The polynomial $p(x)$ is defined by $$ p(x) = ax^3 - ax^2 - 15x + 18, $$ where $a$ is a constant. It is given that $(x + 2)$ is a factor of $p(x)$. (a) Find the value of $a$. (b) Hence factorise $p(x)$ completely. (c) Solve the equation $p(\text{cosec}^2\theta) = 0$ for $-90^\circ < \theta < 90^\circ$.

(a) a = 4 · (b) p(x) = (x + 2)(2x - 3)^2 · (c) θ = 54.7° and θ = -54.7°

A curve has parametric equations $$ x = \frac{e^{2t} - 2}{e^{2t} + 1}, \quad y = e^{3t} + 1. $$ (a) Find an expression for $\frac{dy}{dx}$ in terms of $t$. (b) Find the exact gradient of the curve at the point where the curve crosses the $y$-axis.

(a) dy/dx = (1/2)e^t(e^(2t)+1)^2 · (b) 9√2/2

9709/23 Oct/Nov 2024 — Mathematics Solved Past Paper

Q1MediumLogarithmic and Exponential Functions

The variables $x$ and $y$ satisfy the equation $a^{2y} = e^{3x+k}$ , where $a$ and $k$ are constants. The graph of $y$ against $x$ is a straight line.

(a)

Use logarithms to show that the gradient of the straight line is $\frac{3}{2\ln a}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

Take natural logarithms of both sides of $a^{2y} = e^{3x+k}$ , use the power law of logarithms to bring the exponent $2y$ down, then rearrange into the form $y = mx + c$ to read off the gradient.

Working

Take natural logarithms of both sides:

\ln(a^{2y}) = \ln(e^{3x+k})

Using the power law of logarithms, $\ln(a^{2y}) = 2y\ln a$ , and since $\ln(e^u) = u$ :

2y\ln a = 3x + k

Divide both sides by $2\ln a$ :

y = \frac{3}{2\ln a}x + \frac{k}{2\ln a}

This is in the form $y = mx + c$ , so the gradient is the coefficient of $x$ .

Answer

The gradient of the straight line is $\frac{3}{2\ln a}$ , as required.

Final answer

The gradient is 3/(2 ln a)

Detailed explanation

Walkthrough

The equation $a^{2y} = e^{3x+k}$ is exponential in form. To show that the graph of $y$ against $x$ is a straight line, we need to express $y$ as a linear function of $x$ . Taking natural logarithms of both sides is the key step: it converts the exponential equation into a linear one, because logarithms turn exponents into coefficients.

On the left-hand side, $\ln(a^{2y})$ simplifies using the power law of logarithms: $\ln(a^{2y}) = 2y\ln a$ . This brings the exponent $2y$ down as a coefficient. On the right-hand side, $\ln(e^{3x+k})$ simplifies to $3x+k$ because the natural logarithm and the exponential function are inverse operations.

This gives the linear equation $2y\ln a = 3x + k$ . Dividing through by $2\ln a$ isolates $y$ :

y = \frac{3}{2\ln a}x + \frac{k}{2\ln a}

This is in the standard slope-intercept form $y = mx + c$ , where the gradient $m$ is the coefficient of $x$ , namely $\frac{3}{2\ln a}$ .

Key Takeaways

The essential skill is recognising that taking logarithms converts an exponential relationship into a linear one. The power law $\ln(a^b) = b\ln a$ is the tool that brings exponents down. The inverse relationship $\ln(e^u) = u$ simplifies the right-hand side directly. Once in the form $y = mx + c$ , the gradient and intercept can be read off immediately.

Common Mistakes

Forgetting to apply the power law: writing $\ln(a^{2y})$ without simplifying it to $2y\ln a$ .
Confusing $\ln(a^{2y})$ with $(\ln a)^{2y}$ — the power law applies to the exponent, not to the logarithm itself.
Stopping at $2y\ln a = 3x + k$ without isolating $y$ — the gradient is the coefficient of $x$ only after $y$ is the subject.

Things to Be Careful About

The mark scheme notes "AG – necessary detail needed", meaning the intermediate step $2y\ln a = 3x + k$ must be shown explicitly; simply stating the final gradient without working would not earn the mark.
The gradient is the coefficient of $x$ after $y$ is isolated, so the rearrangement to $y = \ldots$ form is essential.

Techniques used

take natural logarithms of both sidesapply the power law of logarithms to bring down exponentsrearrange into the slope-intercept form y = mx + c

(b)

Given that the straight line passes through the points $(0.4, 0.95)$ and $(3.3, 3.80)$ , find the values of $a$ and $k$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Compute the gradient of the line through the two given points and equate it to $\frac{3}{2\ln a}$ from part (a). Solve for $a$ , then substitute one of the points into the linear equation $2y\ln a = 3x + k$ to find $k$ .

Working

The gradient of the line through $(0.4, 0.95)$ and $(3.3, 3.80)$ is:

\frac{3.80 - 0.95}{3.3 - 0.4} = \frac{2.85}{2.9}

From part (a), the gradient is also $\frac{3}{2\ln a}$ , so:

\frac{3}{2\ln a} = \frac{2.85}{2.9}

Solve for $\ln a$ :

\ln a = \frac{3 \times 2.9}{2 \times 2.85} = \frac{8.7}{5.7} = \frac{29}{19}

Therefore:

a = e^{29/19} \approx 4.6

Now use the equation $2y\ln a = 3x + k$ with the point $(0.4, 0.95)$ :

2(0.95)\ln a = 3(0.4) + k

1.9 \times \frac{29}{19} = 1.2 + k

\frac{19}{10} \times \frac{29}{19} = 1.2 + k

\frac{29}{10} = 1.2 + k

2.9 = 1.2 + k

k = 1.7

Answer

a = e^{29/19} \approx 4.6, \quad k = 1.7

Final answer

a = e^(29/19) ≈ 4.6, k = 1.7

Detailed explanation

Walkthrough

From part (a), we know the gradient of the line is $\frac{3}{2\ln a}$ . We are given two points on the line, $(0.4, 0.95)$ and $(3.3, 3.80)$ , so we can compute the gradient directly from the points:

\frac{3.80 - 0.95}{3.3 - 0.4} = \frac{2.85}{2.9}

Equating this to the gradient from part (a) gives an equation in $\ln a$ only:

\frac{3}{2\ln a} = \frac{2.85}{2.9}

Cross-multiplying and solving:

\ln a = \frac{3 \times 2.9}{2 \times 2.85} = \frac{8.7}{5.7} = \frac{29}{19}

Exponentiating both sides gives $a = e^{29/19} \approx 4.6$ .

With $a$ known, we substitute one of the given points into the linear equation $2y\ln a = 3x + k$ to find $k$ . Using $(0.4, 0.95)$ :

2(0.95) \times \frac{29}{19} = 3(0.4) + k

1.9 \times \frac{29}{19} = 1.2 + k

\frac{29}{10} = 1.2 + k

2.9 = 1.2 + k

k = 1.7

Key Takeaways

This question ties together several ideas: the gradient of a line through two points, the linear form obtained by taking logarithms, and solving for unknown constants by substitution. It shows how the linear form from part (a) connects the geometry of the line (its gradient) to the constants $a$ and $k$ .

