Mathematics 9709/22 — February/March 2024

Walkthrough

First, we identify the vertex of the modulus graph $y = |3x - 7|$. The vertex occurs where the expression inside the modulus is zero, giving $3x - 7 = 0$, so $x = 7/3$. This is the x-intercept.

Next, we find the y-intercept by substituting $x = 0$ into the equation: $y = |3(0) - 7| = |-7| = 7$. This gives the point $(0, 7)$.

The graph of $y = |3x - 7|$ is V-shaped. The right arm ($x \geq 7/3$) has gradient 3, and the left arm ($x < 7/3$) has gradient -3. We sketch this V-shape with the vertex at $(7/3, 0)$ and the left arm passing through $(0, 7)$.

Key Takeaways

• The graph of $y = |ax + b|$ is V-shaped with its vertex on the x-axis at $x = -b/a$.
• The y-intercept is found by evaluating $|b|$.
• The two arms have gradients $a$ and $-a$.

Common Mistakes

• Forgetting to take the absolute value when finding the y-intercept (getting $-7$ instead of $7$).
• Incorrectly placing the vertex on the y-axis instead of the x-axis.

Things to Be Careful About

• Ensure the sketch clearly shows the V-shape and the correct intercepts. The mark scheme awards marks for the shape and the coordinates.

Walkthrough

The equation $|3x - 7| = k(x - 4)$ can be interpreted graphically as finding the intersection points of $y = |3x - 7|$ and $y = k(x - 4)$.

The graph $y = |3x - 7|$ is the V-shaped graph from part (a), with vertex at $(7/3, 0)$, left arm gradient $-3$, and right arm gradient $3$.

The equation $y = k(x - 4)$ represents a straight line with gradient $k$ passing through the fixed point $(4, 0)$. Since $(4, 0)$ is on the x-axis to the right of the vertex, we analyse how the line intersects the V-shape.

For the line to intersect the V-shape in exactly two points, it must cross both arms. Since the V-shape is above or on the x-axis, and the line passes through $(4, 0)$ on the x-axis, the line must go upwards to the left (negative gradient) to intersect the left arm. Thus $k < 0$.

The left arm has gradient $-3$. If the line's gradient $k$ is less than or equal to $-3$ (i.e., $k \leq -3$), the line is parallel to or steeper than the left arm, meaning it will not intersect the left arm (or will only touch at the vertex). For the line to intersect the left arm, it must be less steep than the left arm, so $k > -3$.

If $k = 0$, the line is the x-axis, intersecting only at the vertex (one root). If $k > 0$, the line goes downwards to the left, where the V-shape is positive, so no intersection occurs for $x < 4$. Thus $k < 0$.

Combining these conditions, we get $-3 < k < 0$.

Key Takeaways

• Interpreting equations as intersections of graphs is a powerful technique for solving modulus equations.
• Fixed points on lines (like $(4, 0)$ here) help visualise the family of lines.
• Comparing gradients determines the number of intersections.

Common Mistakes

• Forgetting that the line must have a negative gradient to intersect both arms.
• Including $k = -3$ or $k = 0$ in the answer, which give fewer than two roots.

Things to Be Careful About

• Use strict inequalities ($<$) not inclusive inequalities ($\leq$), as the boundary cases give exactly one root.
• Ensure the reasoning clearly explains why $k$ must be between $-3$ and $0$.

Walkthrough

We are told that $(2x - 1)$ is a factor of $p(x)$. By the factor theorem, if $(2x - 1)$ is a factor then the corresponding root is $x = \frac{1}{2}$ and $p\left(\frac{1}{2}\right) = 0$. Substituting $x = \frac{1}{2}$ into the cubic gives an equation involving only $a$. Simplify the fractions and solve to find $a = 31$. Then replace $a$ in $p(x)$ and divide by the known factor $(2x - 1)$. The quotient is a quadratic, which we factorise into two linear factors. This gives the complete factorisation.

Key Takeaways

• The factor theorem states that if $(mx - n)$ is a factor of a polynomial $f(x)$, then $f\left(\frac{n}{m}\right) = 0$.
• Polynomial division by a linear factor reduces a cubic to a quadratic.
• A cubic with three linear factors can be written as the product of three binomials.

