(a) Show that $$ 3\tan 2\theta + \tan(\theta + 45^{\circ}) \equiv \frac{\tan^2\theta + 8\tan\theta + 1}{1 - \tan^2\theta} $$ (b) Hence solve the equation $3\tan 2\theta + \tan(\theta + 45^{\circ}) = 4$ for $0^{\circ} < \theta < 180^{\circ}$.

(a) Identity shown: (tan^2 θ + 8 tan θ + 1)/(1 - tan^2 θ) · (b) θ = 17.4°, 117.6°

A curve has equation $y = \frac{1 + e^{2x}}{1 + 3x}$. The curve has exactly one stationary point $P$. (a) Find $\frac{dy}{dx}$ and hence show that the $x$-coordinate of $P$ satisfies the equation $x = \frac{1}{6} + \frac{1}{2}e^{-2x}$. (b) Show by calculation that the $x$-coordinate of $P$ lies between 0.35 and 0.45. (c) Use an iterative formula based on the equation in part (a) to find the $x$-coordinate of $P$ correct to 3 significant figures. Give the result of each…

(a) x = 1/6 + (1/2)e^(-2x) · (b) f(0.35) 0, so the x-coordinate of P lies between 0.35 and 0.45 · (c) x = 0.394

9709/21 May/June 2024 — Mathematics Solved Past Paper

Q13MMedium-EasyDifferentiation

A curve has equation $y = 2\tan x - 5\sin x$ for $0 \le x < \frac{1}{2}\pi$ .

Find the $x$ -coordinate of the stationary point of the curve. Give your answer correct to 3 significant figures.

DifficultyMedium-Easy

Worked solution

Approach

Differentiate $y$ with respect to $x$ using the standard derivatives $\frac{d}{dx}(\tan x) = \sec^2 x$ and $\frac{d}{dx}(\sin x) = \cos x$ . Set the derivative equal to zero because stationary points occur where $\frac{dy}{dx} = 0$ . Then use $\sec x = \frac{1}{\cos x}$ to obtain an equation in $\cos x$ , and take the inverse cosine.

Working

Differentiate:

\frac{dy}{dx} = 2\sec^2 x - 5\cos x

At a stationary point, $\frac{dy}{dx} = 0$ :

2\sec^2 x - 5\cos x = 0

2\sec^2 x = 5\cos x

Use $\sec^2 x = \frac{1}{\cos^2 x}$ :

\frac{2}{\cos^2 x} = 5\cos x

Multiply by $\cos^2 x$ (valid because $0 \leq x < \frac{\pi}{2}$ , so $\cos x \neq 0$ ):

2 = 5\cos^3 x

\cos^3 x = \frac{2}{5} = 0.4

\cos x = \sqrt[3]{0.4}

Therefore:

x = \cos^{-1}\left(\sqrt[3]{0.4}\right)

x = 0.74246\ldots

Correct to 3 significant figures:

x = 0.742

Answer

x = 0.742

Final answer

x = 0.742

Detailed explanation

Walkthrough

The first step is to differentiate the curve. The derivative of $\tan x$ is $\sec^2 x$ , and the derivative of $\sin x$ is $\cos x$ , so

\frac{dy}{dx} = 2\sec^2 x - 5\cos x.

A stationary point occurs where the gradient is zero, so set this derivative equal to zero:

2\sec^2 x - 5\cos x = 0.

To solve this, rewrite $\sec^2 x$ as $\frac{1}{\cos^2 x}$ . This turns the equation into one involving only $\cos x$ :

\frac{2}{\cos^2 x} = 5\cos x.

Multiplying through by $\cos^2 x$ gives $2 = 5\cos^3 x$ , so $\cos^3 x = 0.4$ . Taking the cube root gives $\cos x = \sqrt[3]{0.4}$ . Finally, apply $\cos^{-1}$ to find the angle in radians:

x = \cos^{-1}\left(\sqrt[3]{0.4}\right) = 0.74246\ldots

Rounded to 3 significant figures, the $x$ -coordinate is $0.742$ .

Key Takeaways

The standard derivatives $\frac{d}{dx}(\tan x) = \sec^2 x$ and $\frac{d}{dx}(\sin x) = \cos x$ are essential.
Stationary points are found by solving $\frac{dy}{dx} = 0$ .
The identity $\sec x = \frac{1}{\cos x}$ lets us rewrite a derivative equation as a purely trigonometric equation.
In calculus, angles are measured in radians unless the question says otherwise.

Common Mistakes

Forgetting the constant factors: the derivative is $2\sec^2 x - 5\cos x$ , not $\sec^2 x - \cos x$ .
Trying to solve $2\sec^2 x = 5\cos x$ without converting $\sec^2 x$ to $1/\cos^2 x$ .
Taking a square root instead of a cube root after obtaining $\cos^3 x = 0.4$ .
Giving the answer in degrees. The mark scheme allows $42.5^{\circ}$ as an intermediate step, but the final answer must be in radians: $0.742$ .

Things to Be Careful About

The domain $0 \leq x < \frac{\pi}{2}$ ensures $\cos x > 0$ , so multiplying by $\cos^2 x$ is safe and no extraneous solutions are introduced.
Use the cube root, not the square root, when solving $\cos^3 x = 0.4$ .
Round only at the final step. The unrounded value is $0.74246\ldots$ , so to 3 significant figures it is $0.742$ .
Make sure your calculator is in radian mode before evaluating $\cos^{-1}\left(\sqrt[3]{0.4}\right)$ .

Techniques used

differentiate tan x and sin x using standard derivativesset the derivative equal to zero to locate stationary pointsrewrite sec^2 x in terms of cos xsolve the resulting trigonometric equationevaluate inverse cosine and round to 3 significant figures

Approach

Differentiate the given equation implicitly with respect to $x$ , using the product rule for the term $x^2\ln y$ and the chain rule for terms involving $y$ . Then substitute the coordinates of the given point and solve for $\frac{dy}{dx}$ .

Working

Differentiate each term with respect to $x$ :

\frac{d}{dx}\left(x^2\ln y\right) = 2x\ln y + x^2 \cdot \frac{1}{y}\frac{dy}{dx}

\frac{d}{dx}\left(y^2\right) = 2y\frac{dy}{dx}

\frac{d}{dx}(4x) = 4, \qquad \frac{d}{dx}(9) = 0

So the differentiated equation is

2x\ln y + \frac{x^2}{y}\frac{dy}{dx} + 2y\frac{dy}{dx} + 4 = 0

Substitute $x = 2$ and $y = 1$ :

2(2)\ln 1 + \frac{2^2}{1}\frac{dy}{dx} + 2(1)\frac{dy}{dx} + 4 = 0

Since $\ln 1 = 0$ , this becomes

4\frac{dy}{dx} + 2\frac{dy}{dx} + 4 = 0

6\frac{dy}{dx} = -4

\frac{dy}{dx} = -\frac{2}{3}

Answer

\frac{dy}{dx} = -\frac{2}{3}

Final answer

-2/3

Detailed explanation

Walkthrough

We are given a curve defined implicitly by an equation in $x$ and $y$ , and we need its gradient at a point. Since $y$ is not written explicitly as a function of $x$ , we differentiate both sides with respect to $x$ and treat $y$ as a function of $x$ . Every time we differentiate an expression involving $y$ , we must multiply by $\frac{dy}{dx}$ .