Common Mistakes

Using the gradient formula incorrectly, e.g. $\frac{x_2 - x_1}{y_2 - y_1}$ instead of $\frac{y_2 - y_1}{x_2 - x_1}$ .
Arithmetic errors when simplifying $\frac{2.85}{2.9}$ or when computing $\frac{3 \times 2.9}{2 \times 2.85}$ .
Forgetting to substitute the point back into the correct equation — the linear form $2y\ln a = 3x + k$ (or equivalently the original exponential form, as in the mark scheme's alternative method).

Things to Be Careful About

The mark scheme says "Allow greater accuracy", so both $a = 4.6$ (rounded to 2 significant figures) and $a = e^{29/19}$ (exact) are acceptable.
Either point can be used to find $k$ ; both should give $k = 1.7$ . It is good practice to check with the second point.
The mark scheme's alternative method works directly with the exponential form: $a^{1.9} = e^{1.2+k}$ and $a^{7.6} = e^{9.9+k}$ , leading to the same values of $a$ and $k$ .

Techniques used

compute the gradient of a line through two pointsequate gradient expressions and solve for the unknown basesubstitute a known point to determine the intercept constant

Q24MMediumAlgebra

Solve the inequality $|x - 7| > 4x + 3$ .

DifficultyMedium

Worked solution

Approach

Because the expression inside the modulus changes sign at $x = 7$ , split the inequality into two cases. In each case, remove the modulus sign, solve the resulting linear inequality, and intersect the solution with the case condition.

Working

For $x \geq 7$ , $|x - 7| = x - 7$ . The inequality becomes

x - 7 > 4x + 3

Solve:

\begin{aligned} x - 7 &> 4x + 3 \\ -10 &> 3x \\ x &< -\frac{10}{3} \end{aligned}

This contradicts $x \geq 7$ , so this case gives no solutions.

For $x < 7$ , $|x - 7| = 7 - x$ . The inequality becomes

7 - x > 4x + 3

Solve:

\begin{aligned} 7 - x &> 4x + 3 \\ 4 &> 5x \\ x &< \frac{4}{5} \end{aligned}

Since $\frac{4}{5} < 7$ , the condition $x < 7$ is automatically satisfied, so this case gives $x < \frac{4}{5}$ .

Combining the cases, the only valid solutions come from the second case.

Answer

x < \frac{4}{5}

Equivalently, $x \in (-\infty, \frac{4}{5})$ .

Final answer

x < 4/5

Detailed explanation

Walkthrough

The modulus $|x-7|$ is defined differently depending on whether $x-7$ is non-negative or negative. It changes at $x=7$ , so we split the problem there.

For $x \geq 7$ , the expression inside the modulus is non-negative, so $|x-7| = x-7$ . Substituting this into the inequality gives $x - 7 > 4x + 3$ . Solving this linear inequality gives $x < -\frac{10}{3}$ . However, we are working under the assumption $x \geq 7$ , and no number can be both $\geq 7$ and $< -\frac{10}{3}$ . Therefore this case contributes no solutions.

For $x < 7$ , the expression inside the modulus is negative, so $|x-7| = 7-x$ . The inequality becomes $7 - x > 4x + 3$ . Solving gives $4 > 5x$ , so $x < \frac{4}{5}$ . Since $\frac{4}{5} < 7$ , every number with $x < \frac{4}{5}$ automatically satisfies $x < 7$ . Thus this case gives the full solution set.

Combining the two cases, the solution is $x < \frac{4}{5}$ , which can also be written as $(-\infty, \frac{4}{5})$ .

Key Takeaways

A modulus inequality must be handled by considering the sign of the expression inside the modulus.
After solving each case, intersect the solution with the case's domain.
A case can produce no solutions; this must be stated clearly.
The final answer can be given as an inequality or interval notation; both are accepted.

Common Mistakes

Forgetting to split at $x=7$ and treating $|x-7|$ as $x-7$ everywhere.
Solving a case but failing to check whether the result is compatible with the case condition.
Squaring both sides of the inequality without considering the sign of $4x+3$ . This can introduce an incorrect lower bound such as $x > -\frac{10}{3}$ , which is not part of the true solution set.
Not showing the method; the mark scheme requires an attempt at solving the inequality, not just a final answer.

Things to Be Careful About

At $x=7$ , the modulus is 0 and the right-hand side is $31$ , so $x=7$ is not a solution.
The boundary $x = \frac{4}{5}$ comes from the case $x<7$ ; it is less than 7, so the case condition does not remove any part of the solution.
The solution set is unbounded below: any sufficiently negative $x$ makes the right-hand side negative, so the inequality is automatically true.
The mark scheme allows the interval form $(-\infty, \frac{4}{5})$ as an equivalent answer.

Techniques used

split the modulus into casessolve linear inequalities in each caseintersect solutions with case conditionscombine valid cases

Approach

Differentiate $f(x) = \tan^2\left(\frac{1}{2}x\right)$ using the chain rule, then substitute $x = \frac{2}{3}\pi$ and evaluate using exact trigonometric values.

Working

Write $f(x)$ as a composite function and apply the chain rule:

f'(x) = 2\tan\left(\frac{1}{2}x\right) \cdot \sec^2\left(\frac{1}{2}x\right) \cdot \frac{1}{2}

Simplify:

f'(x) = \tan\left(\frac{1}{2}x\right)\sec^2\left(\frac{1}{2}x\right)

Substitute $x = \frac{2}{3}\pi$ :

\frac{1}{2}x = \frac{1}{2} \cdot \frac{2}{3}\pi = \frac{\pi}{3}

So:

f'\left(\frac{2}{3}\pi\right) = \tan\left(\frac{\pi}{3}\right)\sec^2\left(\frac{\pi}{3}\right)

Using $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$ and $\sec\left(\frac{\pi}{3}\right) = 2$ :

f'\left(\frac{2}{3}\pi\right) = \sqrt{3} \cdot 2^2 = 4\sqrt{3}

Answer

f'\left(\frac{2}{3}\pi\right) = 4\sqrt{3}

Final answer

4√3

Detailed explanation

Walkthrough

We are asked to differentiate $f(x) = \tan^2\left(\frac{1}{2}x\right)$ and evaluate the derivative at $x = \frac{2}{3}\pi$ . The function is a composite: the outer operation is squaring, applied to $\tan\left(\frac{1}{2}x\right)$ , which is itself a composite of $\tan$ with $\frac{1}{2}x$ .

The chain rule requires us to differentiate from the outside in. First, differentiate the square: $2\tan\left(\frac{1}{2}x\right)$ . Keep the inner function unchanged. Next, differentiate $\tan\left(\frac{1}{2}x\right)$ : the derivative of $\tan u$ is $\sec^2 u$ , so this gives $\sec^2\left(\frac{1}{2}x\right)$ . Finally, differentiate $\frac{1}{2}x$ , which gives $\frac{1}{2}$ . Multiplying these three factors gives $f'(x) = 2\tan\left(\frac{1}{2}x\right)\sec^2\left(\frac{1}{2}x\right) \cdot \frac{1}{2} = \tan\left(\frac{1}{2}x\right)\sec^2\left(\frac{1}{2}x\right)$ .

