A curve has equation $y = \frac{4\sin x}{3 + \cos 2x}$ for values of $x$ such that $0 \leq x \leq 2\pi$. (a) Find $\frac{dy}{dx}$. (b) Hence find the coordinates of the stationary points of the curve.

(a) dy/dx = (4cos x(3+cos 2x) + 8 sin x sin 2x)/(3+cos 2x)^2 · (b) (pi/2, 2) and (3pi/2, -2)

(a) Sketch on the same diagram the graphs of $y = 2x - 3 $ and $y = \ln(x + 1)$. (b) The $x$-coordinates of the points where the graphs intersect are denoted by $\alpha$ and $\beta$, where $\alpha < \beta$. Show that $\alpha = 1.5 - 0.5\ln(\alpha + 1)$. (c) Use an iterative formula, based on the equation in part (b), to find the value of $\alpha$ correct to 3 significant figures. Give the result of each iteration to 5 significant figures. (d) Show by calculation that $2.055…

(a) Sketch of y = ln(x+1) and y = 2x-3 with two intersection points · (b) alpha = 1.5 - 0.5 ln(alpha + 1) · (c) 1.12 (3 significant figures) · (d) 2.055 < beta < 2.065

(a) Find the quotient and remainder when $18x^3 - 6x^2 - 30x + 4$ is divided by $(3x - 1)$. (b) Hence find $$ \int_1^5 \frac{18x^3 - 6x^2 - 30x + 4}{3x - 1} \,dx $$ Give your answer in the form $a - \ln b$, where $a$ and $b$ are integers.

(a) Quotient = 6x^2 - 10; remainder = -6 · (b) 208 - ln 49

A curve has equation $3e^{2x}y + 4e^{3x} + y^3 = 18$. (a) Show that $\frac{dy}{dx} = \frac{-2e^{2x}y - 4e^{3x}}{e^{2x} + y^2}$. (b) Show that the curve has no stationary points.

(a) dy/dx = (-2e^(2x)y - 4e^(3x))/(e^(2x) + y^2) · (b) No stationary points, since -10e^(3x) = 18 is impossible for real x.

9709/22 Feb/Mar 2025 — Mathematics Solved Past Paper

Q13MMedium-EasyLogarithmic and Exponential Functions

Solve the equation $\ln(3x + 1) - \ln(x - 5) = \ln 7$ .

DifficultyMedium-Easy

Worked solution

Approach

Use the logarithm law $\ln a - \ln b = \ln\left(\frac{a}{b}\right)$ to combine the left-hand side into a single logarithm. Since $\ln$ is a one-to-one function, equate the arguments and solve the resulting linear equation. Finally, check that the value satisfies the domain requirements of the original logarithms.

Working

\ln(3x + 1) - \ln(x - 5) = \ln 7

Apply the quotient law:

\ln\left(\frac{3x+1}{x-5}\right) = \ln 7

Since the logarithms are equal, their arguments are equal:

\frac{3x+1}{x-5} = 7

Multiply both sides by $x-5$ :

3x + 1 = 7(x - 5)

3x + 1 = 7x - 35

Collect like terms:

36 = 4x

x = 9

Check the domain: $3(9)+1=28>0$ and $9-5=4>0$ , so both original logarithms are defined.

Answer

x = 9

Final answer

x = 9

Detailed explanation

Walkthrough

We start with an equation containing two logarithms on the left. The expression $\ln(3x+1) - \ln(x-5)$ has the form $\ln a - \ln b$ , so the quotient law of logarithms lets us rewrite it as $\ln\left(\frac{3x+1}{x-5}\right)$ . This is the key step because it reduces the equation to a single logarithm on each side.

Once both sides are single logarithms with the same base, we can use the fact that the natural logarithm function is one-to-one: if $\ln A = \ln B$ , then $A = B$ . This gives $\frac{3x+1}{x-5} = 7$ .