The first term $x^2\ln y$ is a product of $x^2$ and $\ln y$ , so we use the product rule:

\frac{d}{dx}(x^2\ln y) = 2x\ln y + x^2\cdot\frac{1}{y}\frac{dy}{dx}

The derivative of $\ln y$ is $\frac{1}{y}\frac{dy}{dx}$ by the chain rule.

The second term $y^2$ differentiates to $2y\frac{dy}{dx}$ , again by the chain rule.

The term $4x$ differentiates to $4$ , and the constant $9$ differentiates to $0$ . Collecting these gives the implicit derivative equation.

Then we substitute the given point $(2,1)$ . Because $\ln 1 = 0$ , the first term disappears, and the equation simplifies to a linear equation in $\frac{dy}{dx}$ . Solving it gives the gradient.

Key Takeaways

Implicit differentiation: differentiate both sides with respect to $x$ and multiply each derivative of a $y$ -expression by $\frac{dy}{dx}$ .
The product rule is often needed for terms that contain both $x$ and $y$ .
After differentiating, substitute the given coordinates and solve for $\frac{dy}{dx}$ .
Remember $\ln 1 = 0$ ; this often simplifies substitution.

Common Mistakes

Forgetting the $\frac{dy}{dx}$ factor when differentiating $\ln y$ or $y^2$ .
Applying the product rule incorrectly to $x^2\ln y$ ; the first term is $2x\ln y$ , not $2x$ alone.
Forgetting to include the derivative of $4x$ (which is $4$ ) on the left-hand side.
Substituting the point before differentiating.
Making an arithmetic error when solving $6\frac{dy}{dx} = -4$ ; the answer must be exact, e.g. $-\frac{2}{3}$ , and not obtained from wrong working.

Things to Be Careful About

The derivative of $\ln y$ with respect to $x$ is $\frac{1}{y}\frac{dy}{dx}$ , not just $\frac{1}{y}$ .
The derivative of $y^2$ is $2y\frac{dy}{dx}$ , not $2y$ .
The right-hand side is a constant, so its derivative is $0$ ; keep the equation equal to $0$ .
When substituting, $\ln 1 = 0$ , so the $2x\ln y$ term vanishes.
The mark scheme requires an exact equivalent answer and does not allow an answer obtained from wrong working, so show all steps clearly.

Techniques used

differentiate implicitly with respect to xapply the product rulesubstitute given coordinatessolve for dy/dx

Approach

Sketch the V-shaped modulus graph by locating its vertex and the gradients of its two branches, then draw the line $y = 5 - x$ on the same axes, marking the two intersections.

Working

For $y = |3x - 8|$ :

Vertex: $3x - 8 = 0$ , so $x = \frac{8}{3}$ and $y = 0$ .
For $x \geq \frac{8}{3}$ , $y = 3x - 8$ (gradient $3$ ).
For $x \leq \frac{8}{3}$ , $y = 8 - 3x$ (gradient $-3$ ).

The line $y = 5 - x$ has gradient $-1$ and intercept $5$ .

Find the intersection points. For $x < \frac{8}{3}$ :

-(3x - 8) = 5 - x

-3x + 8 = 5 - x

x = \frac{3}{2}

For $x \geq \frac{8}{3}$ :

3x - 8 = 5 - x

4x = 13

x = \frac{13}{4}

So the graphs meet at $\left(\frac{3}{2}, \frac{7}{2}\right)$ and $\left(\frac{13}{4}, \frac{7}{4}\right)$ .

Answer

Sketch the V-shaped graph $y = |3x - 8|$ with vertex $\left(\frac{8}{3}, 0\right)$ , and the line $y = 5 - x$ , intersecting at $x = \frac{3}{2}$ and $x = \frac{13}{4}$ .

Final answer

V-shaped graph with vertex (8/3, 0) and line y = 5 - x intersecting at x = 3/2 and x = 13/4

Detailed explanation

Walkthrough

The graph of $y=|3x-8|$ is not a single straight line. The modulus changes the sign of $3x-8$ when $3x-8=0$ , i.e. at $x=\frac{8}{3}$ . To the right of this point $3x-8$ is positive, so the graph is $y=3x-8$ ; to the left it is negative, so the graph is $y=-(3x-8)=8-3x$ . This creates a V shape with vertex at $\left(\frac{8}{3},0\right)$ .

The line $y=5-x$ is a straight line with slope $-1$ and $y$ -intercept $5$ . To place it correctly, find where it meets each arm of the V. On the left arm use $8-3x=5-x$ , giving $x=\frac{3}{2}$ . On the right arm use $3x-8=5-x$ , giving $x=\frac{13}{4}$ . These are the two intersection points required by the mark scheme.

Key Takeaways

A modulus expression $|ax+b|$ can be graphed as two linear pieces meeting at the point where $ax+b=0$ . The graph is V-shaped. Solving $|ax+b|=cx+d$ requires solving two linear equations, one for each branch.

Common Mistakes

Drawing the vertex at the wrong point; the vertex is where $3x-8=0$ , not at the $y$ -intercept.
Drawing only one arm of the V.
Placing the line $y=5-x$ so that it does not cross both arms.
Forgetting that the two intersections are needed to show the line is correctly positioned.

Things to Be Careful About

Use the correct branch conditions: $x \geq \frac{8}{3}$ for the right arm and $x \leq \frac{8}{3}$ for the left arm. The intersection on the left arm is $\left(\frac{3}{2},\frac{7}{2}\right)$ and on the right arm is $\left(\frac{13}{4},\frac{7}{4}\right)$ . Label these points on the sketch.

Techniques used

locate the vertex of a modulus graphsketch the two linear branches of a modulus functionfind intersection points of modulus and linear graphs

(b)

Solve the inequality $|3x - 8| < 5 - x$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Solve the boundary equation $|3x - 8| = 5 - x$ by considering the two possible signs of $3x - 8$ , then use a test value to decide which interval satisfies the strict inequality.

Working

The boundary values satisfy $|3x - 8| = 5 - x$ .

Case 1: $x \geq \frac{8}{3}$

3x - 8 = 5 - x

4x = 13

x = \frac{13}{4}

Case 2: $x < \frac{8}{3}$

-(3x - 8) = 5 - x

-3x + 8 = 5 - x

-2x = -3

x = \frac{3}{2}

Test $x = 2$ :

|3(2) - 8| = 2, \quad 5 - 2 = 3

Since $2 < 3$ , the interval between $\frac{3}{2}$ and $\frac{13}{4}$ satisfies the inequality. The endpoints are excluded because the inequality is strict.

Answer

\frac{3}{2} < x < \frac{13}{4}

Final answer

3/2 < x < 13/4

Detailed explanation

Walkthrough

To solve $|3x-8|<5-x$ , first solve the corresponding equality because the boundary of the inequality occurs where the two sides are equal. The modulus requires two cases: when $3x-8\ge 0$ , the modulus sign can be removed; when $3x-8<0$ , it is replaced by a negative sign.

For $x\ge\frac{8}{3}$ , the equation becomes $3x-8=5-x$ , giving $x=\frac{13}{4}$ . For $x<\frac{8}{3}$ , it becomes $-(3x-8)=5-x$ , giving $x=\frac{3}{2}$ . These are the two boundary values.

To decide which region satisfies the strict inequality, test a point between the boundaries, such as $x=2$ . Since $|3(2)-8|=2<3=5-2$ , the interval between the boundaries is correct. The endpoints are not included because the inequality is strict.