Now substitute $x = \frac{2}{3}\pi$ . Then $\frac{1}{2}x = \frac{\pi}{3}$ . We need the exact values $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$ and $\sec\left(\frac{\pi}{3}\right) = \frac{1}{\cos(\pi/3)} = \frac{1}{1/2} = 2$ . Squaring the secant gives $4$ , so the derivative evaluates to $\sqrt{3} \times 4 = 4\sqrt{3}$ .

Key Takeaways

The chain rule applies layer by layer to composite functions; here there are two layers inside the square.
Exact trigonometric values at standard angles such as $\pi/3$ are needed to give an exact final answer.
$\sec x = 1/\cos x$ , so $\sec^2(\pi/3) = 4$ .

Common Mistakes

Dropping the factor of $\frac{1}{2}$ from the innermost derivative, which yields $2\tan\left(\frac{1}{2}x\right)\sec^2\left(\frac{1}{2}x\right)$ instead of the correct $\tan\left(\frac{1}{2}x\right)\sec^2\left(\frac{1}{2}x\right)$ .
Forgetting to square the secant when computing $\sec^2(\pi/3)$ — $\sec(\pi/3) = 2$ , so $\sec^2(\pi/3) = 4$ .
Mixing up radians and degrees when evaluating the trig functions.

Things to Be Careful About

The mark scheme allows unsimplified forms, but the exact value $4\sqrt{3}$ is the expected final answer.
The domain $0 \le x < \pi$ guarantees $x = \frac{2}{3}\pi$ is valid and that $\frac{1}{2}x$ lies in $(0, \frac{\pi}{2})$ , where $\tan$ and $\sec$ are defined and positive.

Techniques used

apply the chain rule to differentiate a composite functionevaluate the derivative at a given pointuse exact trigonometric values

(b)

Find the exact value of $\int_0^{\frac{1}{2}\pi} (f(x) + \sin x) \,dx$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Rewrite the integrand using the Pythagorean identity $\tan^2\theta = \sec^2\theta - 1$ , integrate term by term, then apply the limits.

Working

Use the identity $\tan^2\left(\frac{1}{2}x\right) = \sec^2\left(\frac{1}{2}x\right) - 1$ :

\int_0^{\frac{1}{2}\pi} \left(\tan^2\left(\frac{1}{2}x\right) + \sin x\right)\,dx = \int_0^{\frac{1}{2}\pi} \left(\sec^2\left(\frac{1}{2}x\right) - 1 + \sin x\right)\,dx

Integrate term by term:

\int \sec^2\left(\frac{1}{2}x\right)\,dx = 2\tan\left(\frac{1}{2}x\right), \quad \int 1\,dx = x, \quad \int \sin x\,dx = -\cos x

So:

\int_0^{\frac{1}{2}\pi} \left(\sec^2\left(\frac{1}{2}x\right) - 1 + \sin x\right)\,dx = \left[2\tan\left(\frac{1}{2}x\right) - x - \cos x\right]_0^{\frac{1}{2}\pi}

Evaluate at the upper limit $x = \frac{1}{2}\pi$ :

2\tan\left(\frac{\pi}{4}\right) - \frac{\pi}{2} - \cos\left(\frac{\pi}{2}\right) = 2(1) - \frac{\pi}{2} - 0 = 2 - \frac{\pi}{2}

Evaluate at the lower limit $x = 0$ :

2\tan(0) - 0 - \cos(0) = 0 - 0 - 1 = -1

Subtract:

\left(2 - \frac{\pi}{2}\right) - (-1) = 3 - \frac{\pi}{2}

Answer

\int_0^{\frac{1}{2}\pi} \left(f(x) + \sin x\right)\,dx = 3 - \frac{\pi}{2}

Final answer

3 - pi/2

Detailed explanation

Walkthrough

We need to evaluate $\int_0^{\frac{1}{2}\pi} \left(\tan^2\left(\frac{1}{2}x\right) + \sin x\right)\,dx$ .

The integrand contains $\tan^2\left(\frac{1}{2}x\right)$ , which has no direct standard integral. The Pythagorean identity $\tan^2\theta = \sec^2\theta - 1$ rewrites it as $\sec^2\left(\frac{1}{2}x\right) - 1$ . Now every term is integrable: $\sec^2\left(\frac{1}{2}x\right)$ integrates to $2\tan\left(\frac{1}{2}x\right)$ (because the derivative of $\tan\left(\frac{1}{2}x\right)$ is $\frac{1}{2}\sec^2\left(\frac{1}{2}x\right)$ , so we multiply by the reciprocal $2$ ), the constant $1$ integrates to $x$ , and $\sin x$ integrates to $-\cos x$ .

The antiderivative is therefore $2\tan\left(\frac{1}{2}x\right) - x - \cos x$ . We evaluate this at the limits.

At $x = \frac{1}{2}\pi$ : $2\tan\left(\frac{\pi}{4}\right) - \frac{\pi}{2} - \cos\left(\frac{\pi}{2}\right) = 2(1) - \frac{\pi}{2} - 0 = 2 - \frac{\pi}{2}$ .

At $x = 0$ : $2\tan(0) - 0 - \cos(0) = 0 - 0 - 1 = -1$ .

Subtracting the lower value from the upper value: $\left(2 - \frac{\pi}{2}\right) - (-1) = 3 - \frac{\pi}{2}$ .

Key Takeaways

The Pythagorean identity $\tan^2\theta = \sec^2\theta - 1$ is the standard tool for integrating $\tan^2$ terms.
$\int \sec^2(ax+b)\,dx = \frac{1}{a}\tan(ax+b)$ .
Definite integrals are evaluated by substituting both limits and subtracting, carefully handling signs.

Common Mistakes

Writing $\int \sec^2\left(\frac{1}{2}x\right)\,dx = \tan\left(\frac{1}{2}x\right)$ and forgetting the factor $2$ — the correct antiderivative is $2\tan\left(\frac{1}{2}x\right)$ .
Using the wrong sign for $\int \sin x\,dx$ ; it is $-\cos x$ , not $+\cos x$ .
Forgetting that $\cos(0) = 1$ , so the lower-limit substitution contributes $-1$ , and the subtraction becomes $+1$ .

Things to Be Careful About

The mark scheme awards B1 for writing $\sec^2\frac{1}{2}x - 1 + \sin x$ and M1 for the integration step — both must be shown explicitly.
Leave the answer in exact form $3 - \frac{1}{2}\pi$ ; do not convert to a decimal.
The limits are in radians; evaluate the trig functions in radian mode.

Techniques used

rewrite the integrand using a Pythagorean identityintegrate standard trigonometric functions term by termevaluate a definite integral by substituting limits

Q4MediumAlgebraTrigonometry

The polynomial $p(x)$ is defined by

p(x) = ax^3 - ax^2 - 15x + 18,

where $a$ is a constant. It is given that $(x + 2)$ is a factor of $p(x)$ .

