Mathematics 9709 / 21 — May/June 2025

DifficultyMedium-Easy

Worked solution

Approach

Solve the boundary equation $|3x + 8| = 9$ to find the two critical values. Since the inequality is strict and less than, the solution is the open interval between these values.

Working

The boundary points satisfy:

3x + 8 = 9 \quad \text{or} \quad 3x + 8 = -9

First equation:

3x + 8 = 9 \Rightarrow 3x = 1 \Rightarrow x = \frac{1}{3}

Second equation:

3x + 8 = -9 \Rightarrow 3x = -17 \Rightarrow x = -\frac{17}{3}

Since $|3x + 8| < 9$ means the expression lies between $-9$ and $9$ :

-\frac{17}{3} < x < \frac{1}{3}

Answer

-\frac{17}{3} < x < \frac{1}{3}

Final answer

-17/3 < x < 1/3

Detailed explanation

Walkthrough

A modulus inequality $|u| < 9$ is equivalent to $-9 < u < 9$ . We first find where $3x + 8$ equals $\pm 9$ ; these are the endpoints of the interval. Solving the two linear equations gives $x = -\frac{17}{3}$ and $x = \frac{1}{3}$ . Because the inequality is strict, the solution is the open interval between these endpoints.

Key Takeaways

For $|ax + b| < c$ , solve $ax + b = \pm c$ and take the interval between the two solutions. The same boundary method also works for $|ax + b| \leq c$ , with closed endpoints.

Common Mistakes

A common mistake is to write two separate inequalities $3x + 8 < 9$ and $3x + 8 < -9$ instead of $\pm$ . Another common mistake is to use the wrong inequality direction or to include the endpoints when the inequality is strict.

Things to Be Careful About

Keep the exact fractions $\pm\frac{17}{3}$ and $\frac{1}{3}$ ; converting to decimals too early can lead to rounding errors. The final interval must be written with $x$ in the middle.

Techniques used

solve the modulus boundary equationcombine the two boundary values into an interval

(c)

Hence state the integers that satisfy both of the inequalities in parts (a) and (b).

1M Similar

Q3MediumLogarithmic and Exponential FunctionsTrigonometryNumerical Solution of Equations

(a)

Sketch, on a single diagram, the graphs of $y = 3e^{-2x}$ and $y = \sec x$ for values of $x$ such that $0 \le x < \frac{1}{2}\pi$ .

DifficultyMedium-Easy

Worked solution

Approach

Recognise the two curve shapes. The curve $y = 3e^{-2x}$ is a positive exponential decay: it starts at $(0,3)$ and tends to $0$ as $x$ increases. The curve $y = \sec x$ is the reciprocal of $\cos x$ : it starts at $(0,1)$ and tends to $+\infty$ as $x \to \frac{\pi}{2}^-$ , with a vertical asymptote at $x = \frac{\pi}{2}$ . Draw both on the same axes for $0 \le x < \frac{\pi}{2}$ .

Working

For $y = 3e^{-2x}$ :

At $x = 0$ , $y = 3$ .
As $x$ increases, $y$ decreases towards $0$ but never reaches it.

For $y = \sec x = \frac{1}{\cos x}$ :

At $x = 0$ , $y = 1$ .
As $x \to \frac{\pi}{2}^-$ , $\cos x \to 0^+$ , so $\sec x \to +\infty$ .
There is a vertical asymptote at $x = \frac{\pi}{2}$ .

The two curves cross once in the interval, near $x \approx 0.487$ .

Answer

A single sketch showing the decreasing exponential curve and the increasing secant curve, with the vertical asymptote at $x = \frac{\pi}{2}$ and one intersection point.

Final answer

Sketch of y = 3e^{-2x} and y = sec x on the same axes for 0 ≤ x < π/2, with one intersection near x = 0.487

Detailed explanation

Walkthrough

The question asks for a sketch, not an accurate plot. Start by recalling the two standard shapes.

For $y = 3e^{-2x}$ , the coefficient $3$ gives the value at $x=0$ : $y=3$ . The factor $-2$ in the exponent makes the curve decay rapidly: as $x$ increases, $3e^{-2x}$ gets closer and closer to $0$ but never touches the $x$ -axis. So the graph is a decreasing curve above the $x$ -axis.

For $y = \sec x$ , remember that $\sec x = \frac{1}{\cos x}$ . On $0 \le x < \frac{\pi}{2}$ , $\cos x$ starts at $1$ and decreases to $0$ ; therefore $\sec x$ starts at $1$ and increases without bound as $x$ approaches $\frac{\pi}{2}$ . This gives a vertical asymptote at $x = \frac{\pi}{2}$ .

Place both curves on the same axes. The exponential curve starts higher at $3$ , while the secant curve starts at $1$ and rises steeply near the asymptote. They cross once; the intersection is approximately at $x = 0.487$ .

Key Takeaways

Know the basic shape of $y = e^{kx}$ : growth if $k>0$ , decay if $k<0$ .
Know that $y = \sec x$ has vertical asymptotes where $\cos x = 0$ and is always at least $1$ in magnitude where defined.
A sketch should show intercepts, asymptotes and the general trend, not precise coordinates.

Common Mistakes

Drawing $y = 3e^{-2x}$ as if it crosses the $x$ -axis; it only approaches the axis.
Drawing $y = \sec x$ as a cosine wave; it is the reciprocal and has an asymptote.
Forgetting that the domain stops just before $\frac{\pi}{2}$ , so the asymptote is at the right-hand boundary.

Things to Be Careful About

Mark the point $(0,3)$ for the exponential curve and $(0,1)$ for the secant curve.
The vertical asymptote at $x = \frac{\pi}{2}$ should be clearly indicated, usually with a dashed line.
The curves must be on the same diagram, with the intersection shown once.