Key Takeaways

Modulus inequalities can be solved by first solving the associated equality, then testing intervals. The solution can also be read from the graph: the modulus graph must lie below the line $y=5-x$ .

Common Mistakes

Including the endpoints $\frac{3}{2}$ and $\frac{13}{4}$ ; the inequality is strict, so they must be excluded.
Choosing the wrong interval; always test a point.
Forgetting the branch condition when removing the modulus sign.
Reversing the inequality when dividing by a negative number.

Things to Be Careful About

When $x<\frac{8}{3}$ , $3x-8$ is negative, so $|3x-8|=8-3x$ . The two boundary values are $x=\frac{3}{2}$ and $x=\frac{13}{4}$ . The final answer may be written as $\frac{3}{2}<x<\frac{13}{4}$ or $x>\frac{3}{2}$ and $x<\frac{13}{4}$ .

Techniques used

split the modulus equation into sign casessolve linear equations or inequalitiescombine intervals and test a point

(c)

Hence determine the largest integer $N$ satisfying the inequality $|3e^{0.1N} - 8| < 5 - e^{0.1N}$ .

2M Similar

DifficultyMedium

Worked solution

Approach

Set $x = e^{0.1N}$ . The inequality has exactly the same form as part (b), so $e^{0.1N}$ must lie between the two boundary values. Use natural logarithms to isolate $N$ , then choose the largest integer that satisfies both bounds.

Working

From part (b), the inequality $|3x - 8| < 5 - x$ is equivalent to

\frac{3}{2} < x < \frac{13}{4}

Substitute $x = e^{0.1N}$ :

\frac{3}{2} < e^{0.1N} < \frac{13}{4}

Since $e^{0.1N}$ is increasing, take natural logarithms:

\ln\left(\frac{3}{2}\right) < 0.1N < \ln\left(\frac{13}{4}\right)

For the upper bound:

N < 10\ln\left(\frac{13}{4}\right) \approx 10(1.17865) = 11.7865

The lower bound gives $N > 10\ln\left(\frac{3}{2}\right) \approx 4.055$ , so the largest integer is $N = 11$ .

Check:

e^{0.1(11)} = e^{1.1} \approx 3.004

and

\frac{3}{2} < 3.004 < \frac{13}{4} = 3.25

so $N = 11$ is valid.

Answer

N = 11

Final answer

N = 11

Detailed explanation

Walkthrough

The key idea is to notice that the inequality in part (c) is the same as part (b) with $x$ replaced by $e^{0.1N}$ . Therefore the solution of part (b) applies directly: $e^{0.1N}$ must be between $\frac{3}{2}$ and $\frac{13}{4}$ .

Since the exponential function is increasing, we can take natural logarithms of all three parts without changing the direction of the inequalities. This gives $0.1N$ between $\ln(3/2)$ and $\ln(13/4)$ . Multiplying by $10$ isolates $N$ .

The upper bound gives $N<11.7865$ , so the largest possible integer is $11$ . The lower bound gives $N>4.055$ , so $N=11$ is certainly within the interval. We check by substituting back: $e^{1.1}\approx3.004$ , which lies between $1.5$ and $3.25$ .

Key Takeaways

Recognising a repeated inequality structure allows substitution. Logarithms are used to solve inequalities where the unknown is in an exponent. The monotonicity of $e^x$ means taking logs preserves the inequality direction.

Common Mistakes

Rounding $11.7865$ up to $12$ ; the inequality is strict, so $N$ must be less than $11.7865$ , making $11$ the largest integer.
Forgetting the lower bound and not checking that $N=11$ satisfies both sides.
Using the wrong base for the logarithm; here natural logarithms are appropriate because the expression is $e^{0.1N}$ .

Things to Be Careful About

The inequality is strict on both sides. The upper bound is the one that limits the largest integer. Always verify the chosen integer by substituting back into the original exponential inequality.

Techniques used

substitute an exponential expression into a known inequalityapply natural logarithms to isolate the exponentselect the largest integer satisfying the bounds

Q4MediumTrigonometry

(a)

Show that

3\tan 2\theta + \tan(\theta + 45^{\circ}) \equiv \frac{\tan^2\theta + 8\tan\theta + 1}{1 - \tan^2\theta}

4M Similar

DifficultyMedium

Worked solution

Approach

Use the double angle formula for $\tan 2\theta$ and the compound angle formula for $\tan(\theta + 45^\circ)$ to write the left-hand side entirely in terms of $\tan\theta$ , then combine the two terms over a common denominator.

Working

Let $t = \tan\theta$ . Then:

3\tan 2\theta = 3\left(\frac{2t}{1 - t^2}\right) = \frac{6t}{1 - t^2}

Also, since $\tan 45^\circ = 1$ :

\tan(\theta + 45^\circ) = \frac{t + 1}{1 - t}

To combine the fractions, write the second term with denominator $1 - t^2 = (1-t)(1+t)$ :

\frac{t+1}{1-t} = \frac{(t+1)^2}{(1-t)(1+t)} = \frac{(t+1)^2}{1 - t^2}

Therefore:

\frac{6t}{1 - t^2} + \frac{(t+1)^2}{1 - t^2} = \frac{6t + (t+1)^2}{1 - t^2}

\frac{6t + (t+1)^2}{1 - t^2} = \frac{t^2 + 8t + 1}{1 - t^2}

Replacing $t$ with $\tan\theta$ gives the required identity.

Answer

3\tan 2\theta + \tan(\theta + 45^\circ) \equiv \frac{\tan^2\theta + 8\tan\theta + 1}{1 - \tan^2\theta}

Final answer

Identity shown: (tan^2 θ + 8 tan θ + 1)/(1 - tan^2 θ)

Detailed explanation

Walkthrough

The left-hand side of the identity contains two trigonometric terms, while the right-hand side is a single fraction in $\tan\theta$ . To prove the identity, express both terms on the left in terms of $\tan\theta$ and combine them.

First, use the double angle formula for tangent:

\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}

Multiplying by $3$ gives $\frac{6\tan\theta}{1 - \tan^2\theta}$ .

Next, use the compound angle formula for tangent:

\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}

With $A = \theta$ and $B = 45^\circ$ , and using $\tan 45^\circ = 1$ , this becomes:

\tan(\theta + 45^\circ) = \frac{\tan\theta + 1}{1 - \tan\theta}

The two terms now have different denominators. Since $1 - \tan^2\theta = (1 - \tan\theta)(1 + \tan\theta)$ , multiply the numerator and denominator of the second fraction by $1 + \tan\theta$ :

\frac{\tan\theta + 1}{1 - \tan\theta} = \frac{(\tan\theta + 1)^2}{1 - \tan^2\theta}

Now both fractions have the same denominator, so add their numerators:

\frac{6\tan\theta}{1 - \tan^2\theta} + \frac{(\tan\theta + 1)^2}{1 - \tan^2\theta} = \frac{6\tan\theta + \tan^2\theta + 2\tan\theta + 1}{1 - \tan^2\theta}

Simplify the numerator to $\tan^2\theta + 8\tan\theta + 1$ , which is exactly the right-hand side.

Key Takeaways

The double angle formula for $\tan$ expresses $\tan 2\theta$ in terms of $\tan\theta$ .
The compound angle formula for $\tan$ expresses $\tan(\theta + 45^\circ)$ in terms of $\tan\theta$ .
Rational expressions with related denominators can be combined by rewriting one denominator as a product and scaling the other fraction.
When an identity is given, all algebraic steps must be shown; the final answer cannot be assumed.