(a)

Find the value of $a$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $(x+2)$ is a factor of $p(x)$ , the factor theorem gives $p(-2)=0$ . Substitute $x=-2$ into the polynomial and solve for $a$ .

Working

Substitute $x=-2$ :

p(-2) = a(-2)^3 - a(-2)^2 - 15(-2) + 18

Simplify:

p(-2) = -8a - 4a + 30 + 18 = -12a + 48

Since $x+2$ is a factor, $p(-2)=0$ :

-12a + 48 = 0

Solve:

a = 4

Answer

a = 4

Final answer

a = 4

Detailed explanation

Walkthrough

The factor theorem says that if $(x+2)$ is a factor of $p(x)$ , then $p(-2)=0$ . This is because a factor $(x-c)$ corresponds to a root $x=c$ . Here $c=-2$ . Substitute $x=-2$ into $p(x)=ax^3-ax^2-15x+18$ :

p(-2)=a(-2)^3-a(-2)^2-15(-2)+18=-8a-4a+30+18=-12a+48

Set this equal to zero and solve:

-12a+48=0 \Rightarrow a=4

This single equation determines the unknown coefficient.

Key Takeaways

The factor theorem links factors and roots: $(x+2)$ being a factor means $x=-2$ is a root.
Substituting a known root into a polynomial gives an equation that can determine an unknown coefficient.

Common Mistakes

Using $x=2$ instead of $x=-2$ .
Sign errors when evaluating $(-2)^3=-8$ and $(-2)^2=4$ .
Incorrectly combining $-8a-4a=-12a$ .

Things to Be Careful About

Always write $p(-2)=0$ , not $p(2)=0$ .
Check the final value by substituting back into the polynomial if time allows.

Techniques used

apply the factor theoremsubstitute the known rootsolve a linear equation

(b)

Hence factorise $p(x)$ completely.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the known factor $x+2$ to divide $p(x)$ by $x+2$ . The quotient is a quadratic, which can then be factorised completely.

Working

With $a=4$ , the polynomial is

p(x) = 4x^3 - 4x^2 - 15x + 18

Divide by $x+2$ using synthetic division:

\begin{array}{c|cccc} -2 & 4 & -4 & -15 & 18 \\ & & -8 & 24 & -18 \\ \hline & 4 & -12 & 9 & 0 \end{array}

So the quotient is $4x^2 - 12x + 9$ and the remainder is $0$ :

4x^3 - 4x^2 - 15x + 18 = (x+2)(4x^2 - 12x + 9)

Factorise the quadratic:

4x^2 - 12x + 9 = (2x - 3)^2

Therefore:

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

Answer

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

Final answer

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

Detailed explanation

Walkthrough

With $a=4$ , the polynomial becomes $p(x)=4x^3-4x^2-15x+18$ . Since $(x+2)$ is a factor, divide the cubic by $x+2$ . Synthetic division with $-2$ gives:

\begin{array}{c|cccc} -2 & 4 & -4 & -15 & 18 \\ & & -8 & 24 & -18 \\ \hline & 4 & -12 & 9 & 0 \end{array}

The quotient is $4x^2-12x+9$ and the remainder is $0$ , so

p(x)=(x+2)(4x^2-12x+9)

The quadratic is a perfect square:

4x^2-12x+9=(2x-3)^2

Hence the complete factorisation is

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

Key Takeaways

Polynomial division reduces a cubic to a quadratic once a linear factor is known.
Recognising perfect squares helps complete factorisation quickly.
The complete factorisation must include repeated factors with their powers.

Common Mistakes

Arithmetic errors in synthetic division, especially signs when multiplying by $-2$ .
Forgetting to factorise the quadratic quotient completely.
Expanding $(2x-3)^2$ incorrectly.

Things to Be Careful About

Check the factorisation by expanding: $(x+2)(2x-3)^2$ should return the original cubic.
The remainder must be $0$ because $(x+2)$ is a factor.

Techniques used

divide polynomial by a linear factorfactorise a quadraticwrite the complete factorisation

(c)

Solve the equation $p(\text{cosec}^2\theta) = 0$ for $-90^\circ < \theta < 90^\circ$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Let $x = \operatorname{cosec}^2\theta$ . Substitute into the factorised form of $p(x)$ . Since $\operatorname{cosec}^2\theta \geq 1$ , the factor $x+2$ cannot be zero, so only the squared factor can vanish. Then use $\operatorname{cosec}\theta = 1/\sin\theta$ to solve for $\theta$ in the given interval.

Working

Using the factorisation from part (b),

p\left(\operatorname{cosec}^2\theta\right) = \left(\operatorname{cosec}^2\theta + 2\right)\left(2\operatorname{cosec}^2\theta - 3\right)^2 = 0

Since $\operatorname{cosec}^2\theta \geq 1$ , the first factor is never zero:

2\operatorname{cosec}^2\theta - 3 = 0

So

\operatorname{cosec}^2\theta = \frac{3}{2}

Using $\operatorname{cosec}\theta = \frac{1}{\sin\theta}$ :

\frac{1}{\sin^2\theta} = \frac{3}{2}

Therefore

\sin^2\theta = \frac{2}{3}

and

\sin\theta = \pm \sqrt{\frac{2}{3}}

For $-90^\circ < \theta < 90^\circ$ , the solutions are

\theta = \sin^{-1}\left(\sqrt{\frac{2}{3}}\right) \quad \text{or} \quad \theta = -\sin^{-1}\left(\sqrt{\frac{2}{3}}\right)

Evaluating:

\theta \approx 54.7^\circ \quad \text{or} \quad \theta \approx -54.7^\circ

Answer

\theta = 54.7^\circ \text{ or } \theta = -54.7^\circ

Final answer

θ = 54.7° and θ = -54.7°

Detailed explanation

Walkthrough

We substitute $x=\operatorname{cosec}^2\theta$ into the factorised polynomial. Since $\operatorname{cosec}^2\theta \geq 1$ , the factor $\operatorname{cosec}^2\theta+2$ is always positive and cannot be zero. Therefore the only way the product is zero is

2\operatorname{cosec}^2\theta - 3 = 0

so $\operatorname{cosec}^2\theta = \frac{3}{2}$ . Using the reciprocal identity $\operatorname{cosec}\theta = 1/\sin\theta$ :

\frac{1}{\sin^2\theta} = \frac{3}{2} \Rightarrow \sin^2\theta = \frac{2}{3}

Taking square roots gives $\sin\theta = \pm\sqrt{2/3}$ . On the interval $-90^\circ < \theta < 90^\circ$ , sine is strictly increasing, so each value occurs exactly once: one positive angle and one negative angle. These are $\theta \approx 54.7^\circ$ and $\theta \approx -54.7^\circ$ .

Key Takeaways

A polynomial equation can be solved after substituting a trigonometric expression by using the factorised form.
$\operatorname{cosec}^2\theta \geq 1$ , so factors like $\operatorname{cosec}^2\theta+2$ never vanish.
Solving $\sin^2\theta = k$ requires considering both positive and negative square roots.

