes from inverting the cross-multiplication: reading $\dfrac{8}{1-r} = 12$ as $1 - r = \dfrac{12}{8}$, which gives $r = -\tfrac{1}{2}$ and $T_4 = 8\left(-\tfrac{1}{2}\right)^3 = -1$. A negative $r$ here should be a warning: it would make the series alternate, and $8 - 4 + 2 - 1 + \dots$ sums to $\tfrac{16}{3}$, not 12.

D) $\dfrac{8}{81}$ is $8r^4$, which is $T_5$ — the same off-by-one in the other direction, from using $r^n$ instead of $r^{n-1}$.

Takeaway: $S_\infty = \dfrac{a}{1-r}$ rearranges to $1 - r = \dfrac{a}{S_\infty}$ — the first term over the sum, not the other way up. Then $T_n = ar^{n-1}$, and the fourth term carries $r^3$.

---

Answer: D

Explanation

Step 1 — Find the point of contact.

$$y = (1)^3 - 4(1) = -3$$

so the tangent touches the curve at $(1, -3)$.

Step 2 — Find the gradient there.

$$\frac{dy}{dx} = 3x^2 - 4 \quad\Longrightarrow\quad m = 3(1)^2 - 4 = -1$$

A negative gradient — the curve is falling at $x = 1$, even though $x^3$ dominates further out.

Step 3 — Write the tangent's equation.

$$y - y_1 = m(x - x_1) \quad\Longrightarrow\quad y + 3 = -1(x - 1) \quad\Longrightarrow\quad y = -x - 2$$

Step 4 — Set $y = 0$.

$$0 = -x - 2 \quad\Longrightarrow\quad x = -2$$

Why the distractors are wrong:

A) $1$ is the $x$-coordinate of the point of contact — where the tangent touches the curve, not where it meets the axis.

B) $-3$ is the $y$-coordinate of the point of contact. It is also the tangent's $y$-intercept... no it is not: the tangent's $y$-intercept is $-2$. $-3$ is simply the height of the touching point, and it answers no part of the question.

C) $4$ is where the normal crosses the axis. The normal at that point has gradient $+1$ (the negative reciprocal of $-1$), giving $y = x - 4$ and an intercept at $x = 4$. Tangent and normal are easy to swap when the gradient is $\pm 1$, because the arithmetic looks almost identical.

Takeaway: Point, then gradient, then equation, then intercept — four steps, and each one produces a number that is not the answer. Write down what each number is as you get it.

---

Answer: B

Explanation

Step 1 — Set up the area with one variable.

The parabola is symmetric about the $y$-axis, so the rectangle is too. Call the right-hand corner $x = w$. Then the left-hand corner is at $-w$ and:

• the width is from $-w$ to $w$, which is $2w$ — not $w$
• the height is the value of the parabola there, $12 - w^2$

$$A = 2w\left(12 - w^2\right) = 24w - 2w^3$$

Step 2 — Differentiate and set to zero.

$$\frac{dA}{dw} = 24 - 6w^2 = 0 \quad\Longrightarrow\quad w^2 = 4 \quad\Longrightarrow\quad w = 2$$

(taking the positive root, since $w$ is a length)

Step 3 — Substitute back into the AREA, not into the derivative.

$$A = 2(2)\left(12 - 4\right) = 4 \times 8 = 32$$

So the rectangle is 4 wide and 8 tall.

Why the distractors are wrong:

A) $16$ uses $w$ as the whole width instead of half of it: $2 \times 8 = 16$. The parabola's symmetry is what makes the width $2w$, and it is the easiest feature of the diagram to miss when there is no diagram.

C) $8$ is the height of the rectangle, $12 - 2^2$. It is a real quantity in the problem and not the one asked for.

D) $48$ is $24w$ at $w = 2$ — the first term of the area expression with the $-2w^3$ dropped. It is also, revealingly, larger than the true maximum, which is impossible for any actual rectangle.

Takeaway: In an optimisation question, write the quantity in one variable before differentiating, and check what the variable measures — half a width behaves very differently from a whole one. Substitute the stationary value back into the original quantity, never into the derivative.

---

Answer: C

Explanation

Step 1 — Establish the similarity.

$DE \parallel BC$ makes $\angle ADE = \angle ABC$ and $\angle AED = \angle ACB$ (corresponding angles), and angle $A$ is shared. So $\triangle ADE \;|||\; \triangle ABC$.

Step 2 — Get the right length ratio. This is the trap.

You are given $AD : DB = 2 : 3$ — part to part. Similarity compares $AD$ to the whole side $AB$:

$$AB = AD + DB = 2 + 3 = 5 \quad\Longrightarrow\quad AD : AB = 2 : 5$$

Step 3 — Square it for areas.

