Additional Mathematics 4037/21 — May/June 2026

Walkthrough

The cubic is given in factorised form, which is a gift: each factor set to zero gives one $x$-intercept immediately. Setting $2x - 1 = 0$ gives $x = 0.5$, setting $x + 3 = 0$ gives $x = -3$, and setting $2x + 5 = 0$ gives $x = -2.5$. For the $y$-intercept we substitute $x = 0$ into the whole expression, giving $\frac{1}{3}(-1)(3)(5) = -5$, so the curve crosses the $y$-axis at $(0, -5)$.

To get the shape right, note the leading term: $\frac{1}{3} \times 2x \times x \times 2x = \frac{2}{3}x^3$, a positive cubic. A positive cubic starts low on the left and ends high on the right. With three distinct real roots it must cross the axis three times, so between consecutive roots it bulges: a local maximum between $x = -3$ and $x = -2.5$ (above the axis, hence in the second quadrant), then a local minimum between $x = -2.5$ and $x = 0.5$. Since the $y$-intercept is $-5$ (negative), that minimum dips below the axis, placing it in the third quadrant. The curve must extend clearly beyond the outermost intercepts — it must not stop at $x = -3$ or $x = 0.5$.

Key Takeaways

• A factorised polynomial hands you its $x$-intercepts directly; set each factor to zero.
• The $y$-intercept is always found by substituting $x = 0$.
• The sign of the leading coefficient fixes the overall cubic shape (positive: up on the right; negative: down on the right).
• The sign of the $y$-intercept tells you which side of the axis the middle region sits on.

Common Mistakes

• Stopping the curve at the outermost intercepts — the mark scheme explicitly requires the 'ends' to extend beyond the $x$-axis.
• Putting the minimum on the $y$-axis (B0) instead of between $-2.5$ and $0.5$.
• Forgetting to solve $2x + 5 = 0$ properly and writing the intercept as $-5$ instead of $-2.5$.
• Marking intercept values as table points rather than on the axes; the scheme condones values written beside the graph but they must clearly be intercepts.
• Getting the shape upside down (treating it as a negative cubic).

Things to Be Careful About

• The intercepts must be labelled on the sketch (or clearly beside it), not just listed in a table.
• $x = -\frac{5}{2} = -2.5$: keep the exact value; either form is accepted.
• The curve must be a genuine cubic attempt — a parabola or a curve with only one turning point scores B0 for the intercepts.

Walkthrough

An inequality of the form (cubic) > 0 asks: where is the curve above the $x$-axis? The sketch from part (a) answers this directly. The curve is above the axis in two places: the small hump between the first two roots, $x = -3$ and $x = -2.5$, and everything to the right of the last root, $x = 0.5$. Because the inequality is strict ($>$, not $\geq$), the roots themselves — where the curve equals zero — are not included, so we write strict inequalities: $-3 < x < -\frac{5}{2}$ and $x > \frac{1}{2}$.

Key Takeaways

• A polynomial inequality is solved by sketching (or sign-charting) the polynomial and reading where it is above or below the axis.
• Strict inequalities exclude the roots; inclusive ones include them.
• A positive cubic with three roots is positive on the first gap, negative on the second, and positive again beyond the last root.

Common Mistakes

• Using $\leq$ and $\geq$ instead of strict inequalities — the mark scheme awards SC1 only if 0 was otherwise scored, so the strict form is what earns the marks.
• Reversing an interval, e.g. writing $-\frac{5}{2} < x < -3$, which is empty and wrong (though the reversed-but-correct form $-\frac{5}{2} > x > -3$ is accepted).
• Missing one of the two intervals, giving only $x > \frac{1}{2}$ or only $-3 < x < -\frac{5}{2}$.
• Solving by expanding the cubic instead of using the sketch — unnecessary work that invites arithmetic errors.

Things to Be Careful About

• Both intervals are required for full marks (B1 each).
• Any conjunction ('and', 'or', comma) between the two inequalities is accepted.
• The answer must be exact: $-\frac{5}{2}$ (or $-2.5$) and $\frac{1}{2}$ (or $0.5$).

