Additional Mathematics 4037/13 — October/November 2024

Walkthrough

The curve is given as a product of two linear factors. Expanding gives $y = x^2 - x - 12$, a quadratic whose derivative is easy to find: $\frac{\mathrm{d}y}{\mathrm{d}x} = 2x - 1$. A stationary point occurs where the gradient is zero, so $2x - 1 = 0$ gives $x = \frac{1}{2}$. Substituting this back into the expanded quadratic gives $y = \frac{1}{4} - \frac{1}{2} - 12 = -\frac{49}{4}$. (Equivalently, one could complete the square: $y = \left(x - \frac{1}{2}\right)^2 - \frac{49}{4}$, which shows the minimum directly.)

Key Takeaways

• A stationary point is found by setting $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$.
• For a quadratic, completing the square gives the stationary point immediately.
• Always substitute the $x$-value back into the original curve to get the $y$-coordinate.

Common Mistakes

• Forgetting to substitute back to find the $y$-coordinate (giving only $x = \frac{1}{2}$ scores only 2 of the 3 marks).
• Sign slips when expanding $(x+3)(x-4)$ — the constant term is $-12$, not $+12$.
• Arithmetic errors in $\frac{1}{4} - \frac{1}{2} - 12$; write all terms over a common denominator of 4.

Things to Be Careful About

• The answer must be a coordinate pair; the exact form $-\frac{49}{4}$ (or the equivalent decimal $-12.25$) is accepted ("oe" in the scheme).
• This is a non-calculator paper, so keep the working in exact fractions.

Walkthrough

The modulus $|f(x)|$ takes every negative output of $f$ and makes it positive, which geometrically means reflecting the below-axis portion of the graph in the $x$-axis. The parabola $y = (x+3)(x-4)$ has roots $x = -3$ and $x = 4$ and lies below the axis between them (its minimum is $-\frac{49}{4}$ from part (a)). So only the middle arc is reflected: it becomes an upward arch peaking at $\left(\frac{1}{2}, \frac{49}{4}\right)$, while the outer arms are unchanged. At the roots the reflected arc meets the unchanged arms, creating sharp cusps on the $x$-axis. The $y$-intercept is $|{-12}| = 12$.

Key Takeaways

• $y = |f(x)|$ reflects the negative parts of $y = f(x)$ in the $x$-axis; positive parts are unchanged.
• Roots of $f$ become cusps (sharp corners) on the $x$-axis.
• Intercepts are found by substituting $x = 0$ (for the $y$-intercept) and solving $f(x) = 0$ (for $x$-intercepts).

Common Mistakes

• Drawing the whole parabola reflected (an upside-down U) instead of only the below-axis part — the outer arms must still rise.
• Drawing smooth turning points at the roots instead of cusps.
• Giving the $y$-intercept as $-12$ (forgetting the modulus makes it $+12$).
• Marking the maximum below the axis or mislabelling it — the scheme ignores incorrect labels on the maximum, but the shape must be right.

Things to Be Careful About

• The scheme awards B1 for the correct shape (cusps on the $x$-axis, maximum in the first quadrant) and B1 for the correct intercepts, and the intercepts mark depends on the shape being correct.
• The intercepts must be stated: $(-3, 0)$, $(4, 0)$ and $(0, 12)$.
• This is a sketch, not a plot: correct shape and key features matter, not plotted accuracy.

Walkthrough

Solving $|(x+3)(x-4)| = k$ means finding where the horizontal line $y = k$ meets the W-shaped graph from part (b). For small positive $k$ the line crosses four branches (two outer arms plus the arch twice). As $k$ increases past the peak value $\frac{49}{4}$, the arch is no longer reached, and only the two outer arms are cut — giving exactly 2 distinct real roots. So the condition is $k > \frac{49}{4}$.

Key Takeaways

• The number of real roots of $f(x) = k$ equals the number of intersections of $y = f(x)$ with the horizontal line $y = k$.
• Critical values of $k$ occur at turning points of the graph.