Common Mistakes

• Forgetting to set $p\left(\frac{1}{2}\right) = 0$; substituting without equating to zero gives no equation for $a$.
• Arithmetic errors when combining fractions such as $\frac{3}{4}$, $\frac{3}{2}$ and $10$.
• Sign errors during polynomial division, especially when subtracting.
• Stopping after finding the quadratic factor instead of factorising it completely.

Things to Be Careful About

• The factor is $(2x - 1)$, so the root is $x = \frac{1}{2}$, not $x = -\frac{1}{2}$ or $x = 1$.
• When dividing, keep the divisor as $2x - 1$ and check the quotient by multiplying back.
• Include the constant term $-10$ correctly in the division.

Walkthrough

Use the factorised form of $p(x)$ from part (a). Replacing $x$ with $\operatorname{cosec} \theta$ means each factor can be zero, giving three possible values for $\operatorname{cosec} \theta$: $\frac{1}{2}$, $-\frac{2}{3}$, and $-5$. However, $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$, and since $\sin \theta$ is always between $-1$ and $1$, the reciprocal must have magnitude at least $1$. Therefore $\frac{1}{2}$ and $-\frac{2}{3}$ are impossible. Only $\operatorname{cosec} \theta = -5$ remains. Solving $\sin \theta = -\frac{1}{5}$ in the range $-90^\circ < \theta < 90^\circ$ gives $\theta = -11.5^\circ$.

Key Takeaways

• The range of $\operatorname{cosec} \theta$ is $(-\infty, -1] \cup [1, \infty)$.
• When solving $p(\operatorname{cosec} \theta) = 0$, check which roots are attainable by $\operatorname{cosec} \theta$.
• Inverse sine returns the principal value in $[-90^\circ, 90^\circ]$, which matches the given range.

Common Mistakes

• Trying to solve $\sin \theta = 2$ or $\sin \theta = -\frac{3}{2}$ from invalid roots.
• Forgetting that $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$, so writing $\theta = \operatorname{cosec}^{-1}(-5)$ without converting.
• Giving only the positive angle or missing the negative sign.

Things to Be Careful About

• The range $-90^\circ < \theta < 90^\circ$ excludes the endpoints, but our answer lies inside the interval.
• Ensure the calculator is in degrees when evaluating $\sin^{-1}\left(-\frac{1}{5}\right)$.
• Use the correct root from part (a); only $-5$ is valid because $|\operatorname{cosec} \theta| \ge 1$.

Walkthrough

The trapezium rule approximates the area under a curve by replacing small sections of the curve with straight-line segments (trapeziums). With 3 intervals across $[0, 6]$, each interval has width $h = \frac{6 - 0}{3} = 2$. The $x$-values are therefore $0, 2, 4, 6$.

At each of these $x$-values we substitute into $y = \sqrt{1 + e^{0.5x}}$. Notice that $0.5 \times 0 = 0$, so $e^0 = 1$, giving $\sqrt{2}$. At $x = 2$, we get $e^1 = e$, so $y_1 = \sqrt{1 + e}$. At $x = 4$, $e^2$, and at $x = 6$, $e^3$. These exact forms are useful for the B1 mark; the mark scheme also accepts decimal equivalents.

The trapezium rule formula is $\frac{h}{2}[y_0 + 2(y_1 + y_2) + y_3]$, with the end values unweighted and the interior values doubled. Since $h = 2$, the $\frac{h}{2}$ factor is $1$, which simplifies the calculation considerably — we just sum the four values with the interior ones doubled.

Substituting the decimal approximations gives $1.41421 + 9.64862 + 4.59200 = 15.65483$, which rounds to $15.7$ to 3 significant figures.

Key Takeaways

• The trapezium rule approximates the area under a curve using straight-line segments.
• For $n$ intervals across $[a, b]$, the width is $h = \frac{b - a}{n}$ and the formula is $\frac{h}{2}[y_0 + 2(y_1 + \cdots + y_{n-1}) + y_n]$.
• The end ordinates are unweighted, but every interior ordinate is doubled.
• Always keep more digits during calculation and round only at the end to the required precision.

