Mathematics 9709/21 — October/November 2025

Walkthrough

Start by recognising that the absolute value of two expressions is equal when the expressions themselves are equal or when they are negatives of one another. In Case 1, set $2x - 3 = 5x + 2$, solve for $x$ and obtain $x = -\frac{5}{3}$. In Case 2, set $2x - 3 = -(5x + 2)$, which gives $x = \frac{1}{7}$. Both values can be substituted back into the original modulus equation to confirm they work. This case-splitting method avoids the need to square both sides and keeps the working simple.

Key Takeaways

A modulus equation of the form $|A| = |B|$ can be solved by writing $A = B$ or $A = -B$. Each linear equation gives a candidate solution. This is one of the core techniques for solving modulus equations.

Common Mistakes

A common error is to solve only $2x - 3 = 5x + 2$ and forget the opposite-sign case $2x - 3 = -(5x + 2)$. Another common error is to make a sign error when expanding $-(5x + 2)$. It is also easy to state approximate decimals when exact fractions are expected.

Things to Be Careful About

Always use both modulus cases. After solving, check each candidate by substitution, although for simple linear cases and no squaring both candidates are normally valid. Keep the answer in exact form, $x = -\frac{5}{3}$ or $x = \frac{1}{7}$, unless an approximation is requested.

Walkthrough

The original equation in part (b) is exactly the same modulus equation as in part (a) if we set $x = \sec\theta$. Therefore the possible values of $\sec\theta$ are $\sec\theta = -\frac{5}{3}$ and $\sec\theta = \frac{1}{7}$. Since $\sec\theta = \frac{1}{\cos\theta}$, these become $\cos\theta = -\frac{3}{5}$ and $\cos\theta = 7$. The second is impossible because cosine must lie between $-1$ and $1$. For $\cos\theta = -\frac{3}{5}$, take the reference angle $\alpha = \cos^{-1}\left(\frac{3}{5}\right) \approx 0.9273$ radians. In the interval $\pi < \theta < 2\pi$, cosine is negative only in the third quadrant, so the required angle is $\theta = \pi + \alpha$, which is approximately $4.0688$ radians, rounded to $4.07$.

Key Takeaways

When an equation is obtained by substituting a trigonometric function for a variable, the same algebraic roots are used. The reciprocal identity $\sec\theta = \frac{1}{\cos\theta}$ changes the equation into a standard cosine equation. A solution such as $\cos\theta = 7$ must be rejected using the range of cosine. The interval $\pi < \theta < 2\pi$ determines which quadrant gives the final angle.

Common Mistakes

A common mistake is to try to solve $\cos\theta = 7$ instead of rejecting it. Another is to forget that the interval is $\pi < \theta < 2\pi$ and give the principal solution $2.21$ radians, which is not in the interval. Some students also finish with an exact expression instead of a value correct to 3 significant figures, or give the angle in degrees. The mark scheme condones degrees only for the process mark, but the final answer must be in radians.

Things to Be Careful About

Use the result from part (a) correctly: $x$ stands for $\sec\theta$, so the candidates are $\sec\theta = -\frac{5}{3}$ and $\sec\theta = \frac{1}{7}$. Check the range of $\cos\theta$ before continuing. Determine the correct quadrant: in $\pi < \theta < 2\pi$, $\cos\theta$ is negative only in the third quadrant. Round only at the end, to 3 significant figures, giving $4.07$.

Walkthrough

Point $B$ is the $x$-intercept of the curve, so we need the value of $x$ that makes $y = 0$. We begin by setting the equation $8e^{-\frac{1}{2}x} - 1 = 0$ and rearranging until the exponential stands alone on one side. This gives $e^{-\frac{1}{2}x} = \frac{1}{8}$. To peel away the exponential we take the natural logarithm of both sides; the rule $\ln(e^u) = u$ lets us strip the left side down to $-\frac{1}{2}x$, and on the right we have $\ln\frac{1}{8} = -\ln 8$. Multiplying through by $-2$ then gives $x = 2\ln 8$. Finally, the power law for logarithms, $\ln(2^3) = 3\ln 2$, converts $\ln 8$ to $3\ln 2$, so $x = 6\ln 2$.