Walkthrough

The graph of $y = \cos x$ oscillates between $-1$ and $1$. Multiplying by $4$ stretches it vertically so it oscillates between $-4$ and $4$; adding $2$ shifts it up but does not change how far it swings either side of its midline. That swing, from midline to maximum, is the amplitude, which is simply the multiplier $4$ in front of the cosine. The vertical shift $+2$ and the horizontal scaling $\frac{x}{8}$ have no effect on the amplitude.

Key Takeaways

• For $y = A\cos(bx) + c$ (or sine), the amplitude is $|A|$.
• The vertical translation $c$ moves the midline but never changes the amplitude.

Common Mistakes

• Giving $6$ (the maximum value) instead of $4$: the amplitude is measured from the midline $y = 2$, not from the $x$-axis.
• Confusing amplitude with period or with the range.

Things to Be Careful About

• This is a one-mark B1: no working is needed, just the value $4$. The answer is exact, so no rounding issues arise.

Walkthrough

Replacing $x$ by $\frac{x}{8}$ stretches the cosine graph horizontally by a factor of $8$: every feature that used to happen at angle $θ$ now happens at $8θ$. Since ordinary cosine completes one full cycle over $360°$, this stretched version takes $8 \times 360° = 2880°$ to complete one cycle. Equivalently, using the formula period $= \frac{360°}{b}$ with $b = \frac{1}{8}$ gives $\frac{360°}{1/8} = 2880°$.

Key Takeaways

• Period of $y = A\cos(bx)$ in degrees is $\frac{360°}{b}$.
• Dividing $x$ by a number $k$ stretches the graph horizontally by factor $k$, multiplying the period by $k$.

Common Mistakes

• Writing $45°$ (i.e. computing $\frac{360°}{8}$ instead of $360° \times 8$): dividing $x$ by 8 makes the period longer, not shorter.
• Leaving the answer as $8\pi$ or otherwise mixing radians into a degrees question.

Things to Be Careful About

• The question asks for the period in degrees, so the answer must be $2880°$, not any radian equivalent.

Walkthrough

First isolate the cosine: subtract $2$ and divide by $4$ to get $\cos\left(\frac{x}{8}\right) = -\frac{1}{2}$. Cosine equals $+\frac{1}{2}$ at $60°$, so cosine equals $-\frac{1}{2}$ at $180° - 60° = 120°$ (second quadrant) and at $-120°$ (third quadrant). These give $\frac{x}{8} = \pm 120°$, hence $x = \pm 960°$. Because the function has period $2880°$, adding or subtracting $2880°$ from either solution leaves the interval $[-1080°, 1080°]$, so these two are the only solutions — which matches the sketch in part (d), where the curve crosses the axis once on each side between $900°$ and $1080°$.

Key Takeaways

• Solve $\cos u = k$ by finding the base angle from $\arccos|k|$ and reflecting into the second/third quadrants.
• When the argument is a multiple like $\frac{x}{8}$, solve for the argument first, then scale up.
• Use the period to check whether further solutions fall inside the required interval.

Common Mistakes

• Taking $\frac{x}{8} = 60°$ instead of $120°$, forgetting the negative sign of the cosine.
• Stopping at $\frac{x}{8} = 120°$ and forgetting to multiply by $8$.
• Finding only $x = 960°$ and missing the negative solution $x = -960°$.
• Adding $360°$ instead of $2880°$ when hunting for extra solutions, generating false extras inside the interval.

Things to Be Careful About

• The mark scheme awards M1 for reaching $\frac{x}{8} = 120°$ (or $-120°$), then A2 for both $x = \pm 960°$ with no extras — one correct answer alone implies the M1 but only earns one A mark.
• Answers are exact here ($\pm 960°$); no rounding is needed.
• All solutions in $-1080° \leq x \leq 1080°$ are required, not just one.