Techniques used

sketch an exponential decay curvesketch a secant curve with a vertical asymptoteidentify the intersection on a single set of axes

(b)

Show that the $x$ -coordinate of the point of intersection of the two graphs satisfies the equation

x = \frac{1}{2}\ln(3\cos x).

DifficultyMedium-Easy

Worked solution

Approach

At the point of intersection, set the two expressions equal. Use $\sec x = \frac{1}{\cos x}$ , rearrange to isolate $e^{2x}$ , then take natural logarithms.

Working

At the intersection:

3e^{-2x} = \sec x = \frac{1}{\cos x}

Divide by 3 and take the reciprocal:

e^{-2x} = \frac{1}{3\cos x}

so

e^{2x} = 3\cos x

Taking $\ln$ of both sides:

2x = \ln(3\cos x)

Therefore

x = \frac{1}{2}\ln(3\cos x)

as required.

Answer

x = \frac{1}{2}\ln(3\cos x)

Final answer

x = 1/2 ln(3 cos x)

Detailed explanation

Walkthrough

At the intersection, the two $y$ -values are equal, so write

3e^{-2x} = \sec x.

Replace $\sec x$ by $\frac{1}{\cos x}$ . This is the key reciprocal identity.

Now isolate the exponential term. Dividing by $3$ gives $e^{-2x} = \frac{1}{3\cos x}$ . Taking the reciprocal of both sides gives $e^{2x} = 3\cos x$ . Taking natural logarithms removes the exponential: $\ln(e^{2x}) = 2x$ , so $2x = \ln(3\cos x)$ . Finally divide by $2$ to obtain $x = \frac{1}{2}\ln(3\cos x)$ .

This is an "answer given" proof, so every rearrangement must be shown.

Key Takeaways

The intersection of two graphs is found by equating their equations.
$\sec x = \frac{1}{\cos x}$ .
To solve an equation involving $e^{ax}$ , isolate $e^{ax}$ and then take natural logarithms.

Common Mistakes

Taking logarithms before isolating $e^{2x}$ ; this leads to incorrect expressions.
Forgetting the factor $\frac{1}{2}$ when dividing by $2$ .
Writing $\ln(3\cos x)$ as $\ln 3 + \ln(\cos x)$ and then mishandling it; here it is left as a single logarithm.

Things to Be Careful About

This is a show-that question, so the final line must be reached with all intermediate algebra shown.
On the given interval, $\cos x > 0$ , so $3\cos x > 0$ and the logarithm is defined.
Use radian mode throughout.

Techniques used

equate the two graph expressionsuse the reciprocal identity for secanttake natural logarithms to solve for x

(c)

Use an iterative formula, based on the equation in part (b), to find the $x$ -coordinate of the point of intersection correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

DifficultyMedium

Worked solution

Approach

Use the equation from part (b) in the iterative form

x_{n+1} = \frac{1}{2}\ln(3\cos x_n)

with the calculator in radian mode. Start from a convenient value such as $x_0 = 0.5$ and iterate until successive values agree to 5 decimal places.

Working

Let

x_{n+1} = \frac{1}{2}\ln(3\cos x_n)

Starting with $x_0 = 0.5$ :

$n$	$x_n$
0	0.50000
1	0.48401
2	0.48830
3	0.48717
4	0.48747
5	0.48739
6	0.48741
7	0.48740
8	0.48740

Since $x_7$ and $x_8$ both round to $0.48740$ to 5 decimal places, the root is $x = 0.487$ correct to 3 decimal places.

Answer

x = 0.487 \quad (\text{3 decimal places})

Final answer

x = 0.487

Detailed explanation

Walkthrough

The equation from part (b) is already written as $x$ equal to an expression involving $x$ . This makes it natural to iterate:

x_{n+1} = \frac{1}{2}\ln(3\cos x_n).

Choose a starting value. A sensible choice is $x_0 = 0.5$ , because the intersection is near $0.5$ . Substitute $x_0$ into the right-hand side to get $x_1$ , then repeat.

Each iteration should be recorded to 5 decimal places so that convergence can be seen. The sequence settles at $0.48740$ to 5 decimal places, so the root is $0.487$ to 3 decimal places.

Key Takeaways

An equation of the form $x = F(x)$ can be solved by iteration $x_{n+1} = F(x_n)$ .
Recording values to more decimal places than required lets you see when the sequence has converged.
Two consecutive iterations agreeing to the required precision is a practical stopping criterion.

Common Mistakes

Using degree mode on the calculator; the formula uses radians because the original functions are in radians.
Quoting only the final answer $0.487$ without showing iterations; the mark scheme requires evidence.
Rounding each iteration to 3 decimal places too early, which can hide convergence or introduce error.

Things to Be Careful About

The final answer must be given to 3 decimal places: $0.487$ .
Intermediate iterations should be shown to 5 decimal places as requested.
If using a sign-change check instead, it must be in an interval such as $[0.4865, 0.4875]$ to justify 3 decimal places.

Techniques used

rearrange the equation into iterative formapply successive substitutionround successive approximations to the required precision

Q4MediumDifferentiationIntegration

Fig. 4.1

The diagram shows the curve with equation $y = 6e^{2x} - e^{3x}$ . The shaded region is bounded by the axes and the curve.

(a)

Find the exact $x$ -coordinate of the maximum point.

DifficultyMedium-Easy

Worked solution

Approach

Differentiate $y = 6e^{2x} - e^{3x}$ with respect to $x$ , set $\frac{dy}{dx} = 0$ , and solve for $x$ to find the x-coordinate of the maximum point.

