Additional Mathematics 4037/12 — October/November 2023

Cambridge O-Level · Calculator · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Trigonometry · Factors of polynomials · Equations, inequalities and graphs · Logarithmic and exponential functions · Calculus · Straight-line graphs · +6 more

Q1Factors of polynomialsEquations, inequalities and graphsFree sample

Fig. 1

The diagram shows the graph of the cubic polynomial $y = \mathrm{f}(x)$ .

(a)

Find an expression for $\mathrm{f}(x)$ in factorised form. Write each linear factor with its coefficients as integers.

DifficultyMedium-Easy

Worked solution

Approach

The curve crosses the $x$ -axis at $x = -\frac{1}{3}$ , $x = 1$ and $x = \frac{5}{2}$ , so these are the roots of $\mathrm{f}(x)$ . Write $\mathrm{f}(x) = k\left(x + \frac{1}{3}\right)(x - 1)\left(x - \frac{5}{2}\right)$ and use the $y$ -intercept $(0, -15)$ to find $k$ , then clear the fractions so each factor has integer coefficients.

Working

\mathrm{f}(x) = k\left(x + \frac{1}{3}\right)(x - 1)\left(x - \frac{5}{2}\right)

Using the point $(0, -15)$ :

-15 = k\left(\frac{1}{3}\right)(-1)\left(-\frac{5}{2}\right) = k \times \frac{5}{6}

k = -15 \times \frac{6}{5} = -18

\mathrm{f}(x) = -18\left(x + \frac{1}{3}\right)(x - 1)\left(x - \frac{5}{2}\right)

Distributing $k$ into the fractional factors to make integer coefficients: $-18\left(x + \frac{1}{3}\right) = -(6)(3x+1)$ and $-18\left(x - \frac{5}{2}\right) = +(3)(2x-5)$ , giving

\mathrm{f}(x) = -3(3x+1)(x-1)(2x-5)

Check at $x = 0$ : $-3(1)(-1)(-5) = -15$ . ✓

Answer

\mathrm{f}(x) = -3(3x+1)(x-1)(2x-5)

Final answer

f(x) = -3(3x+1)(x-1)(2x-5)

Detailed explanation

Walkthrough

The graph gives us all three places where the cubic equals zero: $x = -\frac{1}{3}$ , $x = 1$ and $x = \frac{5}{2}$ . Each root $r$ contributes a factor $(x - r)$ , so the polynomial must be a constant multiple of $\left(x + \frac{1}{3}\right)(x - 1)\left(x - \frac{5}{2}\right)$ . The constant is not yet known because stretching the whole cubic vertically does not move its $x$ -intercepts. To pin it down we use the one other piece of information on the graph: the curve passes through $(0, -15)$ . Substituting $x = 0$ into the factored form gives $k \times \frac{5}{6} = -15$ , so $k = -18$ . Finally the question insists each linear factor has integer coefficients, so we absorb the fractions: multiplying out $k = -18$ across the factors as $-3 \times 6 \times 3$ turns $\left(x + \frac{1}{3}\right)$ into $(3x+1)$ and $\left(x - \frac{5}{2}\right)$ into $(2x-5)$ , leaving $-3(3x+1)(x-1)(2x-5)$ . A quick check at $x=0$ confirms $-15$ .

Key Takeaways

A cubic's $x$ -intercepts give its linear factors; the leading constant is fixed by any additional point such as the $y$ -intercept.
Fractions inside factors can always be cleared by folding the scale factor into them.
Always verify with the given point after clearing fractions.

Common Mistakes

Writing $(x - \frac{1}{3})$ instead of $(x + \frac{1}{3})$ — sign slips when converting a negative root into a factor are the most common error here.
Forgetting the vertical stretch constant entirely (giving only B1 for $k(x+\frac{1}{3})(x-1)(x-\frac{5}{2})$ ).
Finding $m(3x+1)(x-1)(2x-5)$ but not evaluating $m$ (capped at B2 in the mark scheme).
Sign errors when distributing $k = -18$ into the factors; the final answer must give $-15$ at $x = 0$ .

Things to Be Careful About

The answer must be fully factorised with integer coefficients in every linear factor — the form $-18(x+\frac{1}{3})(x-1)(x-\frac{5}{2})$ scores only partial credit per the scheme.
Check your final expression reproduces $(0, -15)$ before moving on.

Techniques used

read the roots of the cubic from its graphwrite the cubic as a product of linear factors with an unknown scale factordetermine the scale factor using the y-intercept

(b)

Write down the values of $x$ such that $\mathrm{f}(x) < 0$ .

DifficultyEasy

Worked solution

Approach

$\mathrm{f}(x) < 0$ wherever the curve is below the $x$ -axis. Read those intervals straight off the sketch.

Working

From , the curve is below the $x$ -axis between the first two roots and again after the third root:

-\frac{1}{3} < x < 1 \quad \text{and} \quad x > \frac{5}{2}

Strict inequalities are used because $\mathrm{f}(x) = 0$ at the roots themselves, which is not less than zero.

Answer

-\frac{1}{3} < x < 1 \quad \text{or} \quad x > \frac{5}{2}

Final answer

-1/3 < x < 1 or x > 5/2

Detailed explanation

Walkthrough

An inequality like $\mathrm{f}(x) < 0$ asks: for which $x$ is the curve below the horizontal axis? Looking at the sketch, the curve dips below the axis immediately after crossing at $x = -\frac{1}{3}$ , stays below through the minimum near $y = -15$ , and comes back above the axis at $x = 1$ . That gives the interval $-\frac{1}{3} < x < 1$ . Then the curve rises to its local maximum and falls again, crossing the axis at $x = \frac{5}{2}$ and staying below thereafter, giving $x > \frac{5}{2}$ . Both pieces together form the complete answer. Strict inequality signs are correct because at the roots the function equals exactly zero, which does not satisfy "less than zero".

Key Takeaways

Solving $\mathrm{f}(x) < 0$ graphically means identifying the intervals where the curve sits below the $x$ -axis.
Roots are excluded when the inequality is strict ( $<$ or $>$ ).

Common Mistakes

Using $\leq$ instead of $<$ at the endpoints — the function is zero there, not negative.
Including the middle interval $1 < x < \frac{5}{2}$ , where the curve is actually above the axis.
Giving answers in terms of $y$ or describing points rather than writing intervals in terms of $x$ — the mark scheme explicitly requires the answer in terms of $x$ .
Missing the second interval beyond $x = \frac{5}{2}$ entirely.

Things to Be Careful About

Both intervals must be stated; each carries its own B1 mark.
Write the answer using strict inequalities in terms of $x$ , matching the scheme exactly.

Techniques used

read intervals where a curve lies below the x-axis from its graphexpress solution sets of inequalities in terms of x

The rest of this paper

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Q9Equations, inequalities and graphs4M
Q10Circular measure · Trigonometry7M
Q11Vectors in two dimensions9M
Q12Calculus6M

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