Walkthrough

Everything needed for the sketch has been established in the earlier parts. The amplitude $4$ and the $+2$ shift mean the curve swings between $-2$ and $6$ about the line $y = 2$. The period $2880°$ means that within the window $-1080°$ to $1080°$ we see less than half a cycle: exactly one hump. The peak sits at $x = 0$ with height $6$. From part (c) the curve crosses the $x$-axis at $x = \pm 960°$. At the edges of the window, $x = \pm 1080°$, the height is $4\cos 135° + 2 = 2 - 2\sqrt{2} \approx -0.83$, so the curve finishes just under the axis. Joining these features with a smooth, even curve gives the required sketch.

Key Takeaways

• A sketch is built from amplitude, midline, period, intercepts and end behaviour — all obtainable algebraically before drawing.
• When the interval shown is shorter than one period, the sketch shows only part of the wave, and the ends need not be turning points.

Common Mistakes

• Drawing more than one hump (forgetting the huge period $2880°$).
• Putting the maximum at $y = 4$ instead of $y = 6$ (ignoring the $+2$ shift).
• Placing intercepts at $\pm 120°$ (forgetting to scale by $8$) or at $\pm 1080°$ exactly.
• Ending the curve on the $x$-axis rather than slightly below it at $x = \pm 1080°$.
• Drawing a jagged or flicked-out curve rather than a smooth cosine shape.

Things to Be Careful About

• The mark scheme gives B1 for the correct shape with a maximum at $y = 6$ and only one turning point, and B2 additionally requires the $x$-intercepts strictly between $900°$ and $1080°$ (and $-900°$ and $-1080°$), with the ends at approximately $(\pm 1080°, y)$ where $0 > y > -2$.
• Poor curvature or slight feathering at the ends is condoned, but the curve must be smooth and symmetric with exactly one turning point.
• Anything drawn outside $-1080° \leq x \leq 1080°$ is ignored.

Walkthrough

The first condition compares two terms of the progression. Using the formula for the $n$th term, $u_n = a + (n-1)d$, the 6th term is $a + 5d$ and the 3rd term is $a + 2d$. Setting the first equal to 1.5 times the second and expanding gives $a + 5d = 1.5a + 3d$, which rearranges to $2d = 0.5a$, i.e. $a = 4d$.

The second condition uses the sum formula $S_n = \frac{n}{2}(2a + (n-1)d)$. With $n = 10$ this is $\frac{10}{2}(2a + 9d) = 255$, so $2a + 9d = 51$.

We now have two linear equations. Substituting $a = 4d$ into the second gives $17d = 51$, so $d = 3$ and then $a = 12$.

Key Takeaways

• The $n$th term of an AP is $u_n = a + (n-1)d$; the sum of the first $n$ terms is $S_n = \frac{n}{2}(2a + (n-1)d)$.
• Worded conditions about terms and sums translate directly into simultaneous equations.
• Elimination by substitution solves these quickly when one equation gives a simple relation like $a = 4d$.

Common Mistakes

• Writing the 6th term as $a + 6d$ instead of $a + 5d$ (off-by-one error in $(n-1)$).
• Misreading 255 as 225 — the mark scheme has a special follow-through route for that error but it costs marks.
• Arithmetic slips when expanding $1.5(a + 2d)$; note $1.5 \times 2d = 3d$.
• Not showing the elimination step — the M1 requires solving their equations as far as finding $a$ or $d$.

Things to Be Careful About

• Both equations must be formed correctly to earn the B1 marks; the scheme accepts equivalent forms ('oe').
• The A2 for $a = 12$, $d = 3$ depends on a correct method being shown — answers from wrong working score nothing.
• Keep exact values ($d = 3$, $a = 12$); no rounding issues arise here.

Walkthrough

We need the smallest $n$ such that the sum of the first $n$ terms exceeds 40 000. Substituting $a = 12$ and $d = 3$ into $S_n = \frac{n}{2}(2a + (n-1)d)$ gives $\frac{n}{2}(24 + 3(n-1)) > 40\,000$.

Multiplying through by 2 and expanding the bracket: $n(21 + 3n) > 80\,000$, i.e. $3n^2 + 21n - 80\,000 > 0$.