Working

Differentiate using the chain rule:

\frac{dy}{dx} = 12e^{2x} - 3e^{3x}

Set the derivative equal to zero:

12e^{2x} - 3e^{3x} = 0

Factor out $3e^{2x}$ :

3e^{2x}(4 - e^x) = 0

Since $e^{2x} > 0$ for all real $x$ , the factor $3e^{2x}$ cannot be zero. Therefore:

4 - e^x = 0

e^x = 4

x = \ln 4

This can also be written as $x = 2\ln 2$ .

Answer

x = \ln 4

Final answer

x = ln 4

Detailed explanation

Walkthrough

Step 1: Differentiate. The function is $y = 6e^{2x} - e^{3x}$ . Using the chain rule, the derivative of $e^{kx}$ is $ke^{kx}$ . So $\frac{d}{dx}(6e^{2x}) = 12e^{2x}$ and $\frac{d}{dx}(e^{3x}) = 3e^{3x}$ , giving $\frac{dy}{dx} = 12e^{2x} - 3e^{3x}$ .

Step 2: Set derivative to zero. At a maximum (or minimum) point, the gradient is zero. So we solve $12e^{2x} - 3e^{3x} = 0$ .

Step 3: Factor. Both terms share a factor of $3e^{2x}$ . Factoring gives $3e^{2x}(4 - e^x) = 0$ .

Step 4: Solve. Since $e^{2x}$ is always positive and never zero, we must have $4 - e^x = 0$ , giving $e^x = 4$ , so $x = \ln 4$ .

Key Takeaways

The derivative of $e^{kx}$ is $ke^{kx}$ , found using the chain rule.
At a stationary point, $\frac{dy}{dx} = 0$ .
Exponential functions $e^{kx}$ are always positive, so they can be safely divided out or ignored when solving equations of the form $e^{kx} \cdot f(x) = 0$ .

Common Mistakes

Forgetting the chain rule and writing $\frac{dy}{dx} = 6e^{2x} - e^{3x}$ instead of $12e^{2x} - 3e^{3x}$ .
Attempting to solve $6e^{2x} - e^{3x} = 0$ (setting $y = 0$ ) instead of setting the derivative to zero.
Writing the answer as a decimal (e.g., $x = 1.386$ ) when the question asks for an exact answer.

Things to Be Careful About

The question asks for the exact x-coordinate, so leave the answer in logarithmic form: $\ln 4$ or $2\ln 2$ .
Always check that $e^{2x} \neq 0$ before concluding that the other factor must be zero.

Techniques used

differentiate exponential function using chain ruleset derivative equal to zero for stationary pointsolve exponential equation by factoring

(b)

Find the area of the shaded region. Give your answer in the form $\frac{p}{q}$ , where $p$ and $q$ are integers.

DifficultyMedium

Worked solution

Approach

The shaded region is bounded by the curve, the y-axis ( $x = 0$ ), and the x-axis. Find where the curve meets the x-axis to determine the upper limit of integration, then evaluate the definite integral $\int_0^{x_0} (6e^{2x} - e^{3x})\,dx$ .

Working

Find the x-intercept: Set $y = 0$ :

6e^{2x} - e^{3x} = 0

Factor out $e^{2x}$ :

e^{2x}(6 - e^x) = 0

Since $e^{2x} > 0$ , we have:

6 - e^x = 0 \implies e^x = 6 \implies x = \ln 6

Integrate:

\int (6e^{2x} - e^{3x})\,dx = 3e^{2x} - \frac{1}{3}e^{3x} + c

Apply the limits from $x = 0$ to $x = \ln 6$ :

\text{Area} = \left[3e^{2x} - \frac{1}{3}e^{3x}\right]_0^{\ln 6}

Upper limit ( $x = \ln 6$ ):

3e^{2\ln 6} - \frac{1}{3}e^{3\ln 6}

Using the index law $e^{k\ln a} = (e^{\ln a})^k = a^k$ :

3e^{2\ln 6} = 3 \cdot 6^2 = 3 \cdot 36 = 108

\frac{1}{3}e^{3\ln 6} = \frac{1}{3} \cdot 6^3 = \frac{1}{3} \cdot 216 = 72

So the upper limit gives $108 - 72 = 36$ .

Lower limit ( $x = 0$ ):

3e^0 - \frac{1}{3}e^0 = 3 - \frac{1}{3} = \frac{8}{3}

Subtract:

\text{Area} = 36 - \frac{8}{3} = \frac{108}{3} - \frac{8}{3} = \frac{100}{3}

Answer

\frac{100}{3}

Final answer

100/3

Detailed explanation

Walkthrough

Step 1: Find the x-intercept. The shaded region is bounded by the curve, the y-axis, and the x-axis. The y-axis corresponds to $x = 0$ . To find where the curve meets the x-axis, set $y = 0$ : $6e^{2x} - e^{3x} = 0$ . Factor out $e^{2x}$ to get $e^{2x}(6 - e^x) = 0$ . Since $e^{2x}$ is never zero, we solve $6 - e^x = 0$ , giving $e^x = 6$ and $x = \ln 6$ . This is our upper limit of integration.

Step 2: Integrate. The integral of $6e^{2x}$ is $3e^{2x}$ (divide by the coefficient of $x$ in the exponent). The integral of $e^{3x}$ is $\frac{1}{3}e^{3x}$ . So the antiderivative is $3e^{2x} - \frac{1}{3}e^{3x}$ .

Step 3: Evaluate at the upper limit. At $x = \ln 6$ , we need $e^{2\ln 6}$ and $e^{3\ln 6}$ . Using the identity $e^{k\ln a} = a^k$ , we get $e^{2\ln 6} = 6^2 = 36$ and $e^{3\ln 6} = 6^3 = 216$ . So $3(36) - \frac{1}{3}(216) = 108 - 72 = 36$ .