Common Mistakes

Trying to solve $\operatorname{cosec}^2\theta + 2 = 0$ , which has no real solution.
Forgetting the negative solution when taking the square root.
Giving angles outside the required interval, such as $125.3^\circ$ or $-125.3^\circ$ .

Things to Be Careful About

The domain $-90^\circ < \theta < 90^\circ$ is open, so $\pm 90^\circ$ are not included.
Since sine is one-to-one on this interval, there are exactly two solutions.
Use sufficient accuracy: $54.7^\circ$ is accepted, and the exact form $\sin^{-1}\sqrt{2/3}$ is also acceptable.

Techniques used

substitute a trigonometric expression into a factorised polynomialuse the reciprocal identity for cosecantsolve a trigonometric equation within a given domain

Q5MediumIntegrationLogarithmic and Exponential FunctionsNumerical Solution of Equations

It is given that $\int_a^{a^3} \frac{10}{2x + 1} \,dx = 7$ , where $a$ is a constant greater than 1.

(a)

Show that $a = \sqrt[3]{0.5e^{1.4}(2a + 1) - 0.5}$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Integrate the given function, apply the limits, set the result equal to 7, then use the logarithm property $\ln A - \ln B = \ln\left(\frac{A}{B}\right)$ and exponentiate to solve for $a^3$ .

Working

Since $\frac{d}{dx}\left[\ln(2x+1)\right] = \frac{2}{2x+1}$ , we have

\int \frac{10}{2x+1}\,dx = 5\ln(2x+1) + C

Apply the limits $a$ to $a^3$ :

5\left[\ln(2a^3+1) - \ln(2a+1)\right] = 7

Use $\ln A - \ln B = \ln\left(\frac{A}{B}\right)$ :

5\ln\left(\frac{2a^3+1}{2a+1}\right) = 7

\ln\left(\frac{2a^3+1}{2a+1}\right) = \frac{7}{5} = 1.4

Exponentiate both sides:

\frac{2a^3+1}{2a+1} = e^{1.4}

Multiply through:

2a^3 + 1 = e^{1.4}(2a+1)

Solve for $a^3$ :

2a^3 = e^{1.4}(2a+1) - 1

a^3 = 0.5e^{1.4}(2a+1) - 0.5

Taking cube roots:

a = \sqrt[3]{0.5e^{1.4}(2a+1) - 0.5}

as required.

Answer

a = \sqrt[3]{0.5e^{1.4}(2a+1) - 0.5}

Final answer

a = cube root(0.5 e^1.4 (2a + 1) - 0.5)

Detailed explanation

Walkthrough

The question gives a definite integral equal to 7 and asks you to show that $a$ satisfies a particular equation. The first step is to integrate $\frac{10}{2x+1}$ . Because $\frac{d}{dx}\ln(2x+1) = \frac{2}{2x+1}$ , multiplying by 5 gives exactly $\frac{10}{2x+1}$ , so the antiderivative is $5\ln(2x+1)$ .

Next, apply the limits. Substitute the upper limit $a^3$ and the lower limit $a$ , and subtract:

5\left[\ln(2a^3+1) - \ln(2a+1)\right] = 7

This earns the mark for applying limits correctly and equating to 7.

The key step is to combine the two logarithms. The difference of two logarithms with the same base is the logarithm of the quotient:

\ln A - \ln B = \ln\left(\frac{A}{B}\right)

This is essential because it lets us isolate the fraction $\frac{2a^3+1}{2a+1}$ . The mark scheme explicitly says this detail is necessary for a 'show that' question.

After dividing by 5, we get $\ln\left(\frac{2a^3+1}{2a+1}\right)=1.4$ . To remove the logarithm, exponentiate both sides. Then multiply through by $2a+1$ and rearrange to solve for $a^3$ . Finally, take the cube root to obtain the required form.

Key Takeaways

The integral of $\frac{1}{ax+b}$ is $\frac{1}{a}\ln|ax+b|$ .
The logarithm law $\ln A - \ln B = \ln\left(\frac{A}{B}\right)$ is used to simplify differences of logarithms.
To solve an equation of the form $\ln f(a) = k$ , exponentiate both sides to get $f(a)=e^k$ .
In a 'show that' question, every algebraic step must be shown.

Common Mistakes

Forgetting the factor 5 when integrating $\frac{10}{2x+1}$ . Since the derivative of $\ln(2x+1)$ is $\frac{2}{2x+1}$ , you need to multiply by 5 to obtain $\frac{10}{2x+1}$ .
Not applying the logarithm property to combine the two logs; the mark scheme requires this detail for the 'show that' proof.
Making sign errors when subtracting the lower limit: it is $\ln(2a^3+1) - \ln(2a+1)$ , not the other way round.
Algebraic errors when multiplying by $2a+1$ and dividing by 2.

Things to Be Careful About

Since $a>1$ , both $2a+1$ and $2a^3+1$ are positive, so no absolute value signs are needed.
Keep $e^{1.4}$ exact in the working rather than replacing it with a decimal.
The final answer is given in the question, so the proof must be complete and convincing.

Techniques used

integrate a reciprocal linear expressionapply definite integral limitsuse logarithm laws to combine logsexponentiate and rearrange to isolate a

(b)

Use an iterative formula, based on the equation in part (a), to find the value of $a$ correct to 3 significant figures. Use an initial value of 2 and give the result of each iteration to 5 significant figures.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Rewrite the result of part (a) as an iteration formula

x_{n+1} = \sqrt[3]{0.5e^{1.4}(2x_n+1) - 0.5}

Starting from $x_0 = 2$ , evaluate successive approximations, rounding each to 5 significant figures, until the value is stable to 3 significant figures.

Working

Let

F(x) = \sqrt[3]{0.5e^{1.4}(2x+1) - 0.5}

Use $x_{n+1} = F(x_n)$ .

$n$	$x_n$ (5 sf)
0	2
1	2.1281
2	2.1657
3	2.1765
4	2.1795
5	2.1804
6	2.1807
7	2.1807
8	2.1807
9	2.1808
10	2.1808
11	2.1808

The successive values settle to $2.1808$ to 5 significant figures, so to 3 significant figures:

a \approx 2.18

(Alternatively, substituting $a=2.175$ and $a=2.185$ into $a^3 - 0.5e^{1.4}(2a+1) + 0.5$ gives a sign change, confirming the root lies in $[2.175, 2.185]$ .)

Answer

a \approx 2.18

Final answer

a ≈ 2.18

Detailed explanation

Walkthrough

The equation from part (a) is already written in the form $a = F(a)$ , so it can be used directly as an iteration formula:

x_{n+1} = \sqrt[3]{0.5e^{1.4}(2x_n+1) - 0.5}

Start with the given initial value $x_0 = 2$ . Substitute it into the right-hand side to get $x_1$ , then substitute $x_1$ to get $x_2$ , and so on.

Each value should be recorded to 5 significant figures. This is more precise than the final 3-significant-figure answer and prevents rounding errors from accumulating. The sequence increases towards the root: $2, 2.1281, 2.1657, 2.1765, 2.1795, 2.1804, 2.1807, 2.1807, 2.1807, 2.1808, 2.1808, 2.1808$ .