In similar figures, areas are in the ratio of the squares of corresponding lengths:

$$\frac{\text{area } ADE}{\text{area } ABC} = \left(\frac{2}{5}\right)^2 = \frac{4}{25}$$

$$8 = \frac{4}{25} \times \text{area } ABC \quad\Longrightarrow\quad \text{area } ABC = 8 \times \frac{25}{4} = 50$$

Step 4 — Subtract to get the trapezium.

The trapezium is what is left of the triangle after $ADE$ is removed:

$$50 - 8 = 42 \text{ cm}^2$$

Why the distractors are wrong:

A) $12$ scales the area by the length ratio instead of its square: $8 \times \tfrac{5}{2} = 20$, then $20 - 8 = 12$. Lengths scale by $k$, areas by $k^2$, volumes by $k^3$.

B) $50$ is the area of the whole triangle $ABC$. Correct, and one subtraction short of the answer.

D) $10$ uses $\tfrac{2}{3}$ as the length ratio — the ratio given in the question, taken straight from the page. $AD : DB$ compares two parts; similarity needs $AD : AB$, part to whole.

Takeaway: Convert a part-to-part ratio into part-to-whole before using it in a similarity, then square it for areas. And when the question asks for a trapezium or any other leftover region, finish with the subtraction.

---

Answer: A

Explanation

There is no calculator, and a common denominator of $11 \times 9 \times 14 \times 7$ is not a serious plan. Use the structure instead.

The method: measure each one's distance from 1.

Every fraction here is just under 1. Subtract each from 1 and the comparison inverts — the fraction with the smallest shortfall is the largest fraction:

$$1 - \frac{9}{11} = \frac{2}{11} \qquad 1 - \frac{7}{9} = \frac{2}{9} \qquad 1 - \frac{11}{14} = \frac{3}{14} \qquad 1 - \frac{5}{7} = \frac{2}{7}$$

Now compare the shortfalls. Three of them have numerator 2, and with equal numerators the one with the biggest denominator is smallest:

$$\frac{2}{11} < \frac{2}{9} < \frac{2}{7}$$

That leaves $\tfrac{3}{14}$. Compare it with $\tfrac{2}{11}$ by cross-multiplying: $3 \times 11 = 33$ against $2 \times 14 = 28$, so $\tfrac{3}{14} > \tfrac{2}{11}$.

So $\tfrac{2}{11}$ is the smallest shortfall, and $\dfrac{9}{11}$ is the largest fraction.

Check with decimals: $\tfrac{9}{11} \approx 0.818$, $\tfrac{11}{14} \approx 0.786$, $\tfrac{7}{9} \approx 0.778$, $\tfrac{5}{7} \approx 0.714$. ✓

Why the distractors are wrong:

B) $\dfrac{7}{9}$ falls $\tfrac{2}{9}$ short of 1, and $\tfrac{2}{9} > \tfrac{2}{11}$. It is a tempting choice because 7 and 9 are the smallest numbers on offer, and small numbers feel like a large fraction.

C) $\dfrac{11}{14}$ is the runner-up, $\tfrac{3}{14}$ short. It is the only one whose shortfall does not have numerator 2, which is exactly why it needs the cross-multiplication rather than a glance.

D) $\dfrac{5}{7}$ falls $\tfrac{2}{7}$ short — the largest gap of the four, so the smallest fraction.

Takeaway: To compare fractions close to 1, compare their distances from 1 instead: smallest gap wins. With equal numerators, the larger denominator gives the smaller fraction. Cross-multiplication settles any pair the shortcuts cannot.

---

Answer: B

Explanation

Step 1 — Recognise both patterns.

The numerator is a difference of cubes, since $8 = 2^3$:

$$a^3 - b^3 = (a - b)\left(a^2 + ab + b^2\right) \quad\Longrightarrow\quad x^3 - 8 = (x - 2)\left(x^2 + 2x + 4\right)$$

The denominator is a difference of squares:

$$x^2 - 4 = (x - 2)(x + 2)$$

Step 2 — Cancel the common factor.

$$\frac{(x - 2)\left(x^2 + 2x + 4\right)}{(x - 2)(x + 2)} = \frac{x^2 + 2x + 4}{x + 2}$$

Only $(x - 2)$ is common. Nothing else cancels, and the answer stays a fraction.

Getting the signs right. In $a^3 - b^3$ the first bracket carries the minus sign and every sign in the trinomial is plus. In $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ it is the other way about. The short version: the binomial keeps the sign, the trinomial flips it in the middle.