Common Mistakes

• Giving $k > 12$ (using the $y$-intercept instead of the maximum value $\frac{49}{4}$) — the scheme explicitly excludes this.
• Including $k = \frac{49}{4}$, where the line touches the maximum and gives 3 roots, not 2.
• Forgetting $k > 0$ is given, so no need to discuss $k \leq 0$.

Things to Be Careful About

• The answer is a strict inequality: $k = \frac{49}{4}$ gives 3 roots, so it must be excluded.
• Follow-through is allowed on the magnitude of the candidate's maximum from part (a), but $k > 12$ scores nothing.

Walkthrough

The three terms must become one logarithm to base 2. First, the constant 3 is rewritten as $\log_2 8$, since $2^3 = 8$ — this makes every term a logarithm of the same base. Next, the coefficient 4 in front of $\log_2 a$ is moved inside using the power rule, giving $\log_2 a^4$. Finally, the sum of two logarithms becomes the logarithm of a product, and the subtracted one becomes division: $\log_2 8 + \log_2 a^4 - \log_2 b = \log_2 \frac{8a^4}{b}$.

Key Takeaways

• Any constant $c$ can be written as $\log_b b^c$.
• The power rule $n\log_b x = \log_b x^n$ moves coefficients inside the log.
• Sums and differences of logs of the same base combine into products and quotients.

Common Mistakes

• Writing $4\log_2 a$ as $\log_2 4a$ instead of $\log_2 a^4$.
• Forgetting to convert the 3 into $\log_2 8$ before combining, which loses the 8 in the answer.
• Leaving the answer as a sum of logs rather than a single logarithm, which the question forbids.

Things to Be Careful About

• The mark scheme awards the method mark for correctly using two of the multiplication, division or power rules, and the accuracy mark only if the final logarithm is to base 2 — a base-10 or base-e answer scores nothing.
• The answer must be a single logarithm, so all combining must be completed.

Walkthrough

The equation mixes two different bases: $\lg x$ is base 10, while $\log_x 10$ has base $x$. The change-of-base rule $\log_x 10 = \frac{\lg 10}{\lg x}$ converts it, and since $\lg 10 = 1$, the equation becomes $\lg x = \frac{4}{\lg x}$. Multiplying through by $\lg x$ gives $(\lg x)^2 = 4$, a quadratic in the quantity $\lg x$. Square rooting gives $\lg x = \pm 2$ — both signs must be kept. Finally, undoing the logarithm with base 10 gives $x = 10^2 = 100$ and $x = 10^{-2} = \frac{1}{100}$. Both are positive, so both are valid solutions of the original equation.

Key Takeaways

• The change-of-base rule lets you express any logarithm in terms of one common base.
• An equation like $\lg x = \frac{k}{\lg x}$ is a disguised quadratic in $\lg x$.
• Squaring means two roots, so both $\lg x = 2$ and $\lg x = -2$ must be solved.

Common Mistakes

• Writing $(\lg x)^2$ as $\lg x^2$ — the mark scheme explicitly rejects this; the square is on the whole logarithm.
• Losing the negative root and giving only $x = 100$.
• Giving $x = \pm 100$ instead of $x = \frac{1}{100}$ for $\lg x = -2$.
• Forgetting that $x$ must be positive (both answers here satisfy this).

Things to Be Careful About

• The mark scheme makes each mark dependent on a correct change of base, so the line $\lg x = \frac{4}{\lg x}$ must appear.
• Both solutions are required; giving only one loses the final mark.
• $x = 0.01$ is an acceptable equivalent form for $\frac{1}{100}$.

Walkthrough

The stem defines $\mathrm{p}(x) = ax^3 + bx^2 - 19x + c$ with integer coefficients. Part (a) asks us to show that $7a - 3b = 39$, so we must derive this from the two given facts about $\mathrm{p}$.

First fact: $x + 2$ is a factor. By the factor theorem, if $(x - k)$ is a factor then $\mathrm{p}(k) = 0$; here $k = -2$. Substituting $x = -2$: $(-2)^3 = -8$, $(-2)^2 = 4$, and $-19 \times (-2) = +38$, giving $-8a + 4b + 38 + c = 0$.