Key Takeaways

• An $x$-intercept is found by setting $y = 0$ in the curve's equation.
• Isolating $e^{f(x)}$ and then taking $\ln$ of both sides is the standard way to solve an exponential equation.
• The law $\ln(a^n) = n\ln a$ is used to tidy up integer powers inside the logarithm.

Common Mistakes

• Forgetting to apply the minus sign correctly when writing $\ln\frac{1}{8} = -\ln 8$ instead of $\ln 8$.
• Forgetting the power-of-2 step and leaving the answer as $2\ln 8$ rather than the required $6\ln 2$.
• Dividing by $-\frac{1}{2}$ incorrectly (e.g. dropping the sign) and obtaining a negative $x$.

Things to Be Careful About

• Because the question says "show that", the working must explicitly display the use of $\ln 8 = 3\ln 2$ (or an equivalent such as $64 = 2^6$) — the mark scheme requires this step to be visible for full credit.
• Make sure the chain of equals signs is complete: $-\frac{1}{2}x = \ln\frac{1}{8} = -\ln 8 = -3\ln 2$ is the cleanest form to write.

Walkthrough

First identify the two vertices of the shaded region. Point $A$ is the $y$-intercept: substitute $x = 0$ into the curve to get $y = 8 \cdot 1 - 1 = 7$, so $A = (0,7)$. Point $B$ was found in part (a) as $(6\ln 2, 0)$. The straight line $AB$ together with the two coordinate axes encloses a right-angled triangle $OAB$ whose legs are $7$ and $6\ln 2$, giving triangle area $\frac{1}{2} \cdot 6\ln 2 \cdot 7 = 21\ln 2$.

The shaded region is the small lens-shaped area between the chord $AB$ and the curve. Because the curve $y = 8e^{-x/2} - 1$ is convex (it bends upward), it lies below the straight line $AB$ on $(0, 6\ln 2)$. Therefore the shaded area is the area of triangle $OAB$ minus the area between the curve and the $x$-axis. The latter is found by definite integration of $8e^{-x/2} - 1$ from $0$ to $6\ln 2$.

The integral $\int 8e^{-x/2}\, dx$ uses the standard rule $\int e^{ax}\, dx = \frac{1}{a}e^{ax}$ with $a = -\frac{1}{2}$, giving $-16e^{-x/2}$; combined with $-x$ the antiderivative is $-16e^{-x/2} - x$. Substituting the upper limit $6\ln 2$ produces $e^{-3\ln 2} = e^{\ln(1/8)} = \frac{1}{8}$, so that term becomes $-2$. Combined with $-6\ln 2$ and subtracting the lower-limit value $0 - (-16) = 16$, the area under the curve evaluates to $14 - 6\ln 2$.

Subtracting: $21\ln 2 - (14 - 6\ln 2) = 27\ln 2 - 14$, which has the required form $p\ln 2 - q$ with $p = 27$ and $q = 14$.

Key Takeaways

• For a convex curve beneath a chord, the area between chord and curve equals (area under chord) $-$ (area under curve). Here "area under the chord" is the triangle area because the chord meets the axes.
• The standard integral $\int e^{ax}\, dx = \frac{1}{a}e^{ax}$ is the workhorse; remember the division by the coefficient $a$ in the exponent.
• $e^{n\ln a} = a^n$ (and $e^{-n\ln a} = a^{-n} = 1/a^n$) is the key simplification when limits involve $\ln$ expressions.

Common Mistakes

• Integrating $8e^{-x/2}$ as $8e^{-x/2}$ (forgetting the division by $-\frac{1}{2}$) and getting the wrong antiderivative.
• Treating the shaded area as the area under the curve rather than the area between the curve and the line $AB$, which would give the wrong sign.
• Evaluating $e^{-3\ln 2}$ incorrectly (e.g. as $-3\cdot 2$ instead of $\frac{1}{8}$).
• Forgetting to subtract the lower-limit contribution, dropping the $+16$ term and obtaining $14 - 6\ln 2$ as the final answer instead of just an intermediate value.

Things to Be Careful About

• The mark scheme explicitly states the integration step $-16e^{-x/2} - x$ and the substituted form $-16e^{-3\ln 2} - 6\ln 2 + 16$ must be shown; skipping these intermediate lines loses method marks.
• The final answer must be left in the form $p\ln 2 - q$ with positive integers $p$ and $q$ — a decimal answer is acceptable for working but not for the final box.
• The triangle-area mark (M1) may be earned with decimal equivalents (e.g. $14.56$) but the final simplification must be exact.