Step 4: Evaluate at the lower limit. At $x = 0$ , $e^0 = 1$ , so $3(1) - \frac{1}{3}(1) = 3 - \frac{1}{3} = \frac{8}{3}$ .

Step 5: Subtract. Area $= 36 - \frac{8}{3} = \frac{108 - 8}{3} = \frac{100}{3}$ .

Key Takeaways

To find where a curve meets the x-axis, set $y = 0$ and solve.
When integrating $e^{kx}$ , divide by $k$ : $\int e^{kx}\,dx = \frac{1}{k}e^{kx}$ .
The index law $e^{k\ln a} = a^k$ is essential for simplifying expressions like $e^{2\ln 6}$ .
Always check both limits of a definite integral; forgetting the lower limit is a common error.

Common Mistakes

Using the x-coordinate of the maximum point ( $\ln 4$ ) as the upper limit instead of the x-intercept ( $\ln 6$ ).
Forgetting to evaluate at the lower limit $x = 0$ and just computing the upper limit value.
Incorrectly simplifying $e^{2\ln 6}$ as $2e^6$ or $e^{12}$ instead of $6^2 = 36$ .
Writing the integral as $6e^{2x} - e^{3x}$ without dividing by the coefficients of $x$ in the exponents.

Things to Be Careful About

The question asks for the answer in the form $\frac{p}{q}$ where $p$ and $q$ are integers. $\frac{100}{3}$ satisfies this since $\gcd(100, 3) = 1$ .
The constant of integration $+c$ is not needed for definite integrals, but including it does not lose marks.
Ensure you use the x-intercept ( $x = \ln 6$ ) as the upper limit, not the x-coordinate of the maximum ( $x = \ln 4$ ) from part (a).

Techniques used

find x-intercept by solving exponential equationintegrate exponential functionapply upper and lower limits to definite integralsimplify powers of exponential expressions

Q5MediumAlgebraTrigonometry

The polynomial $p(x)$ is defined by

p(x) = ax^3 + bx^2 - ax - 24,

where $a$ and $b$ are constants. It is given that $(2x - 3)$ is a factor of $p(x)$ and that the remainder is $-15$ when $p(x)$ is divided by $(x + 1)$ .

(a)

Find the values of $a$ and $b$ .

DifficultyMedium

Worked solution

Approach

Use the factor theorem: since $(2x - 3)$ is a factor of $p(x)$ , substituting $x = \frac{3}{2}$ must give $p\left(\frac{3}{2}\right) = 0$ . Use the remainder theorem: since the remainder when $p(x)$ is divided by $(x + 1)$ is $-15$ , substituting $x = -1$ gives $p(-1) = -15$ . This gives two equations in $a$ and $b$ .

Working

Since $(2x - 3)$ is a factor:

p\left(\frac{3}{2}\right) = 0

a\left(\frac{3}{2}\right)^3 + b\left(\frac{3}{2}\right)^2 - a\left(\frac{3}{2}\right) - 24 = 0

\frac{27}{8}a + \frac{9}{4}b - \frac{3}{2}a - 24 = 0

Multiply by 8:

15a + 18b - 192 = 0

5a + 6b = 64

Since the remainder is $-15$ when divided by $(x + 1)$ :

p(-1) = -15

-a + b + a - 24 = -15

b - 24 = -15

b = 9

Substitute $b = 9$ into $5a + 6b = 64$ :

5a + 54 = 64

5a = 10

a = 2

Answer

a = 2, \quad b = 9

Final answer

a = 2, b = 9

Detailed explanation

Walkthrough

We are told that $(2x - 3)$ is a factor of $p(x)$ . By the factor theorem, this means $x = \frac{3}{2}$ is a root, so $p\left(\frac{3}{2}\right) = 0$ . Substituting this value into the polynomial gives the first equation. We are also told that the remainder is $-15$ when $p(x)$ is divided by $(x + 1)$ . By the remainder theorem, the remainder is $p(-1)$ , so $p(-1) = -15$ . Substituting $x = -1$ gives the second equation. Notice that the $a$ terms cancel in this second equation, leaving $b - 24 = -15$ , so $b = 9$ is found immediately. Then substitute $b = 9$ into the first simplified equation $5a + 6b = 64$ to find $a = 2$ .

Key Takeaways

The factor theorem links a factor $(ax - b)$ to the root $x = \frac{b}{a}$ .
The remainder theorem says that division by $(x - c)$ leaves remainder $p(c)$ .
Substituting known roots or remainders is a powerful way to set up equations for unknown coefficients.

Common Mistakes

Using $x = -\frac{3}{2}$ instead of $x = \frac{3}{2}$ for the factor $(2x - 3)$ .
Using $p(1)$ instead of $p(-1)$ for division by $(x + 1)$ .
Making sign errors when substituting $x = -1$ ; the correct substitution gives $-a + b + a - 24$ .
Stopping before simplifying the first equation; the mark scheme requires the powers of $\frac{3}{2}$ to be evaluated.

Things to Be Careful About

The factor $(2x - 3)$ gives root $x = \frac{3}{2}$ , not $x = -\frac{3}{2}$ .
The divisor $(x + 1)$ gives remainder $p(-1)$ , not $p(1)$ .
If using algebraic long division instead of substitution, you must still end with two expressions correctly equated to $0$ and $-15$ .
An unsupported value of $b = 9$ may only receive partial credit if no method is shown.

Techniques used

apply the factor theoremapply the remainder theoremsolve simultaneous equations

(b)

Hence factorise $p(x)$ completely.