Common Mistakes

• Forgetting to double the interior $y$-values (an easy slip when $h/2$ happens to equal $1$).
• Using the wrong number of intervals — with 3 intervals you need 4 $y$-values, not 3.
• Computing $e^{0.5 \cdot 2}$ incorrectly: the exponent is $1$, not $0.5 \cdot 2$ written as $e^{0.5} \cdot e^{2}$.

Things to Be Careful About

• Make sure to read the exact form of each $y$-value correctly, since $0.5 \times 2 = 1$, $0.5 \times 4 = 2$, and $0.5 \times 6 = 3$.
• 3 significant figures means 3 meaningful digits, so $15.65483 \rightarrow 15.7$, not $15.65$.
• Keep the unrounded sum until the very last step to avoid premature rounding error.

Walkthrough

The standard formula for the volume of the solid obtained by rotating the region under a curve $y = f(x)$ from $x = a$ to $x = b$ about the $x$-axis is $V = \pi \int_a^b y^2 \, dx$. Each thin vertical strip of width $dx$ becomes a disc of radius $y$ and area $\pi y^2$, and these discs are stacked to form the solid.

Squaring the given function simplifies the integral significantly: $y^2 = \left(\sqrt{1 + e^{0.5x}}\right)^2 = 1 + e^{0.5x}$. This avoids the awkward $\sqrt{\phantom{x}}$ that would otherwise appear under the integral sign.

We integrate two terms. The first, $\int 1 \, dx = x$, is straightforward. The second uses the standard result $\int e^{kx} \, dx = \frac{1}{k} e^{kx}$. Here $k = 0.5 = \frac{1}{2}$, so the integral of $e^{0.5x}$ is $\frac{1}{0.5} e^{0.5x} = 2e^{0.5x}$. This gives the antiderivative $x + 2e^{0.5x}$.

We now evaluate between the limits. At $x = 6$: $6 + 2e^{0.5 \times 6} = 6 + 2e^3$. At $x = 0$: $0 + 2e^0 = 2$. The difference is $4 + 2e^3$, and multiplying by the outside $\pi$ gives $4\pi + 2\pi e^3$.

Key Takeaways

• The volume formula $V = \pi \int_a^b y^2 \, dx$ is essential for solids of revolution about the $x$-axis.
• Squaring the integrand before integrating can dramatically simplify the calculation.
• For $\int e^{kx} \, dx$, divide by the coefficient $k$ in the exponent.

Common Mistakes

• Forgetting the $\pi$ factor in the volume formula.
• Integrating $e^{0.5x}$ as just $e^{0.5x}$ instead of $\frac{1}{0.5} e^{0.5x} = 2e^{0.5x}$.
• Sign errors when subtracting the lower limit: remember it is $(\text{upper}) - (\text{lower})$.
• Forgetting to multiply by $\pi$ at the end if it was omitted during integration — though the mark scheme accepts $\pi$ introduced at the end.

Things to Be Careful About

• The exact answer must be left in terms of $e$ and $\pi$; do not approximate.
• Check the integration constant: the lower limit gives $0 + 2e^0 = 2$, not $0$.
• Equivalently, the answer can be written as $2\pi(2 + e^3)$ — the mark scheme allows exact equivalents.

Walkthrough

We are given the curve $y = \frac{x^3}{x+2}$ and told that at point $P$ the gradient is 6. We need to show that the $x$-coordinate of $P$ satisfies $x = \sqrt[3]{12x+12}$.

Step 1: Differentiate using the quotient rule. With $u = x^3$ and $v = x+2$, we have $u' = 3x^2$ and $v' = 1$, so:

Expanding the numerator gives $3x^2(x+2) - x^3 = 3x^3 + 6x^2 - x^3 = 2x^3 + 6x^2$, so:

Step 2: Equate to the given gradient. Setting the derivative equal to 6:

Step 3: Rearrange. Multiplying both sides by $(x+2)^2$ and expanding the right-hand side gives:

The $6x^2$ terms cancel on each side, leaving $2x^3 = 24x + 24$. Dividing by 2 gives:

Step 4: Take the cube root. The cube root is the inverse of cubing, so:

This is the required equation.