Once successive values agree to 3 significant figures, the iteration has converged. Here the values settle at $2.1808$ to 5 significant figures, so to 3 significant figures the answer is $2.18$ . The mark scheme also accepts showing a sign change in the interval $[2.175, 2.185]$ as an alternative justification.

Key Takeaways

A fixed-point iteration formula has the form $x_{n+1} = F(x_n)$ .
Starting from a sensible initial value, repeated substitution can produce a sequence converging to the root.
Rounding intermediate results to more significant figures than the final answer (here 5 sf) avoids premature rounding error.
To justify a final answer to $n$ significant figures, either show enough iterations or demonstrate a sign change in a suitable interval.

Common Mistakes

Not showing enough iterations; the mark scheme requires sufficient iterations to 5 significant figures or a sign change in $[2.175, 2.185]$ .
Giving the final answer as $2.1808$ instead of rounding to exactly 3 significant figures, $2.18$ .
Rounding each iteration to 3 significant figures too early, which can shift the converged value.
Using the wrong iteration formula or an incorrect initial value.

Things to Be Careful About

The iteration must be based on the equation from part (a).
Record every iteration to 5 significant figures.
The final answer must be exactly 3 significant figures: $2.18$ .
If using a sign change, the interval must be $[2.175, 2.185]$ and the signs at the two endpoints must be different.

Techniques used

rewrite equation as fixed-point iterationiterate the formula from an initial valueround intermediate values to 5 significant figuresjustify convergence to 3 significant figures

Q6MediumDifferentiationLogarithmic and Exponential Functions

A curve has parametric equations

x = \frac{e^{2t} - 2}{e^{2t} + 1}, \quad y = e^{3t} + 1.

(a)

Find an expression for $\frac{dy}{dx}$ in terms of $t$ .

4M Similar

DifficultyMedium

Worked solution

Approach

This curve is given parametrically, so

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

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

Working

Differentiate $x$ :

\frac{dx}{dt} = \frac{(e^{2t}+1)\cdot 2e^{2t} - (e^{2t}-2)\cdot 2e^{2t}}{(e^{2t}+1)^2}

Simplify the numerator:

\frac{dx}{dt} = \frac{2e^{2t}\left((e^{2t}+1)-(e^{2t}-2)\right)}{(e^{2t}+1)^2} = \frac{6e^{2t}}{(e^{2t}+1)^2}

Differentiate $y$ :

\frac{dy}{dt} = 3e^{3t}

Therefore,

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3e^{3t}}{\frac{6e^{2t}}{(e^{2t}+1)^2}} = \frac{3e^{3t}(e^{2t}+1)^2}{6e^{2t}}

Simplify:

\frac{dy}{dx} = \frac{1}{2}e^{t}(e^{2t}+1)^2

Answer

\frac{dy}{dx} = \frac{1}{2}e^{t}(e^{2t}+1)^2

Final answer

dy/dx = (1/2)e^t(e^(2t)+1)^2

Detailed explanation

Walkthrough

This is a parametric differentiation question. Since both $x$ and $y$ are given in terms of $t$ , the gradient is found using

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

First differentiate $x = \frac{e^{2t}-2}{e^{2t}+1}$ with respect to $t$ . This is a quotient, so apply the quotient rule: multiply the denominator by the derivative of the numerator, subtract the numerator times the derivative of the denominator, and divide by the denominator squared. The derivative of $e^{2t}$ is $2e^{2t}$ , so the numerator becomes $(e^{2t}+1)2e^{2t} - (e^{2t}-2)2e^{2t}$ . Factor out $2e^{2t}$ ; the bracket simplifies to $3$ , giving $\frac{dx}{dt} = \frac{6e^{2t}}{(e^{2t}+1)^2}$ .

Next differentiate $y = e^{3t}+1$ . Since the derivative of $e^{3t}$ is $3e^{3t}$ , we get $\frac{dy}{dt} = 3e^{3t}$ .

Finally divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ . Dividing by a fraction means multiplying by its reciprocal, so

\frac{dy}{dx} = 3e^{3t} \cdot \frac{(e^{2t}+1)^2}{6e^{2t}} = \frac{1}{2}e^{t}(e^{2t}+1)^2.

The simplification uses $e^{3t}/e^{2t} = e^{t}$ and $3/6 = 1/2$ .

Key Takeaways

The main skill is parametric differentiation: when $x$ and $y$ are functions of a parameter $t$ , use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ . This question also tests the quotient rule and differentiation of exponential functions such as $e^{kt}$ , whose derivative is $ke^{kt}$ .

Common Mistakes

Forgetting to use the quotient rule for $x$ and instead differentiating the numerator and denominator separately.
Making a sign error in the quotient rule numerator.
Leaving the answer as a fraction within a fraction; the mark scheme requires a simplified form with no fractions inside fractions.
Cancelling $e^{2t}$ incorrectly when simplifying $\frac{3e^{3t}(e^{2t}+1)^2}{6e^{2t}}$ .

Things to Be Careful About

Show the quotient rule numerator explicitly to earn the method mark.
The derivative of $e^{2t}$ is $2e^{2t}$ , not $e^{2t}$ .
When forming $\frac{dy}{dx}$ , divide by $\frac{dx}{dt}$ , not by $\frac{dy}{dt}$ .
Simplify the final expression fully; an unsimplified equivalent is acceptable for the final accuracy mark, but it must contain no fractions within fractions.

Techniques used

apply the quotient rule to differentiate a quotientdifferentiate exponential functions of the form e^(kt)use parametric relation dy/dx = (dy/dt)/(dx/dt)simplify a compound fraction of exponential terms

(b)

Find the exact gradient of the curve at the point where the curve crosses the $y$ -axis.

3M Similar

DifficultyMedium

Worked solution

Approach

The curve crosses the $y$ -axis where $x = 0$ . Solve $x = 0$ for $t$ , substitute this value into the expression for $\frac{dy}{dx}$ , and simplify using laws of indices.

Working

Set $x = 0$ :

\frac{e^{2t} - 2}{e^{2t} + 1} = 0

The denominator is never zero, so

e^{2t} - 2 = 0 \quad \Rightarrow \quad e^{2t} = 2

Take natural logarithms:

2t = \ln 2 \quad \Rightarrow \quad t = \frac{1}{2}\ln 2

Substitute into $\frac{dy}{dx}$ :

\frac{dy}{dx} = \frac{1}{2}e^{t}(e^{2t}+1)^2

At $t = \frac{1}{2}\ln 2$ ,

e^{t} = e^{\frac{1}{2}\ln 2} = \sqrt{2}, \qquad e^{2t} = e^{\ln 2} = 2

Therefore,

\frac{dy}{dx} = \frac{1}{2}\cdot \sqrt{2} \cdot (2+1)^2 = \frac{1}{2} \cdot \sqrt{2} \cdot 9 = \frac{9\sqrt{2}}{2}

Answer

\frac{dy}{dx} = \frac{9\sqrt{2}}{2}

Final answer

9√2/2

Detailed explanation

Walkthrough

The curve crosses the $y$ -axis at the point where its $x$ -coordinate is $0$ . Therefore set

\frac{e^{2t}-2}{e^{2t}+1} = 0.