Now solve the resulting linear equation. Multiply through by $x-5$ to clear the denominator, expand the right-hand side, and rearrange to isolate $x$ . This gives $x = 9$ .

Finally, check the domain. The original equation requires $3x+1 > 0$ and $x-5 > 0$ . For $x=9$ , both are positive, so the solution is valid.

Key Takeaways

The quotient law of logarithms: $\ln a - \ln b = \ln\left(\frac{a}{b}\right)$ .
The one-to-one property of logarithms: equal logs imply equal arguments.
Always check that a solution lies in the domain of every logarithm in the original equation.

Common Mistakes

Combining logarithms incorrectly, e.g. writing $\ln(3x+1) - \ln(x-5)$ as $\ln\left(\frac{3x+1}{x-5}\right)$ is correct, but writing it as $\ln\left((3x+1)(x-5)\right)$ is wrong.
Forgetting to multiply the whole right-hand side by $x-5$ ; a common error is to write $3x+1 = 7x - 5$ instead of $3x+1 = 7x - 35$ .
Not checking the domain, which can allow an extraneous root.

Things to Be Careful About

The arguments of logarithms must be positive. Here we need $x > 5$ because of $x-5$ .
The mark scheme requires showing the logarithm property before removing the logs; an unsupported jump straight to $x=9$ may not receive full credit.
When cross-multiplying, expand $7(x-5)$ carefully to avoid sign errors.

Techniques used

combine logarithms using the quotient lawequate arguments of equal logarithmssolve the resulting linear equationcheck the domain of the original logarithms

Approach

We integrate $\frac{dy}{dx} = 4\sec^2\left(\frac{1}{2}x\right)$ with respect to $x$ . Since $\frac{d}{dx}\tan\left(\frac{1}{2}x\right) = \frac{1}{2}\sec^2\left(\frac{1}{2}x\right)$ , the integral is $8\tan\left(\frac{1}{2}x\right) + c$ . Then use the point $\left(\frac{\pi}{2},5\right)$ to find $c$ .

Working

Integrate:

y = \int 4\sec^2\left(\frac{1}{2}x\right)\,dx

Let $u = \frac{1}{2}x$ , so $dx = 2\,du$ :

y = 4\int \sec^2 u \cdot 2\,du = 8\tan u + c = 8\tan\left(\frac{1}{2}x\right) + c

Use $y=5$ when $x=\frac{\pi}{2}$ :

5 = 8\tan\left(\frac{\pi}{4}\right) + c = 8(1) + c

So $c = -3$ .

Answer

y = 8\tan\left(\frac{1}{2}x\right) - 3

Final answer

y = 8 tan(1/2 x) - 3

Detailed explanation

Walkthrough

We are told the gradient function $\frac{dy}{dx}$ and one point on the curve. To recover $y$ from $\frac{dy}{dx}$ , integrate. The integrand is $4\sec^2\left(\frac{1}{2}x\right)$ . Recall that $\frac{d}{dx}\tan x = \sec^2 x$ , so by the chain rule $\frac{d}{dx}\tan\left(\frac{1}{2}x\right) = \frac{1}{2}\sec^2\left(\frac{1}{2}x\right)$ . Therefore integrating $\sec^2\left(\frac{1}{2}x\right)$ gives $2\tan\left(\frac{1}{2}x\right)$ , and multiplying by 4 gives $8\tan\left(\frac{1}{2}x\right)$ . Always add a constant $c$ because differentiation loses constants. Then substitute the given point to solve for $c$ . At $x=\frac{\pi}{2}$ , $\frac{1}{2}x = \frac{\pi}{4}$ and $\tan\frac{\pi}{4}=1$ , so $5=8+c$ , giving $c=-3$ . The equation is $y=8\tan\left(\frac{1}{2}x\right)-3$ .

Key Takeaways

Integration of $\sec^2(ax+b)$ : $\int \sec^2(ax+b)\,dx = \frac{1}{a}\tan(ax+b)+c$ . Initial conditions determine the constant. Need know exact trig values like $\tan\frac{\pi}{4}=1$ .