Common Mistakes

Forgetting that $\tan 45^\circ = 1$ .
Using the wrong sign in the denominator of the compound angle formula: the denominator is $1 - \tan\theta$ , not $1 + \tan\theta$ .
Stopping after writing the two terms separately instead of combining them into a single fraction.
Making sign errors when expanding $(\tan\theta + 1)^2$ .

Things to Be Careful About

The identity is only defined where $1 - \tan^2\theta \neq 0$ ; the proof is algebraic and assumes the expressions are defined.
The mark scheme requires the left-hand side to be expressed as a single fraction before the final answer is confirmed.
Use $\tan\theta$ consistently; substituting a letter such as $t$ can make the algebra clearer.

Techniques used

apply the double angle formula for tanapply the compound angle formula for tancombine algebraic fractions over a common denominatorsimplify the numerator to match the given expression

(b)

Hence solve the equation $3\tan 2\theta + \tan(\theta + 45^{\circ}) = 4$ for $0^{\circ} < \theta < 180^{\circ}$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the identity from part (a) to replace the left-hand side of the equation, then solve the resulting quadratic in $\tan\theta$ . Finally use the period of $\tan$ to find all solutions in $0^\circ < \theta < 180^\circ$ .

Working

From part (a):

\frac{\tan^2\theta + 8\tan\theta + 1}{1 - \tan^2\theta} = 4

Let $t = \tan\theta$ . Then:

\frac{t^2 + 8t + 1}{1 - t^2} = 4

Multiply both sides by $1 - t^2$ :

t^2 + 8t + 1 = 4(1 - t^2)

t^2 + 8t + 1 = 4 - 4t^2

5t^2 + 8t - 3 = 0

Using the quadratic formula:

t = \frac{-8 \pm \sqrt{8^2 - 4(5)(-3)}}{2(5)} = \frac{-8 \pm \sqrt{64 + 60}}{10} = \frac{-8 \pm \sqrt{124}}{10}

t = \frac{-8 \pm 2\sqrt{31}}{10} = \frac{-4 \pm \sqrt{31}}{5}

So $\tan\theta = \frac{-4 + \sqrt{31}}{5} \approx 0.3135$ or $\tan\theta = \frac{-4 - \sqrt{31}}{5} \approx -1.9135$ .

For $\tan\theta \approx 0.3135$ :

\theta = \tan^{-1}(0.3135) \approx 17.4^\circ

Since $\tan$ has period $180^\circ$ , the other angle with this tangent is $17.4^\circ + 180^\circ = 197.4^\circ$ , which is outside the interval. So this tangent gives only $\theta \approx 17.4^\circ$ .

For $\tan\theta \approx -1.9135$ :

\theta = \tan^{-1}(-1.9135) \approx -62.4^\circ

Adding $180^\circ$ gives:

\theta \approx -62.4^\circ + 180^\circ = 117.6^\circ

This is in the interval. The next angle would be $117.6^\circ + 180^\circ = 297.6^\circ$ , outside the interval.

Answer

\theta = 17.4^\circ \quad \text{or} \quad \theta = 117.6^\circ

Final answer

θ = 17.4°, 117.6°

Detailed explanation

Walkthrough

Part (b) says "Hence solve", so use the identity proved in part (a) to replace the left-hand side of the equation $3\tan 2\theta + \tan(\theta + 45^\circ) = 4$ . This gives:

\frac{\tan^2\theta + 8\tan\theta + 1}{1 - \tan^2\theta} = 4

Let $t = \tan\theta$ . Multiplying both sides by $1 - t^2$ gives:

t^2 + 8t + 1 = 4 - 4t^2

Rearranging:

5t^2 + 8t - 3 = 0

This is a quadratic in $t$ . Solve it with the quadratic formula:

t = \frac{-8 \pm \sqrt{8^2 - 4(5)(-3)}}{2(5)} = \frac{-8 \pm \sqrt{124}}{10} = \frac{-4 \pm \sqrt{31}}{5}

So the two possible values are $t \approx 0.3135$ and $t \approx -1.9135$ .

Now find the angles. For a positive tangent, the calculator gives the first-quadrant angle $17.4^\circ$ . Because $\tan$ has period $180^\circ$ , the next angle with the same tangent is $197.4^\circ$ , which is outside $0^\circ < \theta < 180^\circ$ , so only $17.4^\circ$ is kept.

For the negative tangent, the calculator gives $-62.4^\circ$ . Adding $180^\circ$ gives $117.6^\circ$ , which lies in the interval. The next angle would be $297.6^\circ$ , outside the interval, so the solutions are $17.4^\circ$ and $117.6^\circ$ .

Key Takeaways

A trigonometric equation can often be reduced to a quadratic by using a proven identity.
The inverse tangent function gives only a principal value; the period of $\tan$ is $180^\circ$ , so add or subtract $180^\circ$ to find all solutions in a given interval.
Positive tangent values correspond to angles in the first and third quadrants; negative tangent values correspond to the second and fourth quadrants.

Common Mistakes

Forgetting the second solution by only taking the principal value of $\tan^{-1}$ .
Including $197.4^\circ$ or $-62.4^\circ$ as answers without checking the interval.
Making a sign error when rearranging $t^2 + 8t + 1 = 4 - 4t^2$ ; the $t^2$ terms combine to give $5t^2$ .
Not using the identity from part (a), which is the intended method for the "Hence".

Things to Be Careful About

The interval is open, $0^\circ < \theta < 180^\circ$ , so endpoints are not included.
The period of $\tan$ is $180^\circ$ , not $360^\circ$ .
Both roots of the quadratic must be considered; one root can give one solution and the other root can give another.
The mark scheme requires a correct method for solving the quadratic and both correct angles with no extra angles in the interval.

Techniques used

substitute the identity from part (a)form and solve a quadratic in tan thetause the period of tan to find all solutions in the interval

Q5MediumDifferentiationNumerical Solution of Equations

A curve has equation $y = \frac{1 + e^{2x}}{1 + 3x}$ . The curve has exactly one stationary point $P$ .

(a)

Find $\frac{dy}{dx}$ and hence show that the $x$ -coordinate of $P$ satisfies the equation $x = \frac{1}{6} + \frac{1}{2}e^{-2x}$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Differentiate $y = \frac{1 + e^{2x}}{1 + 3x}$ using the quotient rule. At the stationary point $P$ , set $\frac{dy}{dx} = 0$ and rearrange the resulting equation to isolate $x$ .

Working

Let $u = 1 + e^{2x}$ and $v = 1 + 3x$ . Then $u' = 2e^{2x}$ and $v' = 3$ .

Using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$ :

\frac{dy}{dx} = \frac{(1 + 3x)(2e^{2x}) - (1 + e^{2x})(3)}{(1 + 3x)^2}

At the stationary point, $\frac{dy}{dx} = 0$ , so the numerator must be zero:

2e^{2x}(1 + 3x) - 3(1 + e^{2x}) = 0

Expand the brackets:

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

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

e^{2x}(6x - 1) = 3

Divide both sides by $e^{2x}$ :

6x - 1 = 3e^{-2x}

6x = 1 + 3e^{-2x}

x = \frac{1}{6} + \frac{1}{2}e^{-2x}

Answer

The $x$ -coordinate of $P$ satisfies

x = \frac{1}{6} + \frac{1}{2}e^{-2x}

Final answer

x = 1/6 + (1/2)e^(-2x)

Detailed explanation

Walkthrough

We are given a curve $y = \frac{1 + e^{2x}}{1 + 3x}$ and told it has exactly one stationary point $P$ . A stationary point occurs where the derivative is zero, so the first task is to differentiate the function.