Since the denominator $e^{2t}+1$ is never zero, this is equivalent to $e^{2t}-2=0$ , so $e^{2t}=2$ . Taking natural logarithms gives $2t = \ln 2$ , hence $t = \frac{1}{2}\ln 2$ .

Now substitute this value into the gradient expression from part (a):

\frac{dy}{dx} = \frac{1}{2}e^{t}(e^{2t}+1)^2.

Using $e^{\frac{1}{2}\ln 2} = \sqrt{2}$ and $e^{2t} = e^{\ln 2} = 2$ , the gradient becomes

\frac{1}{2}\sqrt{2}(2+1)^2 = \frac{1}{2}\sqrt{2} \cdot 9 = \frac{9\sqrt{2}}{2}.

Key Takeaways

This part combines solving an exponential equation with evaluating a parametric derivative. To find where a parametric curve crosses an axis, set the relevant coordinate equal to zero and solve for the parameter. Exact answers require using the laws of logarithms and indices rather than decimals.

Common Mistakes

Setting $y = 0$ instead of $x = 0$ when finding the crossing with the $y$ -axis.
Forgetting to take the natural logarithm when solving $e^{2t}=2$ .
Writing $e^{\frac{1}{2}\ln 2}$ as $\frac{1}{2}\ln 2$ instead of $\sqrt{2}$ .
Substituting a decimal approximation for $t$ and losing the exact form required.

Things to Be Careful About

The denominator $e^{2t}+1$ is always positive, so there is no risk of division by zero when setting $x=0$ .
The exact parameter value is $t = \frac{1}{2}\ln 2$ , not $t = \ln 2$ .
Use the expression for $\frac{dy}{dx}$ from part (a); the mark scheme allows substitution into the candidate's own expression.
Give the final gradient as an exact value, $\frac{9\sqrt{2}}{2}$ , not as a decimal.

Techniques used

set x = 0 to locate the crossing of the y-axissolve an exponential equation by taking natural logarithmssubstitute the parameter value into the gradient expressionsimplify using laws of indices

Q7MediumTrigonometry

(a)

Prove that $\cos(\theta + 30^\circ)\cos(\theta + 60^\circ) \equiv \frac{1}{4}\sqrt{3} - \frac{1}{2}\sin 2\theta$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Expand each $\cos(\theta + 30^\circ)$ and $\cos(\theta + 60^\circ)$ using the compound-angle formula, multiply out, then simplify using $\sin^2\theta+\cos^2\theta=1$ and the double-angle formula $\sin 2\theta=2\sin\theta\cos\theta$ .

Working

Using $\cos(A+B)=\cos A\cos B-\sin A\sin B$ :

\begin{aligned} \cos(\theta+30^\circ)&=\cos\theta\cos30^\circ-\sin\theta\sin30^\circ\\ &=\frac{\sqrt3}{2}\cos\theta-\frac12\sin\theta \end{aligned}

\begin{aligned} \cos(\theta+60^\circ)&=\cos\theta\cos60^\circ-\sin\theta\sin60^\circ\\ &=\frac12\cos\theta-\frac{\sqrt3}{2}\sin\theta \end{aligned}

Multiplying:

\begin{aligned} \cos(\theta+30^\circ)\cos(\theta+60^\circ) &=\left(\frac{\sqrt3}{2}\cos\theta-\frac12\sin\theta\right) \left(\frac12\cos\theta-\frac{\sqrt3}{2}\sin\theta\right)\\ &=\frac{\sqrt3}{4}\cos^2\theta-\frac34\sin\theta\cos\theta -\frac14\sin\theta\cos\theta+\frac{\sqrt3}{4}\sin^2\theta\\ &=\frac{\sqrt3}{4}\left(\cos^2\theta+\sin^2\theta\right)-\sin\theta\cos\theta\\ &=\frac{\sqrt3}{4}-\sin\theta\cos\theta \end{aligned}

Since $\sin 2\theta=2\sin\theta\cos\theta$ , we have $\sin\theta\cos\theta=\frac12\sin 2\theta$ . Therefore:

\cos(\theta+30^\circ)\cos(\theta+60^\circ) =\frac{\sqrt3}{4}-\frac12\sin2\theta =\frac14\sqrt3-\frac12\sin2\theta

Answer

\cos(\theta+30^\circ)\cos(\theta+60^\circ)\equiv\frac14\sqrt3-\frac12\sin2\theta

Final answer

cos(theta+30)cos(theta+60) = sqrt(3)/4 - (1/2)sin(2theta)

Detailed explanation

Walkthrough

The identity is an answer-given proof, so every step must be shown. Start by expanding each factor using the compound-angle formula $\cos(A+B)=\cos A\cos B-\sin A\sin B$ . Substitute the exact values $\cos30^\circ=\frac{\sqrt3}{2}$ , $\sin30^\circ=\frac12$ , $\cos60^\circ=\frac12$ , $\sin60^\circ=\frac{\sqrt3}{2}$ . Then multiply the two brackets. There are four products: two give $\frac{\sqrt3}{4}\cos^2\theta$ and $\frac{\sqrt3}{4}\sin^2\theta$ , and the cross terms give $-\frac34\sin\theta\cos\theta-\frac14\sin\theta\cos\theta=-\sin\theta\cos\theta$ . Use $\cos^2\theta+\sin^2\theta=1$ to get $\frac{\sqrt3}{4}-\sin\theta\cos\theta$ . Finally use $\sin2\theta=2\sin\theta\cos\theta$ to replace $\sin\theta\cos\theta$ by $\frac12\sin2\theta$ , giving the required right-hand side.

Key Takeaways

This question tests compound-angle expansions, exact trigonometric values, the Pythagorean identity, and the double-angle formula for sine. It also shows how a product of two cosines can be rewritten as a combination of a constant and $\sin2\theta$ .

Common Mistakes

Using incorrect exact values for $30^\circ$ or $60^\circ$ .
Dropping the minus sign in the compound-angle expansion.
Forgetting that the cross terms combine to $-\sin\theta\cos\theta$ .
Stopping at $\frac{\sqrt3}{4}-\sin\theta\cos\theta$ without converting to $\frac12\sin2\theta$ .

Things to Be Careful About

Since the answer is given, the mark scheme requires all necessary detail; an unsupported conclusion may lose marks.
Keep $\cos^2\theta$ and $\sin^2\theta$ grouped so the Pythagorean identity is clear.
Use $\theta$ consistently; do not mix it with $2\theta$ prematurely.

Techniques used

expand compound-angle cosine expressionssubstitute exact trigonometric valuesuse the Pythagorean identityapply the double-angle formula for sine

(b)

Solve the equation $5\cos(2\alpha + 30^\circ)\cos(2\alpha + 60^\circ) = 1$ for $0^\circ < \alpha < 90^\circ$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Apply the identity from part (a) with $\theta=2\alpha$ to turn the product into a single sine term. Solve for $\sin 4\alpha$ , then find all angles $4\alpha$ in the allowed interval and divide by 4.