DifficultyMedium-Easy

Worked solution

Approach

Use the known factor $(2x - 3)$ to divide $p(x) = 2x^3 + 9x^2 - 2x - 24$ . The quotient will be a quadratic, which can then be factorised completely.

Working

Divide the first term:

\frac{2x^3}{2x} = x^2

Multiply back and subtract:

x^2(2x - 3) = 2x^3 - 3x^2

(2x^3 + 9x^2) - (2x^3 - 3x^2) = 12x^2

Bring down $-2x$ . Divide the next term:

\frac{12x^2}{2x} = 6x

Multiply back and subtract:

6x(2x - 3) = 12x^2 - 18x

(12x^2 - 2x) - (12x^2 - 18x) = 16x

Bring down $-24$ . Divide the next term:

\frac{16x}{2x} = 8

Multiply back and subtract:

8(2x - 3) = 16x - 24

(16x - 24) - (16x - 24) = 0

So the quotient is $x^2 + 6x + 8$ . Factorise this quadratic:

x^2 + 6x + 8 = (x + 2)(x + 4)

Therefore:

p(x) = (2x - 3)(x + 2)(x + 4)

Answer

p(x) = (2x - 3)(x + 2)(x + 4)

Final answer

p(x) = (2x - 3)(x + 2)(x + 4)

Detailed explanation

Walkthrough

After finding $a = 2$ and $b = 9$ , the polynomial is $p(x) = 2x^3 + 9x^2 - 2x - 24$ . Since $(2x - 3)$ is a factor, dividing by it should produce a quadratic quotient with no remainder. The long division is performed term by term: divide the leading term by $2x$ , multiply back, subtract, and bring down the next term. This gives the quotient $x^2 + 6x + 8$ . Finally, factorise the quadratic as $(x + 2)(x + 4)$ , so the complete factorisation is $(2x - 3)(x + 2)(x + 4)$ .

Key Takeaways

Polynomial division reduces a cubic with a known linear factor to a quadratic.
The quotient can then be factorised by inspection or by the quadratic formula.
A complete factorisation must include all linear factors.

Common Mistakes

Sign errors when subtracting after multiplying back.
Forgetting to bring down the next term during long division.
Stopping at the quadratic quotient without factorising it.
If using synthetic division with root $\frac{3}{2}$ , the quotient is $2x^2 + 12x + 16$ ; remember to factor out the $2$ and combine with $(x - \frac{3}{2})$ to obtain $(2x - 3)(x^2 + 6x + 8)$ .

Things to Be Careful About

The mark scheme requires the division to go as far as the term in $x$ ; show enough working.
Check the final factorisation by expanding: $(2x - 3)(x + 2)(x + 4)$ should return $2x^3 + 9x^2 - 2x - 24$ .
One sign error in the division may still earn partial credit, but the final factors must be correct for full marks.

Techniques used

divide polynomial by a linear factorfactorise quadratic quotient

(c)

Hence solve the equation $p(3\operatorname{cosec}\theta) = 0$ for $90^\circ < \theta < 270^\circ$ .

DifficultyMedium

Worked solution

Approach

Use the factorised form of $p(x)$ and substitute $x = 3\operatorname{cosec}\theta = \frac{3}{\sin\theta}$ . Set each factor equal to zero and solve for $\sin\theta$ , rejecting any value outside $-1 \leq \sin\theta \leq 1$ . Then find the angle in the given range.

Working

Since

p(x) = (2x - 3)(x + 2)(x + 4),

we need

(2(3\operatorname{cosec}\theta) - 3)(3\operatorname{cosec}\theta + 2)(3\operatorname{cosec}\theta + 4) = 0.

First factor:

6\operatorname{cosec}\theta - 3 = 0 \quad \Rightarrow \quad \frac{6}{\sin\theta} = 3 \quad \Rightarrow \quad \sin\theta = 2

This is impossible since $\sin\theta$ must lie between $-1$ and $1$ .

Second factor:

3\operatorname{cosec}\theta + 2 = 0 \quad \Rightarrow \quad \frac{3}{\sin\theta} = -2 \quad \Rightarrow \quad \sin\theta = -\frac{3}{2}

This is also impossible.

Third factor:

3\operatorname{cosec}\theta + 4 = 0 \quad \Rightarrow \quad \frac{3}{\sin\theta} = -4 \quad \Rightarrow \quad \sin\theta = -\frac{3}{4}

This is valid. For $90^\circ < \theta < 270^\circ$ , $\sin\theta$ is negative in the third quadrant, so

\theta = 180^\circ + \arcsin\left(\frac{3}{4}\right) \approx 228.6^\circ.

The other solution, $360^\circ - \arcsin\left(\frac{3}{4}\right) \approx 311.4^\circ$ , is outside the range.

Answer

\theta \approx 228.6^\circ \quad (228.590^\circ \text{ to 3 decimal places})

Final answer

θ ≈ 228.6° (228.590°)

Detailed explanation

Walkthrough

We start from the complete factorisation of $p(x)$ . Substituting $x = 3\operatorname{cosec}\theta$ means each factor must be considered separately. Since $\operatorname{cosec}\theta = \frac{1}{\sin\theta}$ , each equation becomes an equation in $\sin\theta$ . The values $\sin\theta = 2$ and $\sin\theta = -\frac{3}{2}$ are impossible because the sine function only takes values between $-1$ and $1$ . Only $\sin\theta = -\frac{3}{4}$ is possible. The range $90^\circ < \theta < 270^\circ$ includes the second and third quadrants; sine is negative only in the third quadrant, so the reference angle $\arcsin\left(\frac{3}{4}\right) \approx 48.6^\circ$ must be added to $180^\circ$ . The solution in the fourth quadrant, $311.4^\circ$ , is outside the range and is ignored.