To find where the quadratic changes sign we solve the equation $3n^2 + 21n - 80\,000 = 0$ with the quadratic formula. The discriminant is $441 + 960\,000 = 960\,441$, whose square root is about $980.02$. The positive root is $(-21 + 980.02)/6 \approx 159.83$.

Because $n$ counts terms, it must be a positive integer, and since $S_{159}$ falls just short while $S_{160}$ exceeds 40 000, the answer is $n = 160$. A quick check confirms $S_{160} = 40\,080$.

Key Takeaways

• 'Least number of terms for the sum to exceed X' leads to a quadratic inequality in $n$.
• Solve the corresponding equation, then interpret: the answer is the next integer above the positive root (round up, never down).
• Reject negative roots — they have no meaning as a count of terms.

Common Mistakes

• Rounding down to $n = 159$: at $n = 159$ the sum is only 39 591, which does not exceed 40 000. The mark scheme explicitly penalises $n = 159$ or $n < 160$.
• Forgetting to multiply by 2 before clearing the fraction, giving a wrong quadratic.
• Sign errors in the discriminant: $c = -80\,000$, so $-4ac = +960\,000$.
• Stating both roots including the negative one incorrectly if a calculator is used — the negative root must be $-166.836\ldots$ if quoted.
• Trial and improvement without showing both boundary values ($n = 159$ giving 39 591 and $n = 160$ giving 40 080) loses marks under the special substitution rule.

Things to Be Careful About

• The final answer must be exactly $n = 160$ reached from correct working ('nfww') — a correct-looking answer from flawed algebra scores nothing.
• $n \approx 160$ is condoned, but $n = 159$ or any inequality like $n < 160$ earns M1 A0 only.
• Follow-through applies if part (a) values were wrong: use your own $a$ and $d$ consistently throughout.

Walkthrough

When non-linear variables are plotted to produce a straight line, the relationship follows the standard straight-line equation $Y = mX + c$. Here, the vertical axis is $Y = \ln 3y$ and the horizontal axis is $X = x^2$.

1. First, calculate the gradient $m$ of the straight line using the coordinates $(X_1, Y_1) = (4.64, 12.24)$ and $(X_2, Y_2) = (6.84, 8.94)$:

1. Next, find the $Y$-intercept $c$ by substituting one of the points and the gradient into $Y = mX + c$:

1. Substitute $m$ and $c$ back into the linear relation:

1. Remove the natural logarithm by taking the exponential of both sides:

1. Use index laws $\mathrm{e}^{A + B} = \mathrm{e}^A \mathrm{e}^B$ and divide by $3$ to isolate $y$:

This gives $k = \frac{\mathrm{e}^{19.2}}{3}$ and $a = -1.5$ as exact constants.

Key Takeaways

• In non-linear reduction questions, treat the plotted variables as capital variables $Y$ and $X$.
• Exponentiating both sides is the inverse of taking the natural logarithm: $\ln A = B \iff A = \mathrm{e}^B$.
• Ensure constants asked for in "exact" form are left with $\mathrm{e}$ rather than converted to rounded decimals.

Common Mistakes

• Confusing $x$ and $y$ with $x^2$ and $\ln 3y$, leading to using standard linear equations in terms of $x$ and $y$.
• Forgetting to divide by $3$ when isolating $y$ from $\ln 3y$.
• Converting $\mathrm{e}^{19.2}$ to a decimal prematurely instead of keeping it in exact form as required.

Things to Be Careful About

• The question specifies exact constants $k$ and $a$, so $k = \frac{\mathrm{e}^{19.2}}{3}$ and $a = -1.5$ (or fraction equivalents $\frac{1}{3}\mathrm{e}^{96/5}$ and $-\frac{3}{2}$) should be retained.

Walkthrough

To find $y$ when $x = 3$, substitute $x = 3$ into the derived equation $y = \frac{\mathrm{e}^{19.2}}{3}\mathrm{e}^{-1.5x^2}$ (or into $\ln 3y = 19.2 - 1.5x^2$):

Rounding to 3 significant figures gives $99.6$.