Second fact: dividing by $x + 1$ leaves remainder 20. By the remainder theorem, $\mathrm{p}(-1) = 20$. Substituting $x = -1$: $(-1)^3 = -1$, $(-1)^2 = 1$, $-19 \times (-1) = 19$, giving $-a + b + 19 + c = 20$.

Both equations contain the unknown $c$, so subtracting them eliminates it: subtracting gives $-8a - (-a) = -7a$, $4b - b = 3b$, $38 - 19 = 19$, and the right side is $0 - 20 = -20$. So $-7a + 3b + 19 = -20$, hence $-7a + 3b = -39$, i.e. $7a - 3b = 39$, exactly the printed target — so the derivation is complete.

Key Takeaways

• The factor theorem: $x + 2$ a factor means $\mathrm{p}(-2) = 0$.
• The remainder theorem: dividing by $x + 1$ leaves remainder equal to $\mathrm{p}(-1)$.
• When two equations share an unwanted unknown ($c$ here), eliminate it by adding or subtracting.

Common Mistakes

• Sign slips when substituting negative values: $(-2)^3 = -8$ not $8$, and $-19 \times (-2) = +38$ not $-38$. The mark scheme allows only one sign error for the M marks.
• Forgetting that "remainder 20" means $\mathrm{p}(-1) = 20$, not $\mathrm{p}(-1) = 0$.
• Subtracting in the wrong order and getting $3b - 7a = -39$ without rearranging to the printed form — the AG mark requires the exact target reached from correct work.

Things to Be Careful About

• This is an AG (Answer Given) part: every algebraic line between the substitutions and $7a - 3b = 39$ must be shown — insufficient detail loses the A1 even if the answer matches.
• Keep all terms on one side consistently to avoid sign confusion when eliminating $c$.

Walkthrough

We now know $7a - 3b = 39$ from part (a). The new information is that dividing $\mathrm{p}'(x)$ by $x - 1$ leaves remainder 1, so by the remainder theorem $\mathrm{p}'(1) = 1$.

Differentiating $\mathrm{p}(x) = ax^3 + bx^2 - 19x + c$ term by term gives $\mathrm{p}'(x) = 3ax^2 + 2bx - 19$ (the constant $c$ differentiates to zero). Substituting $x = 1$ and equating to 1 gives $3a + 2b - 19 = 1$, i.e. $3a + 2b = 20$.

Now we have two linear equations in $a$ and $b$. Eliminating $b$: multiply $3a + 2b = 20$ by 3 to get $9a + 6b = 60$, and multiply $7a - 3b = 39$ by 2 to get $14a - 6b = 78$. Adding removes $b$: $23a = 138$, so $a = 6$. Substituting back: $3(6) + 2b = 20$ gives $2b = 2$, so $b = 1$.

Finally, find $c$ by substituting $a = 6$ and $b = 1$ into any earlier true statement — using $\mathrm{p}(-1) = 20$: $-6 + 1 + 19 + c = 20$, so $c = 6$. (Using $\mathrm{p}(-2) = 0$ instead: $-48 + 4 + 38 + c = 0$ also gives $c = 6$.)

Key Takeaways

• The remainder theorem applies to any polynomial, including a derivative: $\mathrm{p}'(1)$ equals the remainder when $\mathrm{p}'(x)$ is divided by $x - 1$.
• Differentiating a cubic kills the constant term, so $c$ never appears in $\mathrm{p}'(x)$ — that is why $c$ must be recovered afterwards from an original equation.
• Simultaneous linear equations are solved cleanly by matching coefficients before adding or subtracting.

Common Mistakes

• Writing $\mathrm{p}'(x)$ incorrectly, e.g. forgetting to reduce each power or keeping the constant $c$.
• Setting $\mathrm{p}'(1) = 0$ out of habit from factor questions — here the remainder is 1.
• Solving the simultaneous equations with sign errors; the dependent M1 requires correctly solving their equation with $7a - 3b = 39$.
• Stopping after finding $a$ and $b$ and forgetting to go back for $c$.
• Using $\mathrm{p}'(-2) = 0$ or other wrong conditions to find $c$.