Key Takeaways

A product is zero only when one of its factors is zero.
$\operatorname{cosec}\theta = \frac{1}{\sin\theta}$ , so equations involving cosecant can be rewritten in terms of sine.
The range of $\sin\theta$ is $[-1, 1]$ , so any solution outside this interval must be rejected.
The quadrant determines how to convert a reference angle into the required solution.

Common Mistakes

Forgetting to reject $\sin\theta = 2$ or $\sin\theta = -\frac{3}{2}$ .
Trying to solve $3\operatorname{cosec}\theta = -4$ without converting to $\sin\theta$ .
Including $311.4^\circ$ , which is outside $90^\circ < \theta < 270^\circ$ .
Using the calculator's negative principal value $-48.6^\circ$ directly instead of adding $180^\circ$ .

Things to Be Careful About

The question is in degrees, so keep the calculator in degree mode.
The mark scheme accepts $228.6^\circ$ or greater accuracy $228.590\ldots$ .
Solutions outside the given range should be ignored, not written as extra answers.
At $\theta = 180^\circ$ , $\sin\theta = 0$ and $\operatorname{cosec}\theta$ is undefined, but this is not a solution here.

Techniques used

substitute cosecant expression into factorised polynomialreject impossible sine valuessolve sine equation within given range

Q6MediumDifferentiationLogarithmic and Exponential Functions

The parametric equations of a curve are

x = \frac{2t + 1}{3t + 4}, \quad y = 2\ln(3t + 4),

where $t > -\frac{4}{3}$ .

(a)

Show that $\frac{dy}{dx}$ can be expressed in the form $c(3t + 4)$ and state the value of the constant $c$ .

5M Similar

DifficultyMedium

Worked solution

Approach

We have parametric equations, so use

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.

Differentiate $x$ with respect to $t$ using the quotient rule, differentiate $y$ with respect to $t$ using the chain rule, then divide.

Working

Differentiate $x = \frac{2t+1}{3t+4}$ with respect to $t$ :

\frac{dx}{dt} = \frac{2(3t+4) - 3(2t+1)}{(3t+4)^2}.

Simplify the numerator:

\frac{dx}{dt} = \frac{6t + 8 - 6t - 3}{(3t+4)^2} = \frac{5}{(3t+4)^2}.

Differentiate $y = 2\ln(3t+4)$ with respect to $t$ :

\frac{dy}{dt} = \frac{6}{3t+4}.

Now divide:

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{6}{3t+4}}{\frac{5}{(3t+4)^2}}.

Simplify:

\frac{dy}{dx} = \frac{6}{3t+4} \cdot \frac{(3t+4)^2}{5} = \frac{6}{5}(3t+4).

So $c = \frac{6}{5}$ .

Answer

\frac{dy}{dx} = \frac{6}{5}(3t+4), \quad c = \frac{6}{5}.

Final answer

dy/dx = (6/5)(3t + 4), c = 6/5

Detailed explanation

Walkthrough

We are given a curve in parametric form, so the gradient $\frac{dy}{dx}$ is not found directly. Instead we differentiate $x$ and $y$ separately with respect to the parameter $t$ , then use

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.

First differentiate $x = \frac{2t+1}{3t+4}$ . This is a quotient, so apply the quotient rule:

\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}.

Here $u = 2t+1$ and $v = 3t+4$ , so $u' = 2$ and $v' = 3$ . Substituting gives $\frac{5}{(3t+4)^2}$ .

Next differentiate $y = 2\ln(3t+4)$ . The derivative of $\ln(3t+4)$ is $\frac{3}{3t+4}$ by the chain rule, so multiplying by 2 gives $\frac{6}{3t+4}$ .

Finally divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ . Dividing by a fraction is the same as multiplying by its reciprocal, so the $(3t+4)^2$ partially cancels with one factor of $3t+4$ , leaving $\frac{6}{5}(3t+4)$ . This is exactly the form $c(3t+4)$ , so $c = \frac{6}{5}$ .

Key Takeaways

Parametric differentiation uses $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ . The quotient rule is needed for rational functions. The chain rule is needed for composite logarithms. Always simplify fully to match the requested form.

Common Mistakes

Forgetting to square the denominator in the quotient rule.
Making a sign error when expanding $2(3t+4) - 3(2t+1)$ .
Differentiating $\ln(3t+4)$ as $\frac{1}{3t+4}$ without multiplying by the inner derivative $3$ .
Dividing $\frac{dx}{dt}$ by $\frac{dy}{dt}$ instead of the other way round.

Things to Be Careful About

The mark scheme requires the final answer in the form $c(3t+4)$ , so you must simplify $\frac{6}{5}(3t+4)$ and state the constant $c$ . Keep the domain $t > -\frac{4}{3}$ in mind, although part (a) does not need it.

Techniques used

differentiate a rational function with the quotient ruledifferentiate a composite logarithm using the chain rulecombine parametric derivatives to find dy/dx

(b)

It is given that the gradient of the curve at the point $(a, \ln 100)$ is $m$ .

Find the values of $a$ and $m$ .

DifficultyMedium-Easy

Worked solution

Approach

Use $y = 2\ln(3t+4)$ and set it equal to $\ln 100$ to find $t$ . Since $t > -\frac{4}{3}$ , we have $3t+4 > 0$ , so take the positive square root. Then substitute $t$ into the parametric equation for $x$ to find $a$ , and into $\frac{dy}{dx} = \frac{6}{5}(3t+4)$ to find $m$ .

Working

Set $y = \ln 100$ :

2\ln(3t+4) = \ln 100.