Key Takeaways

• The quotient rule is the key tool for differentiating a quotient of two functions.
• Setting the derivative equal to a known gradient and rearranging algebraically is the standard way to find a specific point on a curve.
• The cube root is the inverse of cubing, used here to rewrite the equation in a form suitable for iteration.

Common Mistakes

• Forgetting the minus sign in the quotient rule: it is $\frac{u'v - uv'}{v^2}$, not $\frac{u'v + uv'}{v^2}$.
• Errors expanding $(x+2)^2$ as $x^2 + 4x + 4$.
• Failing to cancel the $6x^2$ terms on both sides after expanding.
• Trying to divide by $x$ at some point, which would be invalid (the root is not $x = 0$).

Things to Be Careful About

• The quotient rule must be applied correctly, with the minus sign between $u'v$ and $uv'$.
• All intermediate algebraic steps must be shown, since the question asks to "show that" the equation holds.

Walkthrough

We use the sign-change (intermediate value) method to locate the root.

Step 1: Define a continuous function whose root we seek. From part (a), the $x$-coordinate of $P$ satisfies $x = \sqrt[3]{12x+12}$. Rearranging gives $x - \sqrt[3]{12x+12} = 0$, so define:

The function $g$ is continuous because it is built from continuous operations (addition, multiplication, cube root).

Step 2: Evaluate $g$ at $x = 3.8$. Compute $\sqrt[3]{12(3.8) + 12} = \sqrt[3]{57.6}$. To estimate, find values whose cubes bracket $57.6$: $3.86^3 = 57.51$ and $3.87^3 = 57.96$. So $\sqrt[3]{57.6}$ lies just above $3.86$, approximately $3.8620$. Thus:

Step 3: Evaluate $g$ at $x = 4.0$. Compute $\sqrt[3]{12(4.0) + 12} = \sqrt[3]{60}$. Bracketing: $3.91^3 = 59.78$ and $3.92^3 = 60.24$, so $\sqrt[3]{60} \approx 3.9149$. Thus:

Step 4: Apply the intermediate value theorem. Since $g$ is continuous and changes sign on the interval $[3.8, 4.0]$, there is a root in the open interval $(3.8, 4.0)$.

Key Takeaways

• The sign-change method relies on the intermediate value theorem: a continuous function that changes sign over an interval must have a root there.
• The choice $g(x) = x - \sqrt[3]{12x+12}$ is natural because the original equation is $x = \sqrt[3]{12x+12}$.
• Estimating cube roots can be done by bracketing: find $a, b$ with $a^3 < \text{target} < b^3$ and interpolate.

Common Mistakes

• Computing only $\sqrt[3]{12x+12}$ without subtracting $x$ to obtain a signed value.
• Computing the cube root inaccurately, leading to wrong sign.
• Failing to invoke the sign-change argument (or intermediate value theorem) to justify the conclusion.
• Using the equivalent function $h(x) = x^3 - 12x - 12$ without showing the sign change at both endpoints.

Things to Be Careful About

• The mark scheme accepts approximate values for $g(3.8)$ and $g(4.0)$ (e.g., $-0.06...$ and $0.08...$), but the sign change must be clear.
• The conclusion follows from continuity of $g$, which is automatic here.
• $12x+12$ is positive for both $x = 3.8$ and $x = 4.0$, so the real cube root is well-defined.

Walkthrough

We use the equation $x = \sqrt[3]{12x+12}$ from part (a) to construct an iterative formula.

Step 1: Construct the iterative formula. Set $F(x) = \sqrt[3]{12x+12}$ and define:

Step 2: Choose a starting value. From part (b), the root lies between 3.8 and 4.0. Start with $x_0 = 3.8$ (any value in the interval works).

Step 3: Iterate. Each step substitutes the previous iterate into the formula. The values are computed to 5 significant figures:

Step 4: Round to 3 significant figures. The sequence converges to $3.8845$, which rounds to $3.88$ to 3 significant figures. From $x_2$ onwards, each iterate rounds to $3.88$ at 3 sig fig, justifying the answer.

Key Takeaways

• The iterative formula $x_{n+1} = F(x_n)$ allows numerical solution of equations that cannot be solved algebraically.
• Each iteration should be computed to 5 significant figures (or more) to ensure accuracy of the rounded final answer.
• Iteration should continue until successive values agree to the required precision.