Walkthrough

We are given $x$ and $y$ in terms of the parameter $\theta$, so we first differentiate both with respect to $\theta$.

For $x = \tan\theta$, the derivative is $\sec^2\theta$.

For $y = \sin\theta - 2\sin^3\theta$, differentiate term by term. The derivative of $\sin\theta$ is $\cos\theta$. For $-2\sin^3\theta$, apply the chain rule:

So $\frac{dy}{d\theta} = \cos\theta - 6\sin^2\theta\cos\theta$.

The parametric formula is

Dividing by $\sec^2\theta$ is the same as multiplying by $\cos^2\theta$, which gives $\cos^3\theta(1 - 6\sin^2\theta)$. To write this in terms of $\cos\theta$ only, replace $\sin^2\theta$ with $1 - \cos^2\theta$. Expanding gives $6\cos^5\theta - 5\cos^3\theta$, as required.

Key Takeaways

This question tests parametric differentiation, the chain rule, and the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$. You should be comfortable differentiating $\tan\theta$, $\sin\theta$ and powers of $\sin\theta$, and then converting a parametric derivative into a single-variable expression.

Common Mistakes

• Forgetting to divide by $\frac{dx}{d\theta}$ when forming $\frac{dy}{dx}$.
• Differentiating $-2\sin^3\theta$ incorrectly, for example omitting the factor $3$ or the factor $\cos\theta$.
• Stopping at $\cos^3\theta(1 - 6\sin^2\theta)$ without using the identity to reach the requested form. Since this is a show that question, the mark scheme requires the necessary simplification detail.

Things to Be Careful About

• The derivative formulae for trigonometric functions assume $\theta$ is measured in radians.
• The range $0 < \theta < \frac{1}{2}\pi$ means $\cos\theta$ is positive, but the simplification here does not depend on that sign.
• Do not quote the final answer without showing the intermediate step $\cos^3\theta(6\cos^2\theta - 5)$.

Walkthrough

The curve crosses the $x$-axis where $y=0$. Factorise $y$ as $\sin\theta(1 - 2\sin^2\theta)$. Since $0<\theta<\frac{\pi}{2}$, $\sin\theta$ is never zero, so the only possibility is $1 - 2\sin^2\theta=0$. This gives $\sin^2\theta=\frac{1}{2}$, and in the given range $\theta=\frac{\pi}{4}$.

Substitute $\theta=\frac{\pi}{4}$ into $x=\tan\theta$ to get $x=1$, so the crossing point is $(1,0)$.

Use the derivative from part (a) to find the gradient of the curve at this point. Since $\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}$, the gradient is $6\left(\frac{1}{\sqrt{2}}\right)^5 - 5\left(\frac{1}{\sqrt{2}}\right)^3 = -\frac{1}{\sqrt{2}}$.

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal, $\sqrt{2}$. Finally, use the point-slope form with $(1,0)$ to get $y=\sqrt{2}x - \sqrt{2}$.

Key Takeaways

This question combines parametric differentiation with curve analysis: finding where a parametric curve crosses an axis, evaluating the gradient at that point, and using perpendicular gradients to find a normal.

Common Mistakes

• Forgetting that $\sin\theta\neq0$ in the given range and incorrectly taking $\theta=0$ as a solution.
• Using degrees for $\theta$ when substituting into the derivative. The mark scheme allows degrees when first solving $y=0$, but the final value must be converted to radians before differentiating.
• Using the curve gradient instead of its negative reciprocal for the normal.
• Forgetting to find the $x$-coordinate of the point.

Things to Be Careful About

• The exact value $\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}$ must be used carefully with powers: $\left(\frac{1}{\sqrt{2}}\right)^5=\frac{1}{4\sqrt{2}}$ and $\left(\frac{1}{\sqrt{2}}\right)^3=\frac{1}{2\sqrt{2}}$.
• The final equation must be in the form $y=mx+c$ with exact constants, so write $y=\sqrt{2}x - \sqrt{2}$ rather than a decimal approximation.
• The normal gradient is $\sqrt{2}$, not $-\sqrt{2}$.