The function is a quotient of two expressions, so the quotient rule is the natural tool. With $u = 1 + e^{2x}$ and $v = 1 + 3x$ , we compute $u' = 2e^{2x}$ (the derivative of $e^{2x}$ is $2e^{2x}$ by the chain rule) and $v' = 3$ .

Substituting into the quotient rule gives the derivative as a single fraction. Setting the numerator equal to zero (since the denominator $(1+3x)^2$ is never zero for the relevant values) yields the stationary-point condition.

Expanding the brackets and collecting like terms gives $e^{2x}(6x - 1) = 3$ . To isolate $x$ , divide both sides by $e^{2x}$ , which is valid since $e^{2x} > 0$ for all real $x$ . This produces $6x - 1 = 3e^{-2x}$ , and a final rearrangement gives the required equation.

Key Takeaways

The quotient rule is essential for differentiating fractions of functions.
At a stationary point, $\frac{dy}{dx} = 0$ .
Dividing by $e^{2x}$ is always safe because exponential functions are strictly positive.
The equation $x = \frac{1}{6} + \frac{1}{2}e^{-2x}$ is an implicit equation for $x$ — it cannot be solved algebraically in closed form, which motivates the numerical methods in parts (b) and (c).

Common Mistakes

Forgetting the minus sign in the quotient rule: the formula is $\frac{u'v - uv'}{v^2}$ , not $\frac{u'v + uv'}{v^2}$ .
Incorrectly differentiating $e^{2x}$ — the chain rule gives $2e^{2x}$ , not $e^{2x}$ .
Dropping the factor of 3 when differentiating $1 + 3x$ .
Mark scheme guidance: since the answer is given in the question, "necessary detail" must be shown — every algebraic step from $\frac{dy}{dx} = 0$ to the final equation must be present.

Things to Be Careful About

The denominator $(1 + 3x)^2$ is positive for all $x \neq -\frac{1}{3}$ , so setting the numerator to zero is fully equivalent to setting $\frac{dy}{dx} = 0$ .
Use exact terms throughout: keep $\frac{1}{6}$ and $\frac{1}{2}$ as fractions rather than decimals to match the required form exactly.

Techniques used

apply the quotient rule to differentiate a rational functionset the derivative equal to zero at a stationary pointrearrange the resulting equation using exponential laws

(b)

Show by calculation that the $x$ -coordinate of $P$ lies between 0.35 and 0.45.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Define $f(x) = x - \frac{1}{6} - \frac{1}{2}e^{-2x}$ . The $x$ -coordinate of $P$ is a root of $f(x) = 0$ . Evaluate $f$ at $x = 0.35$ and $x = 0.45$ and show the signs differ.

Working

Let

f(x) = x - \frac{1}{6} - \frac{1}{2}e^{-2x}

Evaluate at $x = 0.35$ :

f(0.35) = 0.35 - \frac{1}{6} - \frac{1}{2}e^{-0.7}

= 0.35 - 0.166667 - 0.5(0.496585)

= 0.35 - 0.166667 - 0.248293

= -0.064960

Evaluate at $x = 0.45$ :

f(0.45) = 0.45 - \frac{1}{6} - \frac{1}{2}e^{-0.9}

= 0.45 - 0.166667 - 0.5(0.406570)

= 0.45 - 0.166667 - 0.203285

= 0.080048

Since $f(0.35) = -0.064960 < 0$ and $f(0.45) = 0.080048 > 0$ , there is a sign change between $x = 0.35$ and $x = 0.45$ . Therefore the root of $f(x) = 0$ , which is the $x$ -coordinate of $P$ , lies between 0.35 and 0.45.

Answer

$f(0.35) = -0.064960 < 0$ and $f(0.45) = 0.080048 > 0$ , so by the sign-change rule the $x$ -coordinate of $P$ lies between 0.35 and 0.45.

Final answer

f(0.35) < 0 and f(0.45) > 0, so the x-coordinate of P lies between 0.35 and 0.45

Detailed explanation

Walkthrough

From part (a), the $x$ -coordinate of $P$ satisfies $x = \frac{1}{6} + \frac{1}{2}e^{-2x}$ . Rearranging, this is equivalent to finding a root of $f(x) = x - \frac{1}{6} - \frac{1}{2}e^{-2x} = 0$ .

The sign-change rule states that if a continuous function changes sign between two points, it must have a root between them. Here $f$ is continuous because it is built from polynomials and exponentials, which are continuous everywhere.

We evaluate $f$ at the two endpoints of the proposed interval. At $x = 0.35$ we compute $e^{-0.7} \approx 0.496585$ , giving $f(0.35) \approx -0.064960$ , which is negative. At $x = 0.45$ we compute $e^{-0.9} \approx 0.406570$ , giving $f(0.45) \approx 0.080048$ , which is positive.

The sign change from negative to positive confirms that a root lies strictly between 0.35 and 0.45.

Key Takeaways

The sign-change (intermediate value) theorem is the standard tool for locating roots of continuous functions.
The auxiliary function $f(x) = x - \frac{1}{6} - \frac{1}{2}e^{-2x}$ is formed by moving all terms to one side of the equation.
Only the signs of the function values matter for the conclusion — the magnitudes give extra confidence but the sign change is the key observation.

Common Mistakes

Evaluating the wrong function — some students try to substitute into $y$ itself rather than into $f(x) = x - \frac{1}{6} - \frac{1}{2}e^{-2x}$ .
Arithmetic errors in computing $e^{-0.7}$ or $e^{-0.9}$ — use a calculator carefully.
Mark scheme guidance: since the answer is given, "necessary detail" is required — both function values must be shown explicitly with their signs.
Forgetting to state the conclusion explicitly: a sign change means the root lies between the two points.

Things to Be Careful About

Keep enough decimal places in the intermediate calculations so the sign is unambiguous. Values like $-0.064960$ and $0.080048$ are clearly negative and positive respectively.
The function $f$ is continuous on $[0.35, 0.45]$ , which is required for the sign-change argument to be valid.
The mark scheme also accepts evaluating $\frac{dy}{dx}$ from part (a) at the two endpoints instead — the same sign-change logic applies to the derivative.

Techniques used

define an auxiliary function from the root equationevaluate the function at interval endpointsapply the sign-change criterion to locate a root

(c)

Use an iterative formula based on the equation in part (a) to find the $x$ -coordinate of $P$ correct to 3 significant figures. Give the result of each iteration to 5 significant figures.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the iterative formula $x_{n+1} = \frac{1}{6} + \frac{1}{2}e^{-2x_n}$ starting from a value inside the interval found in part (b), say $x_0 = 0.4$ . Iterate until successive values agree to 5 significant figures, then round the final value to 3 significant figures.