Common Mistakes

Forgetting the factor from the chain rule: $\int \sec^2\left(\frac{1}{2}x\right)\,dx$ is $2\tan\left(\frac{1}{2}x\right)$ , not $\tan\left(\frac{1}{2}x\right)$ .
Omitting the constant of integration before using the point.
Arithmetic error when solving $5=8+c$ .
Using degrees instead of radians; the problem uses radians.

Things to Be Careful About

Keep the argument $\frac{1}{2}x$ inside the tangent.
The mark scheme requires showing the integration step and the substitution step; an unsupported final answer may not receive full marks.
The function is undefined where $\cos\left(\frac{1}{2}x\right)=0$ (e.g. $x=\pi$ ), but this does not affect the point given.

Techniques used

integrate sec^2(ax+b) using substitution or reverse chain ruleapply initial condition to determine constant of integrationwrite the equation of the curve

Approach

At the point of intersection both curves have the same $y$ -value, so we set $e^{2x} = 8e^{-x}$ and solve for $x$ using the laws of indices and logarithms.

Working

Equate the two expressions for $y$ :

e^{2x} = 8e^{-x}

Divide both sides by $e^{-x}$ (equivalently, multiply by $e^{x}$ ):

e^{2x} \cdot e^{x} = 8 \quad \Rightarrow \quad e^{3x} = 8

Take the natural logarithm of both sides:

3x = \ln 8 = 3\ln 2

Therefore

x = \ln 2.

Answer

x = \ln 2 \qquad \blacksquare

Final answer

x = ln 2

Detailed explanation

Walkthrough

We want the value of $x$ where the two curves meet, because that will be the right-hand boundary of the shaded region in part (b). At a point of intersection the two expressions for $y$ are equal, so we write $e^{2x} = 8e^{-x}$ .

Using the law of indices $e^{a} \cdot e^{b} = e^{a+b}$ in reverse (i.e. dividing by $e^{-x}$ ), we collect all powers of $e$ on the left:
$e^{2x} \cdot e^{x} = e^{3x} = 8.$

Now $3x$ is the unknown and it sits as the power of $e$ . The natural logarithm is the inverse function of the exponential, so applying $\ln$ to both sides gives
$3x = \ln 8.$

Since $8 = 2^{3}$ , $\ln 8 = 3\ln 2$ , so $3x = 3\ln 2$ and hence $x = \ln 2$ .

Key Takeaways

The intersection of two curves $y = f(x)$ and $y = g(x)$ is found by setting $f(x) = g(x)$ .
Dividing by $e^{-x}$ (i.e. multiplying by $e^{x}$ ) is a clean way to combine exponential terms.
$\ln$ and $e$ are inverse functions, so applying $\ln$ to $e^{3x}$ "brings down" the exponent.
$\ln 8 = \ln(2^{3}) = 3\ln 2$ .

Common Mistakes

Forgetting to take $\ln$ of the right-hand side and writing $3x = 8$ instead of $3x = \ln 8$ .
Dividing incorrectly and writing $e^{x} = 8$ instead of $e^{3x} = 8$ .
Stopping at $3x = 3\ln 2$ and not dividing by $3$ to obtain $x = \ln 2$ .

Things to Be Careful About

The question is a "show that", so the working must clearly demonstrate that $x = \ln 2$ is forced by the algebra — every step needs to be present, not just the final line.

Techniques used

equate the two curvescollect exponential terms onto one sidetake natural logarithm

(b)

Find the area of the shaded region.

3M Similar

DifficultyMedium

Worked solution

Approach

The shaded region lies between the two curves from $x = 0$ (the $y$ -axis) to $x = \ln 2$ (the intersection). Since $8e^{-x} \geq e^{2x}$ throughout this interval, the area is
$A = \int_{0}^{\ln 2} \!\bigl(8e^{-x} - e^{2x}\bigr)\, dx.$
Integrate each term, substitute the limits, and simplify using $e^{\ln 2} = 2$ so that $e^{-\ln 2} = \tfrac{1}{2}$ and $e^{2\ln 2} = 4$ .