Things to Be Careful About

• The final M1 is dependent: it is awarded only for solving the candidate's own equation from this part together with the result $7a - 3b = 39$ from part (a), so the link between the parts must actually be used.
• Check the values satisfy both original conditions as a sanity check: $\mathrm{p}(-2) = -48 + 4 + 38 + 6 = 0$ ✓ and $\mathrm{p}(-1) = -6 + 1 + 19 + 6 = 20$ ✓.

Walkthrough

The equation $y = Ab^{x^2}$ is not linear in $x$ and $y$, so it cannot be tested directly on ordinary graph paper. Taking natural logarithms of both sides gives $\ln y = \ln A + x^2 \ln b$. If we treat $x^2$ as the horizontal variable and $\ln y$ as the vertical variable, this has exactly the form of a straight line $Y = mX + c$ with gradient $\ln b$ and intercept $\ln A$. So the first job is purely mechanical: square each $x$ ($1,\ 2.25,\ 4,\ 6.25,\ 9$) and take $\ln$ of each $y$ ($2.64,\ 3.51,\ 4.72,\ 6.28,\ 8.18$). Then plot each pair carefully on the millimetre grid — accuracy here matters because parts (b) and (c) read their answers off this same line. Finally lay a ruler along the points; they lie almost perfectly straight, so the line of best fit passes essentially through all five.

Key Takeaways

• Taking logs converts an exponential-type relationship into a linear one whose gradient and intercept reveal the constants.
• Here the log is taken of $y$ only, while $x$ is squared first — the choice of which variables to transform comes from matching the equation to $Y = mX + c$.
• Careful plotting pays off twice: the same graph supplies answers to all three parts.

Common Mistakes

• Plotting $x$ instead of $x^2$, or $y$ instead of $\ln y$ — the whole point of the transformation is lost.
• Using $\lg$ (base 10) instead of $\ln$; either would give a straight line, but the mark scheme's values are natural logs, so mixing bases corrupts later readings.
• Joining the points with a zig-zag instead of drawing one straight line of best fit.
• Rounding $\ln y$ too coarsely so the points drift off the true line.

Things to Be Careful About

• The mark scheme awards M1 for plotting with at most one error, so all five points must be plotted accurately to the small grid squares.
• Give $\ln y$ values to 3 significant figures before plotting.
• The line must be genuinely straight and pass through the scatter of points, not forced through the origin.

Walkthrough

First establish why the graph works: applying the log law $\ln(pq) = \ln p + \ln q$ and the power law $\ln(b^{x^2}) = x^2 \ln b$ turns $y = Ab^{x^2}$ into $\ln y = x^2 \ln b + \ln A$. Comparing with $Y = mX + c$, the slope of the plotted line is $\ln b$ and where it cuts the vertical axis is $\ln A$. From the drawn line, the rise over the full run is about $8.18 - 2.64 = 5.54$ over a run of $9 - 1 = 8$, giving a gradient near $0.7$. So $b = \mathrm{e}^{0.7} \approx 2$. Reading where the line meets the $\ln y$ axis gives about $1.95$, so $A = \mathrm{e}^{1.95} \approx 7$. Both answers are wanted to 1 significant figure, which conveniently absorbs any small reading error.

Key Takeaways

• In a log-linear plot of $y = Ab^{x^2}$, gradient $= \ln b$ and intercept $= \ln A$ — the constants come out of the log, so you must exponentiate at the end.
• Always write down the linear form explicitly before reading the graph; mixing up which feature gives which constant is the classic error.
• The question asks for 1 significant figure precisely because graph readings carry uncertainty.

Common Mistakes

• Reporting $b = 0.7$ and $A = 1.95$ — forgetting that these are $\ln b$ and $\ln A$, so the final step $b = \mathrm{e}^{0.7}$, $A = \mathrm{e}^{1.95}$ is essential.
• Swapping the roles: treating the intercept as $\ln b$ and the gradient as $\ln A$.
• Computing the gradient from the original data $(x, y)$ rather than from the transformed $(x^2, \ln y)$ line — the mark scheme insists the gradient comes from the linear graph.
• Giving more significant figures than asked; the answer must be 1 s.f.