Key Takeaways

• Substituting into either the linear transformed equation or the final exponential equation yields the same result.
• Unless exact form is specified, round final numerical answers to 3 significant figures.

Common Mistakes

• Forgetting to square $x = 3$, computing $-1.5(3)$ instead of $-1.5(3^2)$.
• Rounding errors from using an approximate value of $k$.

Things to Be Careful About

• Give the answer to 3 significant figures ($99.6$) or exact form ($\frac{\mathrm{e}^{5.7}}{3}$).

Walkthrough

1. Substitute $y = 10$ into the linear form $\ln 3y = 19.2 - 1.5x^2$:

1. Rearrange the equation to isolate $x^2$:

1. Since the question asks for the "values of $x$", remember to include both the positive and negative square roots:

Key Takeaways

• When solving $x^2 = c$ for $x$, there are two solutions: $x = \pm\sqrt{c}$.
• Working from the linear form $\ln 3y = mx^2 + c$ is often algebraically simpler than rearranging the exponential form.

Common Mistakes

• Giving only the positive square root $x = 3.25$ and missing the negative root $x = -3.25$.
• Forgetting that $\ln 3y$ requires multiplying $y$ by $3$ before taking the natural logarithm (i.e. using $\ln 10$ instead of $\ln 30$).

Things to Be Careful About

• The wording "values of $x$" explicitly indicates multiple solutions.
• Maintain intermediate accuracy before rounding to 3 significant figures.

Walkthrough

The function $y = \tan^3 5x$ is built in layers: the outermost layer is a cube, inside that is $\tan$, and inside that is $5x$. The chain rule peels these off one at a time. Differentiating the cube gives $3(\tan 5x)^2$; differentiating $\tan 5x$ gives $5\sec^2 5x$ (the derivative of $\tan u$ is $\sec^2 u$, multiplied by the inner derivative $5$). Multiplying gives $15\tan^2 5x \sec^2 5x$. Since the question asks for the answer in terms of $\tan 5x$, we use the Pythagorean identity $\sec^2 A = 1 + \tan^2 A$ to replace $\sec^2 5x$, giving $15\tan^2 5x(1 + \tan^2 5x)$.

Key Takeaways

• For $[\mathrm{f}(g(x))]^n$, the chain rule gives $n[\mathrm{f}(g(x))]^{n-1} \times \mathrm{f}'(g(x)) \times g'(x)$ — here two applications are needed because there are three nested layers.
• $\frac{\mathrm{d}}{\mathrm{d}u}(\tan u) = \sec^2 u$, and the multiple angle contributes its coefficient: $\frac{\mathrm{d}}{\mathrm{d}x}(\tan kx) = k\sec^2 kx$.
• The identity $\sec^2 A = 1 + \tan^2 A$ lets you rewrite answers in whichever form a question demands.

Common Mistakes

• Forgetting the factor $5$ from the inner derivative of $5x$ — this loses most of the marks (the scheme awards B3 for $3(\tan 5x)^2(5\sec^2 5x)$ but drops marks for a wrong constant).
• Writing $3\tan^2 5x$ without multiplying by the derivative of $\tan 5x$ at all.
• Leaving the answer as $15\tan^2 5x \sec^2 5x$ when the question explicitly asks for it in terms of $\tan 5x$.
• Confusing $\tan^3 5x$ with $\tan(15x)$ or $(\tan 5x)^3$ misread as $\tan^2 25x$.

Things to Be Careful About

• The mark scheme says "nfww" for the final answer: no follow-through from wrong working, so every constant must be correct.
• "isw" applies: later incorrect simplification after the correct answer is ignored, but the printed target form $15\tan^2 5x(1+\tan^2 5x)$ should be reached.
• An equivalent unsimplified route via the quotient rule on $\frac{\sin^3 5x}{\cos^3 5x}$ giving $\frac{15\sin^2 5x}{\cos^4 5x}$ also earns full credit.