Working

Let $\theta = 2\alpha$ . From part (a):

\cos(2\alpha+30^\circ)\cos(2\alpha+60^\circ) =\frac{\sqrt3}{4}-\frac12\sin4\alpha

Substitute into the equation:

5\left(\frac{\sqrt3}{4}-\frac12\sin4\alpha\right)=1

\frac{5\sqrt3}{4}-\frac52\sin4\alpha=1

\frac52\sin4\alpha=\frac{5\sqrt3}{4}-1=\frac{5\sqrt3-4}{4}

\sin4\alpha=\frac{5\sqrt3-4}{10} =\frac{\sqrt3}{2}-\frac25

Numerically:

\sin4\alpha=0.4660\ldots

Since $0^\circ<\alpha<90^\circ$ , we need $0^\circ<4\alpha<360^\circ$ . The solutions of $\sin\phi=0.4660\ldots$ in this interval are:

\phi=27.8^\circ \quad \text{or} \quad \phi=180^\circ-27.8^\circ=152.2^\circ

Therefore:

\alpha=\frac{27.8^\circ}{4}=6.9^\circ \quad \text{or} \quad \alpha=\frac{152.2^\circ}{4}=38.1^\circ

No further solutions occur in $0^\circ<\alpha<90^\circ$ .

Answer

\alpha=6.9^\circ \quad \text{or} \quad \alpha=38.1^\circ

Final answer

alpha = 6.9 degrees or alpha = 38.1 degrees

Detailed explanation

Walkthrough

Use the identity from part (a) with $\theta=2\alpha$ . This converts the product of cosines into $\frac{\sqrt3}{4}-\frac12\sin4\alpha$ . Multiply by 5 and set equal to 1. Rearrange to isolate $\sin4\alpha$ , obtaining $\sin4\alpha=\frac{\sqrt3}{2}-\frac25\approx0.4660$ . Because $0^\circ<\alpha<90^\circ$ , the angle $4\alpha$ lies in $0^\circ<4\alpha<360^\circ$ . Sine is positive in the first and second quadrants, so take the principal value $\phi=\sin^{-1}(0.4660)\approx27.8^\circ$ and also $\phi=180^\circ-27.8^\circ=152.2^\circ$ . Divide by 4 to get $\alpha\approx6.9^\circ$ and $\alpha\approx38.1^\circ$ . No other angles in the interval satisfy the equation.

Key Takeaways

This part combines a proven identity with solving a trigonometric equation. The key skill is to solve for the inner angle $4\alpha$ first, over its full range, and then divide by 4. It also reinforces that $\sin x=k$ has two solutions in $0^\circ$ to $360^\circ$ when $0<k<1$ .

Common Mistakes

Forgetting to multiply the whole bracket by 5.
Solving for $4\alpha$ but forgetting to divide by 4.
Taking only the principal-value solution and missing the second-quadrant solution.
Giving solutions for $4\alpha$ instead of for $\alpha$ .
Not checking that the final values lie in $0^\circ<\alpha<90^\circ$ .

Things to Be Careful About

The mark scheme requires a correct process to obtain at least one value of $\alpha$ , so show the inverse-sine step and the division by 4.
Use degrees throughout; the range is given in degrees.
The exact form $\frac{\sqrt3}{2}-\frac25$ is equivalent to $\frac{5\sqrt3-4}{10}$ ; either is acceptable.
There are no additional solutions such as $4\alpha=360^\circ+27.8^\circ$ because that exceeds $360^\circ$ .

Techniques used

apply the proven identityisolate the sine termsolve a sine equation over a restricted intervaldivide angles to obtain the variable

(c)

Show that the exact value of $\cos 20^\circ \cos 50^\circ + \cos 40^\circ \cos 70^\circ$ is $\frac{1}{2}\sqrt{3}$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the identity from part (a) twice, choosing $\theta=-10^\circ$ and $\theta=10^\circ$ so that the two sine terms cancel. Then add the two results.

Working

From part (a):

\cos(\theta+30^\circ)\cos(\theta+60^\circ) =\frac{\sqrt3}{4}-\frac12\sin2\theta

Put $\theta=-10^\circ$ :

\cos20^\circ\cos50^\circ =\frac{\sqrt3}{4}-\frac12\sin(-20^\circ) =\frac{\sqrt3}{4}+\frac12\sin20^\circ

Put $\theta=10^\circ$ :

\cos40^\circ\cos70^\circ =\frac{\sqrt3}{4}-\frac12\sin20^\circ

Adding:

\begin{aligned} \cos20^\circ\cos50^\circ+\cos40^\circ\cos70^\circ &=\left(\frac{\sqrt3}{4}+\frac12\sin20^\circ\right) +\left(\frac{\sqrt3}{4}-\frac12\sin20^\circ\right)\\ &=\frac{\sqrt3}{2} \end{aligned}

Since $\sin(-20^\circ)=-\sin20^\circ$ . Therefore:

\cos20^\circ\cos50^\circ+\cos40^\circ\cos70^\circ=\frac12\sqrt3

Answer

\cos20^\circ\cos50^\circ+\cos40^\circ\cos70^\circ=\frac12\sqrt3

Final answer

cos20 cos50 + cos40 cos70 = sqrt(3)/2

Detailed explanation

Walkthrough

The expression is exactly the left-hand side of the identity from part (a) for two different choices of $\theta$ . For the first term, choose $\theta=-10^\circ$ because $\theta+30^\circ=20^\circ$ and $\theta+60^\circ=50^\circ$ . For the second term, choose $\theta=10^\circ$ because $\theta+30^\circ=40^\circ$ and $\theta+60^\circ=70^\circ$ . Substituting gives two expressions involving $\sin20^\circ$ with opposite signs: $\frac{\sqrt3}{4}+\frac12\sin20^\circ$ and $\frac{\sqrt3}{4}-\frac12\sin20^\circ$ . Adding cancels the sine terms, leaving $\frac{\sqrt3}{2}=\frac12\sqrt3$ . The key point is $\sin(-20^\circ)=-\sin20^\circ$ .

Key Takeaways

This shows how a single identity can be reused with different substitutions to evaluate a sum of trigonometric products. It also uses the odd symmetry of sine and exact trigonometric values.

Common Mistakes

Choosing the wrong values of $\theta$ so the angles do not match.
Forgetting that $\sin(-20^\circ)=-\sin20^\circ$ , which would prevent the cancellation.
Writing the final answer as $\frac{\sqrt3}{4}$ instead of $\frac{\sqrt3}{2}$ .
Not showing the substitution steps clearly in an answer-given question.

Things to Be Careful About

The mark scheme awards separate marks for each substitution and for the final addition, so show both substitutions explicitly.
The answer is given; all necessary detail must be shown.
$\frac{\sqrt3}{2}$ and $\frac12\sqrt3$ are the same value; either form is acceptable.

Techniques used

choose suitable values of theta in the identityuse the odd symmetry of sineadd the resulting expressions

Mathematics 9709/23 — October/November 2024

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Common Mistakes

Things to Be Careful About

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