Working

Starting with $x_0 = 0.4$ :

x_1 = \frac{1}{6} + \frac{1}{2}e^{-0.8} = 0.166667 + 0.224664 = 0.391331

To 5 significant figures: $x_1 = 0.39133$ .

x_2 = \frac{1}{6} + \frac{1}{2}e^{-0.782662} = 0.166667 + 0.228604 = 0.395271

To 5 significant figures: $x_2 = 0.39527$ .

x_3 = \frac{1}{6} + \frac{1}{2}e^{-0.790542} = 0.166667 + 0.226786 = 0.393453

To 5 significant figures: $x_3 = 0.39345$ .

x_4 = \frac{1}{6} + \frac{1}{2}e^{-0.786906} = 0.166667 + 0.227600 = 0.394267

To 5 significant figures: $x_4 = 0.39427$ .

x_5 = \frac{1}{6} + \frac{1}{2}e^{-0.788534} = 0.166667 + 0.227211 = 0.393878

To 5 significant figures: $x_5 = 0.39388$ .

x_6 = \frac{1}{6} + \frac{1}{2}e^{-0.787756} = 0.166667 + 0.227389 = 0.394056

To 5 significant figures: $x_6 = 0.39406$ .

x_7 = \frac{1}{6} + \frac{1}{2}e^{-0.788112} = 0.166667 + 0.227304 = 0.393971

To 5 significant figures: $x_7 = 0.39397$ .

x_8 = \frac{1}{6} + \frac{1}{2}e^{-0.787942} = 0.166667 + 0.227344 = 0.394011

To 5 significant figures: $x_8 = 0.39401$ .

x_9 = \frac{1}{6} + \frac{1}{2}e^{-0.788022} = 0.166667 + 0.227374 = 0.394041

To 5 significant figures: $x_9 = 0.39404$ .

x_{10} = \frac{1}{6} + \frac{1}{2}e^{-0.788082} = 0.166667 + 0.227359 = 0.394026

To 5 significant figures: $x_{10} = 0.39403$ .

x_{11} = \frac{1}{6} + \frac{1}{2}e^{-0.788052} = 0.166667 + 0.227364 = 0.394031

To 5 significant figures: $x_{11} = 0.39403$ .

The iterations have converged to 0.39403 to 5 significant figures. Therefore, to 3 significant figures, the $x$ -coordinate of $P$ is 0.394.

Answer

x = 0.394

Final answer

x = 0.394

Detailed explanation

Walkthrough

Part (a) gave the equation $x = \frac{1}{6} + \frac{1}{2}e^{-2x}$ , which is already in the form $x = F(x)$ . This is ideal for an iterative scheme: we guess a starting value $x_0$ , then repeatedly apply $x_{n+1} = F(x_n)$ .

From part (b) we know the root lies between 0.35 and 0.45, so $x_0 = 0.4$ is a sensible starting point.

Each iteration computes $x_{n+1} = \frac{1}{6} + \frac{1}{2}e^{-2x_n}$ . The values oscillate slightly around the true root before settling: 0.39133, 0.39527, 0.39345, 0.39427, 0.39388, 0.39406, 0.39397, 0.39401, 0.39404, 0.39403, 0.39403. The last two values agree to 5 significant figures, so we can confidently state the root to 3 significant figures as 0.394.

The question specifically asks for each iteration to be given to 5 significant figures, which is why we round each intermediate value before using it in the next step (though in practice keeping full precision internally and rounding only for display is also acceptable).

Key Takeaways

An equation of the form $x = F(x)$ can be solved iteratively using $x_{n+1} = F(x_n)$ .
A good starting value (from a sign-change interval) speeds up convergence.
Convergence is confirmed when successive iterates agree to the required precision.
Rounding each displayed iteration to 5 significant figures lets the reader verify the convergence pattern.

Common Mistakes

Using the wrong iterative formula — some students try to iterate $y$ itself rather than $x = \frac{1}{6} + \frac{1}{2}e^{-2x}$ .
Not showing enough iterations. The mark scheme requires "sufficient iterations to 5 sf to justify answer" — at least 5–6 iterations, ideally until two consecutive values agree.
Rounding the final answer incorrectly: 0.39403 to 3 significant figures is 0.394, not 0.3940 or 0.39.
Giving iteration values to the wrong number of significant figures — the question explicitly asks for 5 significant figures.

Things to Be Careful About

The sequence oscillates (values alternate slightly above and below the root) before converging. This is normal for this type of iteration and does not indicate a problem.
To be certain of 3 significant figures, the interval $[0.3935, 0.3945]$ must contain the root; the converged value 0.39403 clearly does.
Keep full calculator precision when computing each $e^{-2x_n}$ value; only round the displayed $x_{n+1}$ to 5 significant figures.

Techniques used

apply an iterative formula to generate successive approximationsround each iteration to the required precisiondetermine convergence to the specified number of significant figures

Q6Medium-HardIntegration

Fig. 6.1

The diagram shows the curve with equation $y = \sqrt{\sin 2x + \sin^2 2x}$ for $0 \le x \le \frac{1}{6}\pi$ . The shaded region is bounded by the curve and the straight lines $x = \frac{1}{6}\pi$ and $y = 0$ .

(a)

Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 significant figures.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the trapezium rule with 2 equal intervals across $\left[0, \frac{1}{6}\pi\right]$ and substitute the function values at $x = 0$ , $x = \frac{1}{12}\pi$ and $x = \frac{1}{6}\pi$ into the formula.

Working

With 2 intervals, the strip width is:

h = \frac{\frac{1}{6}\pi - 0}{2} = \frac{\pi}{12}

The $y$ -values at the three points are:

At $x = 0$ :

y_0 = \sqrt{\sin 0 + \sin^2 0} = \sqrt{0 + 0} = 0

At $x = \frac{\pi}{12}$ :

y_1 = \sqrt{\sin\frac{\pi}{6} + \sin^2\frac{\pi}{6}} = \sqrt{\frac{1}{2} + \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \approx 0.866

At $x = \frac{\pi}{6}$ :

y_2 = \sqrt{\sin\frac{\pi}{3} + \sin^2\frac{\pi}{3}} = \sqrt{\frac{\sqrt{3}}{2} + \frac{3}{4}} \approx \sqrt{1.616} \approx 1.271

Applying the trapezium rule:

\text{Area} \approx \frac{h}{2}\left(y_0 + 2y_1 + y_2\right) = \frac{\pi}{24}\bigl(0 + 2(0.866) + 1.271\bigr)

= \frac{\pi}{24}(3.003) \approx 0.3931

Answer

0.39 (to 2 significant figures)

Final answer

0.39

Detailed explanation

Walkthrough

The trapezium rule estimates the area under a curve by splitting the region into a number of trapezium-shaped strips and adding their areas together. With two intervals, the interval $[0, \frac{1}{6}\pi]$ is divided into two equal strips, each of width $h = \frac{\pi}{12}$ . Three function values are needed: the two endpoints and the midpoint.

The function $y = \sqrt{\sin 2x + \sin^2 2x}$ is evaluated at each of these $x$ -values. At $x = 0$ the value is 0 (the curve passes through the origin, as shown in the diagram). At $x = \frac{\pi}{12}$ , we have $\sin\frac{\pi}{6} = \frac{1}{2}$ so $\sin^2\frac{\pi}{6} = \frac{1}{4}$ and the value becomes $\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$ . At $x = \frac{\pi}{6}$ , $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\sin^2\frac{\pi}{3} = \frac{3}{4}$ , giving approximately $1.271$ .

The trapezium rule formula $\frac{h}{2}(y_0 + 2y_1 + y_2)$ weights the middle value twice because it borders two strips. Substituting the values gives $\frac{\pi}{24} \times 3.003 \approx 0.3931$ . To 2 significant figures this is $0.39$ .