Working

Integrate:

\int \!\bigl(8e^{-x} - e^{2x}\bigr)\, dx = -8e^{-x} - \tfrac{1}{2}e^{2x} + C.

Apply the limits $0$ and $\ln 2$ :

\begin{aligned} A &= \left[ -8e^{-x} - \tfrac{1}{2}e^{2x} \right]_{0}^{\ln 2} \\ &= \left( -8e^{-\ln 2} - \tfrac{1}{2}e^{2\ln 2} \right) - \left( -8e^{0} - \tfrac{1}{2}e^{0} \right). \end{aligned}

Using $e^{\ln 2} = 2$ , we have $e^{-\ln 2} = \tfrac{1}{2}$ and $e^{2\ln 2} = 2^{2} = 4$ , and $e^{0} = 1$ :

\begin{aligned} A &= \left( -8 \cdot \tfrac{1}{2} - \tfrac{1}{2} \cdot 4 \right) - \left( -8 - \tfrac{1}{2} \right) \\ &= \left( -4 - 2 \right) + 8 + \tfrac{1}{2} \\ &= -6 + 8 + \tfrac{1}{2} \\ &= 2 + \tfrac{1}{2} \\ &= \tfrac{5}{2}. \end{aligned}

Answer

A = \tfrac{5}{2}

Final answer

5/2

Detailed explanation

Walkthrough

The shaded region in the diagram is the area enclosed by the $y$ -axis on the left, the curve $y = e^{2x}$ below, and the curve $y = 8e^{-x}$ above, up to where the curves meet. From part (a) the right-hand boundary is at $x = \ln 2$ .

To find the area between two curves we integrate (top curve) $-$ (bottom curve) over the relevant $x$ -interval. At $x = 0$ , $8e^{-x} = 8$ and $e^{2x} = 1$ , so $8e^{-x} > e^{2x}$ ; the two curves only meet at $x = \ln 2$ . So $8e^{-x}$ is the upper curve throughout the interval, giving the integral
$A = \int_{0}^{\ln 2} \!\bigl(8e^{-x} - e^{2x}\bigr)\, dx.$

The standard antiderivative of $e^{kx}$ is $\tfrac{1}{k}e^{kx}$ . With $k = -1$ and $k = 2$ we get $-8e^{-x}$ and $\tfrac{1}{2}e^{2x}$ respectively (note the negative sign when integrating $e^{-x}$ ).

We then substitute the limits. The two key facts are $e^{\ln 2} = 2$ , so $e^{-\ln 2} = 1/2$ and $e^{2\ln 2} = 4$ , and $e^{0} = 1$ . Substituting and combining the four numerical pieces gives $-4 - 2 + 8 + \tfrac{1}{2} = \tfrac{5}{2}$ .

Key Takeaways

The area between two curves from $x = a$ to $x = b$ is $\int_{a}^{b}(\text{top} - \text{bottom})\, dx$ .
$\int e^{kx}\, dx = \tfrac{1}{k}e^{kx} + C$ for any constant $k \neq 0$ .
When evaluating $e^{k\ln 2}$ use $e^{k\ln 2} = 2^{k}$ , e.g. $e^{2\ln 2} = 4$ .
The intersection from part (a) supplies the upper limit of integration.

Common Mistakes

Integrating $e^{-x}$ as $e^{-x}$ instead of $-e^{-x}$ (the $1/k$ rule with $k = -1$ gives a minus sign).
Reversing the order of the curves in the integrand and getting a negative area.
Substituting limits incorrectly — for example using $\ln 2$ where $0$ is needed — or forgetting to subtract the lower-limit value.
Leaving the answer in terms of $e$ and $\ln$ instead of evaluating to a number.