Why the distractors are wrong:

A) $x + 2$ cancels $x^2 + 2x + 4$ against $x + 2$. Those are not the same expression and neither is a factor of the other — cancelling is only allowed between identical factors, never between terms that merely look similar.

C) $\dfrac{x^2 + 2x + 4}{x - 2}$ cancels the wrong bracket in the denominator: it removes $(x + 2)$ although the factor shared with the numerator is $(x - 2)$.

D) $\dfrac{x^2 - 2x + 4}{x + 2}$ uses $x^2 - 2x + 4$, the trinomial belonging to $x^3 + 8$. Substituting $x = 0$ settles it: the original is $\tfrac{-8}{-4} = 2$, and this option gives $\tfrac{4}{2} = 2$ as well — so try $x = 1$: the original is $\tfrac{-7}{-3} = \tfrac{7}{3}$, and this option gives $\tfrac{3}{3} = 1$.

Takeaway: $a^3 \pm b^3$ factorises into a binomial times a trinomial, and the trinomial's middle sign is the opposite of the binomial's. When a rational expression will not simplify, check whether a cube is hiding behind a number like 8, 27, 64 or 125.

---

Answer: D

Explanation

Step 1 — Use the remainder theorem on $x - 1$.

Dividing by $x - 1$ leaves the remainder $p(1)$:

$$p(1) = 2 + a - 5 + b = 4 \quad\Longrightarrow\quad a + b - 3 = 4 \quad\Longrightarrow\quad a + b = 7$$

And that is the answer. The question asks for $a + b$, and the first equation gives it directly. Anyone who solves for $a$ and $b$ separately will get there too, but through considerably more work.

Step 2 — The second condition, for completeness.

$x + 2$ being a factor means $p(-2) = 0$:

$$2(-8) + 4a + 10 + b = 0 \quad\Longrightarrow\quad 4a + b = 6$$

Step 3 — Solve the pair, to check.

Subtracting $a + b = 7$ from $4a + b = 6$:

$$3a = -1 \quad\Longrightarrow\quad a = -\tfrac{1}{3}, \qquad b = 7 + \tfrac{1}{3} = \tfrac{22}{3}$$

And $a + b = -\tfrac{1}{3} + \tfrac{22}{3} = \tfrac{21}{3} = 7$ ✓ — the same answer, reached the long way.

Why the distractors are wrong:

A) $6$ is the right-hand side of the second equation, $4a + b = 6$. It is $4a + b$, not $a + b$, and the two coincide only when $a = 0$.

B) $-\dfrac{1}{3}$ is $a$ on its own.

C) $\dfrac{22}{3}$ is $b$ on its own. Both it and option B are correct values that answer a question nobody asked — and both are considerably uglier than the sum they add up to, which is a hint in itself.

Takeaway: Remainder theorem: dividing by $x - k$ leaves $p(k)$. Factor theorem: $x - k$ is a factor exactly when $p(k) = 0$. Before grinding out both unknowns, look at what the question asked for — a combination like $a + b$ often falls out of one equation.

---

Answer: C

Explanation

Step 1 — Separate the two operations. They do different things.

Standard deviation measures spread — how far the values sit from their own mean. So:

• Multiplying every value by 4 stretches the whole set, gaps included. The spread is multiplied by 4.
• Subtracting 6 slides every value down by the same amount. The gaps between them do not change at all, so the spread is untouched.

Step 2 — Apply them.

$$\text{new standard deviation} = 4 \times 3 = 12$$

The $-6$ contributes nothing.

See it on a small set. Take $5, 5, 11, 11$: the mean is 8 and each value is 3 from it, so the standard deviation is 3. Now apply $4v - 6$: the values become $14, 14, 38, 38$, the mean becomes $4(8) - 6 = 26$, and each value is 12 from the new mean. ✓

The general rule: if $y = mv + k$, then $\text{mean}(y) = m \times \text{mean}(v) + k$ but $\text{sd}(y) = |m| \times \text{sd}(v)$. The mean feels the shift; the spread does not.

Why the distractors are wrong:

A) $3$ leaves the standard deviation unchanged. The $-6$ leaves it unchanged, but the $\times 4$ does not — the set really is four times as spread out afterwards.

B) $6$ applies both operations to the standard deviation: $4 \times 3 - 6$. This is the natural mistake, and it is worth seeing exactly why the $-6$ cannot be there: it moves every value and the mean by the same 6, so every distance from the mean stays as it was.

D) $26$ is the new mean, $4(8) - 6$. It is the right formula applied to the wrong statistic.

Takeaway: Multiplying scales the standard deviation; adding or subtracting does not affect it at all. The mean feels both. If you cannot recall which, write down four numbers, apply the transformation, and look.

---