Key Takeaways

The trapezium rule gives an approximation, not an exact value; the answer must be given to a specified accuracy.
With 2 intervals over $[a, b]$ , the strip width is $h = \frac{b - a}{2}$ .
The formula is $\frac{h}{2}(y_0 + 2y_1 + 2y_2 + \ldots + 2y_{n-1} + y_n)$ : only interior values get the factor 2.

Common Mistakes

Using $h = \frac{b - a}{n+1}$ or $h = \frac{b - a}{n - 1}$ instead of $\frac{b - a}{n}$ .
Forgetting to double the middle $y$ -value.
Reporting too many significant figures (the question specifies 2 s.f.).
Computing $y$ -values from $\sin x + \sin^2 x$ instead of $\sin 2x + \sin^2 2x$ (the argument is $2x$ , not $x$ ).

Things to Be Careful About

The mark scheme permits an answer of $0.393$ but not greater accuracy, so 4-figure or more precision is not allowed.
Work in radians throughout because the limits are given as multiples of $\pi$ .

Techniques used

evaluate function at equally-spaced pointsapply the trapezium rule formularound to the requested accuracy

(b)

The shaded region is rotated completely about the $x$ -axis.

Find the exact volume of the solid produced.

6M Similar

DifficultyMedium-Hard

Worked solution

Approach

The volume of the solid formed when the region is rotated completely about the $x$ -axis is $V = \pi \int_0^{\pi/6} y^2\,dx$ . Square the curve, expand using the identity $\sin^2 2x = \frac{1 - \cos 4x}{2}$ , integrate term by term, and apply the limits exactly.

Working

Set up the volume integral using $y^2 = \sin 2x + \sin^2 2x$ :

V = \pi \int_0^{\frac{1}{6}\pi} \left(\sin 2x + \sin^2 2x\right) dx

Use the double-angle identity $\sin^2 2x = \frac{1 - \cos 4x}{2}$ to rewrite the integrand:

\sin 2x + \sin^2 2x = \sin 2x + \frac{1}{2} - \frac{1}{2}\cos 4x

So:

V = \pi \int_0^{\frac{1}{6}\pi} \left(\sin 2x + \frac{1}{2} - \frac{1}{2}\cos 4x\right) dx

Integrate term by term:

V = \pi \left[-\frac{1}{2}\cos 2x + \frac{1}{2}x - \frac{1}{8}\sin 4x\right]_0^{\frac{1}{6}\pi}

Evaluate at the upper limit $x = \frac{\pi}{6}$ :

-\frac{1}{2}\cos\frac{\pi}{3} + \frac{1}{2}\cdot\frac{\pi}{6} - \frac{1}{8}\sin\frac{2\pi}{3} = -\frac{1}{2}\cdot\frac{1}{2} + \frac{\pi}{12} - \frac{1}{8}\cdot\frac{\sqrt{3}}{2} = -\frac{1}{4} + \frac{\pi}{12} - \frac{\sqrt{3}}{16}

Evaluate at the lower limit $x = 0$ :

-\frac{1}{2}\cos 0 + \frac{1}{2}\cdot 0 - \frac{1}{8}\sin 0 = -\frac{1}{2}

Subtract the lower from the upper value:

V = \pi \left[\left(-\frac{1}{4} + \frac{\pi}{12} - \frac{\sqrt{3}}{16}\right) - \left(-\frac{1}{2}\right)\right] = \pi\left(\frac{1}{4} + \frac{\pi}{12} - \frac{\sqrt{3}}{16}\right)

Distribute the $\pi$ :

V = \frac{1}{4}\pi + \frac{1}{12}\pi^2 - \frac{1}{16}\pi\sqrt{3}

Answer

V = \frac{1}{4}\pi + \frac{1}{12}\pi^2 - \frac{1}{16}\pi\sqrt{3}

Final answer

V = (1/4)π + (1/12)π² − (1/16)π√3

Detailed explanation

Walkthrough

When a region bounded by a curve $y = f(x)$ , the $x$ -axis and two vertical lines is rotated about the $x$ -axis, the volume of the resulting solid is given by $V = \pi \int_a^b y^2\,dx$ . The crucial first step is to square the function: $y^2 = \sin 2x + \sin^2 2x$ .

The expression $\sin^2 2x$ cannot be integrated directly, so we apply the double-angle identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$ with $\theta = 2x$ , giving $\sin^2 2x = \frac{1 - \cos 4x}{2}$ . This converts the integrand into a sum of standard functions: a sine, a constant, and a cosine of $4x$ .

The next step is to integrate each term separately. Recall that $\int \sin(ax)\,dx = -\frac{1}{a}\cos(ax)$ , $\int \cos(ax)\,dx = \frac{1}{a}\sin(ax)$ , and $\int 1\,dx = x$ . Applying these:

$\int \sin 2x\,dx = -\frac{1}{2}\cos 2x$
$\int \frac{1}{2}\,dx = \frac{1}{2}x$
$\int -\frac{1}{2}\cos 4x\,dx = -\frac{1}{2} \cdot \frac{1}{4}\sin 4x = -\frac{1}{8}\sin 4x$

So an antiderivative is $-\frac{1}{2}\cos 2x + \frac{1}{2}x - \frac{1}{8}\sin 4x$ .

Applying the limits, the trickiest values are at $x = \frac{\pi}{6}$ : $\cos\frac{\pi}{3} = \frac{1}{2}$ and $\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}$ . At $x = 0$ everything simplifies cleanly: $\cos 0 = 1$ , $\sin 0 = 0$ . Subtracting the lower value ( $-\frac{1}{2}$ ) from the upper value ( $-\frac{1}{4} + \frac{\pi}{12} - \frac{\sqrt{3}}{16}$ ) gives the bracket $\frac{1}{4} + \frac{\pi}{12} - \frac{\sqrt{3}}{16}$ . Multiplying by the outer $\pi$ produces the final exact volume.

Key Takeaways

Volume of revolution about the $x$ -axis: $V = \pi \int_a^b y^2\,dx$ .
Always square $y$ before substituting: the $\sqrt{\cdot}$ disappears.
The identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$ is the standard tool for handling $\sin^2$ in an integrand.
Exact values of $\cos$ and $\sin$ at multiples of $\frac{\pi}{6}$ and $\frac{\pi}{3}$ must be used, not decimal approximations.

Common Mistakes

Forgetting to square: writing $V = \pi \int y\,dx$ instead of $\pi \int y^2\,dx$ .
Integrating $\sin^2 2x$ as $\frac{\sin^3 2x}{6x}$ (treating it as a chain rule integral).
Using $\sin 2x = 2\sin x\cos x$ instead of the $\cos 4x$ form, which produces an unnecessarily complicated integrand.
Forgetting the chain rule when integrating $\cos 4x$ (must divide by 4, not 1).
Sign errors: $-\frac{1}{2}\cos 4x$ integrates to $-\frac{1}{8}\sin 4x$ , not $+\frac{1}{8}\sin 4x$ .
Using degrees instead of radians at the exact-value step.

Things to Be Careful About

The mark scheme follows through on the candidate's own integrand only if it is of the correct form $k_1 \sin 2x + k_2 + k_3 \cos 4x$ with all three constants non-zero. If the identity is applied incorrectly, subsequent marks are for the candidate's own (wrong) antiderivative.
The lower-limit subtraction must be done carefully: the constant $-\frac{1}{2}$ at $x=0$ is subtracted, so $-\frac{1}{4} - (-\frac{1}{2}) = +\frac{1}{4}$ .
The final answer must be exact, so decimal approximations of $\sqrt{3}$ or $\pi$ are not accepted.