Use the logarithm power rule:

\ln(3t+4)^2 = \ln 100.

Hence:

(3t+4)^2 = 100.

Since $t > -\frac{4}{3}$ , $3t+4 > 0$ , so:

3t+4 = 10.

Therefore:

t = 2.

Find $a$ :

a = \frac{2(2)+1}{3(2)+4} = \frac{5}{10} = \frac{1}{2}.

Find $m$ using part (a):

m = \frac{6}{5}(3(2)+4) = \frac{6}{5}(10) = 12.

Answer

a = \frac{1}{2}, \quad m = 12.

Final answer

a = 1/2, m = 12

Detailed explanation

Walkthrough

We know the point has $y$ -coordinate $\ln 100$ . Since $y = 2\ln(3t+4)$ , set

2\ln(3t+4) = \ln 100.

Using the logarithm power rule, $2\ln(3t+4) = \ln(3t+4)^2$ , so

\ln(3t+4)^2 = \ln 100.

Because $\ln$ is a one-to-one function, the arguments must be equal:

(3t+4)^2 = 100.

The domain $t > -\frac{4}{3}$ means $3t+4 > 0$ , so only the positive square root is valid:

3t+4 = 10 \quad \Rightarrow \quad t = 2.

Now substitute $t=2$ into the parametric equation for $x$ to find $a$ :

a = \frac{2(2)+1}{3(2)+4} = \frac{5}{10} = \frac{1}{2}.

The gradient $m$ is the value of $\frac{dy}{dx}$ at this point. From part (a), $\frac{dy}{dx} = \frac{6}{5}(3t+4)$ . At $t=2$ ,

m = \frac{6}{5}(10) = 12.

Key Takeaways

Logarithmic equations can be solved by using the power rule and the one-to-one property of logarithms. The domain of the parameter is essential for choosing the correct root. A parametric point is found by substituting the parameter value into both coordinate equations, and the gradient is found by substituting into $\frac{dy}{dx}$ .

Common Mistakes

Writing $2\ln(3t+4) = \ln 100$ as $2(3t+4) = 100$ instead of using the logarithm power rule.
Taking the negative square root $(3t+4 = -10)$ and obtaining an invalid $t$ .
Substituting $t$ into $y$ instead of $x$ when finding $a$ .
Forgetting to use the derivative expression from part (a) when finding $m$ .

Things to Be Careful About

The mark scheme allows the value of $t$ to be implied by a single correct value of $a$ , but you should still show the log equation. Only $t=2$ is valid because $3t+4$ must be positive. The final answers are $a = \frac{1}{2}$ and $m = 12$ .

Techniques used

apply logarithm laws to solve for the parametersubstitute the parameter into the parametric x-coordinateevaluate the gradient expression at the parameter value

(c)

State whether the curve represents a decreasing function or an increasing function or neither. Give a reason for your answer.

1M Similar

DifficultyEasy

Worked solution

Approach

The curve is increasing if its gradient is positive for all $t$ in the domain. Use the expression for $\frac{dy}{dx}$ from part (a) and the domain $t > -\frac{4}{3}$ .

Working

From part (a):

\frac{dy}{dx} = \frac{6}{5}(3t+4).

Since $t > -\frac{4}{3}$ :

3t+4 > 0.

Therefore:

\frac{dy}{dx} = \frac{6}{5}(3t+4) > 0

for all $t$ in the domain. A positive gradient means the curve is increasing.

Answer

The curve represents an increasing function, because $\frac{dy}{dx} > 0$ for all $t > -\frac{4}{3}$ .

Final answer

Increasing, because dy/dx > 0 for all t > -4/3

Detailed explanation

Walkthrough

From part (a), the gradient of the curve is

\frac{dy}{dx} = \frac{6}{5}(3t+4).

The domain is $t > -\frac{4}{3}$ , so $3t+4 > 0$ . Since $\frac{6}{5}$ is positive, the product $\frac{6}{5}(3t+4)$ is positive for every allowed value of $t$ . A positive gradient everywhere means the curve is increasing.

Key Takeaways

The sign of $\frac{dy}{dx}$ over the whole domain determines whether a function is increasing, decreasing, or neither. Here the domain makes the gradient strictly positive, so the curve is increasing.

Common Mistakes

Answering "increasing" without giving a reason.
Claiming the curve is decreasing because $3t+4$ appears in the denominator in intermediate steps, without simplifying to $\frac{6}{5}(3t+4)$ .
Forgetting that $t > -\frac{4}{3}$ ensures $3t+4 > 0$ .

Things to Be Careful About

The mark scheme awards the mark for "increasing" with a reference to a positive gradient. You must state that the gradient is positive, not just say "increasing". Since the gradient is never zero or negative on the domain, the curve is strictly increasing.

Techniques used

use the domain to determine the sign of the gradientrelate a positive gradient to an increasing function

Q7MediumTrigonometryIntegration

(a)

Prove that $\sin^2 2x + 4\cos^2 x \cos 2x \equiv 4\cos^4 x$ .

DifficultyMedium-Easy

Worked solution

Approach

Use the double-angle formulae to write both $\sin 2x$ and $\cos 2x$ in terms of $\sin x$ and $\cos x$ , then factorise and simplify.