Techniques used

set up the volume of revolution integralapply the double-angle identity for sin^2integrate term by termevaluate at both limitssimplify to an exact form

Q7MediumAlgebraIntegrationLogarithmic and Exponential Functions

The polynomial $p(x)$ is defined by

p(x) = 9x^3 + 6x^2 + 12x + k,

where $k$ is a constant.

(a)

Find the quotient when $p(x)$ is divided by $(3x + 2)$ and show that the remainder is $(k - 8)$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Divide $p(x)$ by $(3x+2)$ using polynomial long division. The quotient and remainder are read from the final subtraction.

Working

First term of quotient:

\frac{9x^3}{3x} = 3x^2

Multiply back:

(3x+2)(3x^2) = 9x^3 + 6x^2

Subtract from $p(x)$ :

(9x^3 + 6x^2 + 12x + k) - (9x^3 + 6x^2) = 12x + k

Next term of quotient:

\frac{12x}{3x} = 4

Multiply back:

4(3x+2) = 12x + 8

Subtract:

(12x + k) - (12x + 8) = k - 8

So the quotient is $3x^2 + 4$ and the remainder is $k - 8$ .

Answer

Quotient = $3x^2 + 4$ ; remainder = $k - 8$ .

Final answer

Quotient = 3x^2 + 4; remainder = k - 8

Detailed explanation

Walkthrough

We are dividing a cubic polynomial by the linear factor $3x+2$ . Start by comparing leading terms: $9x^3$ divided by $3x$ gives $3x^2$ . Multiplying $3x^2$ by $3x+2$ gives $9x^3+6x^2$ , which exactly removes the first two terms of $p(x)$ , leaving $12x+k$ . Then $12x$ divided by $3x$ gives $4$ . Multiplying $4$ by $3x+2$ gives $12x+8$ , and subtracting leaves $k-8$ . This is the remainder. Since the question says 'show that', the subtraction must be written out clearly; an unsupported answer would not earn the final mark.

An equivalent method is synthetic division using the root $x = -\frac{2}{3}$ . The coefficients $9, 6, 12, k$ give $9, 0, 12, k-8$ , so the quotient is $3x^2 + 4$ and the remainder is $k-8$ .

Key Takeaways

This question tests polynomial division and the relationship between dividend, divisor, quotient and remainder:

p(x) = (3x+2)(3x^2+4) + (k-8).

It also shows that the remainder theorem can be used as a check: substituting $x = -\frac{2}{3}$ into $p(x)$ gives $k-8$ .

Common Mistakes

Not showing enough working. Because the remainder is given in the question, the mark scheme requires the division detail; simply stating the answer loses the final mark.
Sign errors when subtracting, especially with $12x + k - (12x + 8)$ .
In synthetic division, forgetting to use $x = -\frac{2}{3}$ or making a sign error in the multiplier.

Things to Be Careful About

The quotient has no $x$ -term: it is $3x^2 + 4$ , not $3x^2 + 0x + 4$ with a missing term. The remainder is a constant because the divisor is linear. If $k = 8$ , the division would be exact.

Techniques used

perform polynomial long divisionmultiply quotient terms by the divisorsubtract to find the remainder

(b)

It is given that $\int_1^6 \frac{p(x)}{3x + 2} \,dx = a + \ln 64$ , where $a$ is an integer.

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

6M Similar

DifficultyMedium

Worked solution

Approach

Use the quotient and remainder from part (a) to rewrite $\frac{p(x)}{3x+2}$ as a polynomial plus a reciprocal-linear term. Integrate term by term, apply the limits, then use logarithm laws to compare with $a + \ln 64$ .

Working

From part (a):

\frac{p(x)}{3x+2} = 3x^2 + 4 + \frac{k-8}{3x+2}

Integrate:

\int \left(3x^2 + 4 + \frac{k-8}{3x+2}\right)\,dx = x^3 + 4x + \frac{k-8}{3}\ln(3x+2)

Apply the limits $x=1$ to $x=6$ :

\begin{aligned} \int_1^6 \frac{p(x)}{3x+2}\,dx &= \left[x^3 + 4x + \frac{k-8}{3}\ln(3x+2)\right]_1^6 \\ &= \left(216 + 24 + \frac{k-8}{3}\ln 20\right) - \left(1 + 4 + \frac{k-8}{3}\ln 5\right) \\ &= 235 + \frac{k-8}{3}(\ln 20 - \ln 5) \\ &= 235 + \frac{k-8}{3}\ln 4. \end{aligned}

Since $\ln 64 = \ln(4^3) = 3\ln 4$ , the given equality is

235 + \frac{k-8}{3}\ln 4 = a + 3\ln 4.

Equate the constant terms and the coefficients of $\ln 4$ :

a = 235, \qquad \frac{k-8}{3} = 3.

Hence $k - 8 = 9$ , so $k = 17$ .

Answer

$a = 235$ and $k = 17$ .

Final answer

a = 235, k = 17

Detailed explanation

Walkthrough

Part (b) asks for a definite integral, but the integrand is a rational function. The key is to use the result of part (a):

\frac{p(x)}{3x+2} = 3x^2 + 4 + \frac{k-8}{3x+2}.

This splits the integral into a polynomial part and a reciprocal-linear part. The polynomial integrates to $x^3 + 4x$ . For the reciprocal part, recall that

\int \frac{1}{3x+2}\,dx = \frac{1}{3}\ln(3x+2),

so the $k-8$ factor gives $\frac{k-8}{3}\ln(3x+2)$ . On the interval $1 \le x \le 6$ , $3x+2 > 0$ , so no absolute value is needed.

Substitute the limits. At $x=6$ , the polynomial part is $216+24=240$ ; at $x=1$ , it is $1+4=5$ ; the difference is $235$ . The log terms combine as

\frac{k-8}{3}(\ln 20 - \ln 5) = \frac{k-8}{3}\ln 4.

Finally, compare with $a + \ln 64$ . Since $\ln 64 = 3\ln 4$ , the coefficient of $\ln 4$ on the right is $3$ . Therefore $a = 235$ and $\frac{k-8}{3}=3$ , giving $k=17$ .

Key Takeaways

This question combines polynomial division, integration of $\frac{1}{ax+b}$ , definite integration with limits, and logarithm laws. It shows how a previous part can be used to simplify an otherwise difficult integrand.

Common Mistakes

Forgetting the factor $\frac{1}{3}$ when integrating $\frac{1}{3x+2}$ .
Not applying the limits to every term, especially the logarithmic term.
Writing $\ln 20 - \ln 5 = \ln 15$ instead of $\ln 4$ .
Comparing the expression to $\ln 64$ without first writing $\ln 64 = 3\ln 4$ .
Using an incorrect quotient from part (a); the mark scheme follows through on the quotient, but the final $k$ would be wrong.

Things to Be Careful About

The question says to use the answer from part (a); the integration mark depends on using the quotient.
The limits are $x=1$ and $x=6$ , giving $3x+2 = 5$ and $20$ respectively.
When equating $235 + \frac{k-8}{3}\ln 4$ with $a + 3\ln 4$ , the constant and logarithmic parts must match separately.
Since $a$ is stated to be an integer, this confirms that the polynomial part $235$ is exactly $a$ .

Techniques used

rewrite integrand using quotient and remainderintegrate polynomial and reciprocal-linear termsapply limits of integrationcombine logarithmic termsequate coefficients to solve for constants

Mathematics 9709/21 — May/June 2024

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