Working

\sin^2 2x = (2\sin x\cos x)^2 = 4\sin^2 x\cos^2 x

\cos 2x = \cos^2 x - \sin^2 x

So

\sin^2 2x + 4\cos^2 x\cos 2x = 4\sin^2 x\cos^2 x + 4\cos^2 x(\cos^2 x - \sin^2 x)

Factorise $4\cos^2 x$ :

= 4\cos^2 x\left(\sin^2 x + \cos^2 x - \sin^2 x\right)

= 4\cos^2 x \cdot \cos^2 x = 4\cos^4 x

Answer

\sin^2 2x + 4\cos^2 x\cos 2x \equiv 4\cos^4 x

Final answer

sin^2 2x + 4cos^2 x cos 2x = 4cos^4 x

Detailed explanation

Walkthrough

We need to show that the left-hand side can be rearranged to $4\cos^4 x$ . The most reliable way is to write everything in terms of $\sin x$ and $\cos x$ .

Start with $\sin 2x$ . The double-angle formula gives $\sin 2x = 2\sin x\cos x$ , so when we square it we get $4\sin^2 x\cos^2 x$ .

Next, replace $\cos 2x$ by $\cos^2 x - \sin^2 x$ . This is the standard double-angle formula for cosine.

Substitute both results into the left-hand side:

4\sin^2 x\cos^2 x + 4\cos^2 x(\cos^2 x - \sin^2 x)

Now factor the common factor $4\cos^2 x$ :

4\cos^2 x\left(\sin^2 x + \cos^2 x - \sin^2 x\right)

The $\sin^2 x$ terms cancel, leaving $\cos^2 x$ in the bracket. Therefore the expression becomes $4\cos^2 x \cdot \cos^2 x = 4\cos^4 x$ , which is exactly the right-hand side.

Key Takeaways

The double-angle formulae allow us to convert $\sin 2x$ and $\cos 2x$ into expressions involving only $\sin x$ and $\cos x$ .
Factorising common terms is often the key to simplifying a trigonometric expression.
The Pythagorean identity $\sin^2 x + \cos^2 x = 1$ is used implicitly when simplifying $\sin^2 x + \cos^2 x - \sin^2 x$ to $\cos^2 x$ .

Common Mistakes

Forgetting to square the $2$ when expanding $\sin^2 2x$ ; it is $(2\sin x\cos x)^2 = 4\sin^2 x\cos^2 x$ , not $2\sin^2 x\cos^2 x$ .
Using the wrong double-angle formula for $\cos 2x$ (e.g. $1 - 2\sin^2 x$ is also correct, but then the factorisation is less direct).
Dropping the factor $4$ when substituting.

Things to Be Careful About

The identity is an identity, so the final line must use $\equiv$ or state that the two sides are equal for all $x$ .
When factorising, ensure every term is included: $4\sin^2 x\cos^2 x$ and $4\cos^2 x(\cos^2 x - \sin^2 x)$ both share $4\cos^2 x$ .
The mark scheme requires a clear method: write down the identities used and show the intermediate factorised line.

Techniques used

expand sin 2x using the double-angle formulaexpand cos 2x in terms of sin^2 x and cos^2 xfactor out a common factorsimplify using the Pythagorean identity

(b)

Find the set of possible values of the constant $k$ for which the equation

\sin^2 2x + 4\cos^2 x \cos 2x + 5 = k

has no real solutions.

DifficultyMedium-Easy

Worked solution

Approach

Use the identity from part (a) to rewrite the equation as $k = 5 + 4\cos^4 x$ , then use the range of $\cos^4 x$ to find when no real $x$ can satisfy it.

Working

From part (a),

\sin^2 2x + 4\cos^2 x\cos 2x = 4\cos^4 x

So the equation becomes

4\cos^4 x + 5 = k

For real $x$ ,

0 \le \cos^4 x \le 1

so

5 \le 4\cos^4 x + 5 \le 9

Thus real solutions exist exactly when $5 \le k \le 9$ . Therefore the equation has no real solutions when

k < 5 \quad \text{or} \quad k > 9

Answer

k < 5 \text{ or } k > 9

Final answer

k < 5 or k > 9

Detailed explanation

Walkthrough

The equation contains exactly the expression from part (a). Replace $\sin^2 2x + 4\cos^2 x\cos 2x$ by $4\cos^4 x$ . Then the equation becomes $4\cos^4 x + 5 = k$ , or $k = 5 + 4\cos^4 x$ .

For real $x$ , the value of $\cos x$ lies between $-1$ and $1$ , so $\cos^4 x$ lies between $0$ and $1$ . Multiplying by $4$ gives values from $0$ to $4$ , and adding $5$ gives values from $5$ to $9$ . Therefore the left-hand side can equal any number in $[5,9]$ .

If $k$ is outside this interval, no real $x$ can make the equation true. Hence the equation has no real solutions when $k<5$ or $k>9$ .

Key Takeaways

The range of $\cos^4 x$ is $[0,1]$ , not $[-1,1]$ , because the fourth power makes negative values positive.
An equation of the form $f(x)=k$ has no real solutions exactly when $k$ is outside the range of $f$ .
The identity from part (a) is a powerful tool for simplifying the equation.

Common Mistakes

Thinking $\cos^4 x$ ranges over $[-1,1]$ ; it actually ranges over $[0,1]$ .
Forgetting to add $5$ when finding the range of the left-hand side.
Writing $k \le 5$ or $k \ge 9$ instead of strict inequalities; at $k=5$ and $k=9$ real solutions do exist.

Things to Be Careful About

The question asks for values for which there are no real solutions, so the answer is the complement of the range $[5,9]$ .
Use strict inequalities: $k<5$ or $k>9$ .
The mark scheme expects the use of the part (a) identity; an unsupported answer may not receive full credit.

Techniques used

apply the identity from part (a)determine the range of cos^4 xtranslate the range into a condition for no real solutions

(c)

Find the exact value of

\int_{-\frac{1}{3}\pi}^{\frac{1}{3}\pi} \sqrt{\sin^2 t + 4\cos^2(\frac{1}{2}t)\cos t}\,dt.