Things to Be Careful About

The mark scheme awards an M1 only when the limits are applied correctly AND the result is simplified so that $e$ and $\ln$ no longer appear, so the final step of writing $e^{-\ln 2} = \tfrac{1}{2}$ , $e^{2\ln 2} = 4$ is essential.
Always double-check which curve is on top across the whole interval, not just at one point.

Techniques used

set up definite integral as area between curvesintegrate e^{ax} using a = 1/a rulesubstitute limits and simplify

Q4MediumDifferentiationTrigonometry

A curve has equation $y = \frac{4\sin x}{3 + \cos 2x}$ for values of $x$ such that $0 \leq x \leq 2\pi$ .

(a)

Find $\frac{dy}{dx}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the quotient rule to differentiate $y = \frac{4\sin x}{3+\cos 2x}$ . Write the numerator as $u$ and the denominator as $v$ , differentiate each, then substitute into the quotient-rule formula.

Working

Let

u = 4\sin x, \quad v = 3 + \cos 2x

Then

\frac{du}{dx} = 4\cos x, \qquad \frac{dv}{dx} = -2\sin 2x

By the quotient rule,

\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}

\frac{dy}{dx} = \frac{(3+\cos 2x)(4\cos x) - (4\sin x)(-2\sin 2x)}{(3+\cos 2x)^2}

\frac{dy}{dx} = \frac{4\cos x(3+\cos 2x) + 8\sin x\sin 2x}{(3+\cos 2x)^2}

Answer

\frac{dy}{dx} = \frac{4\cos x(3+\cos 2x) + 8\sin x\sin 2x}{(3+\cos 2x)^2}

Final answer

dy/dx = (4cos x(3+cos 2x) + 8 sin x sin 2x)/(3+cos 2x)^2

Detailed explanation

Walkthrough

We need to differentiate a fraction, so the quotient rule is appropriate. Identify $u = 4\sin x$ and $v = 3+\cos 2x$ . The derivative of $4\sin x$ is $4\cos x$ . The derivative of $3+\cos 2x$ is $-2\sin 2x$ because the derivative of $\cos 2x$ is $-2\sin 2x$ by the chain rule. Substitute into $\frac{vu' - uv'}{v^2}$ . The minus sign in front of $uv'$ combines with the minus sign of $v'$ , giving the plus term $+8\sin x\sin 2x$ .

Key Takeaways

This question tests the quotient rule and the chain rule for composite trigonometric functions. It is important to keep the denominator squared and to handle the signs carefully.

Common Mistakes

Forgetting to square the denominator $(3+\cos 2x)^2$ .
Incorrect sign when applying $vu' - uv'$ : the derivative of $\cos 2x$ is negative, so subtracting it gives a plus.
Differentiating $\cos 2x$ as $-\sin 2x$ instead of $-2\sin 2x$ .

Things to Be Careful About

The domain $0 \leq x \leq 2\pi$ is not needed for part (a).
The denominator $3+\cos 2x$ is always positive, so the fraction is defined for all $x$ in the domain.

Techniques used

apply the quotient ruledifferentiate composite trigonometric functionssimplify the numerator

(b)

Hence find the coordinates of the stationary points of the curve.

4M Similar

DifficultyMedium

Worked solution

Approach

Stationary points occur where $\frac{dy}{dx}=0$ . Since the denominator is never zero, set the numerator equal to zero. Use the double-angle identities $\sin 2x = 2\sin x\cos x$ and $\cos 2x = 2\cos^2 x - 1$ to simplify the equation, then solve for $x$ in $0 \leq x \leq 2\pi$ and evaluate $y$ .

Working

At a stationary point,

\frac{dy}{dx}=0

so, since $(3+\cos 2x)^2 > 0$ ,

4\cos x(3+\cos 2x) + 8\sin x\sin 2x = 0

Use $\sin 2x = 2\sin x\cos x$ and $\cos 2x = 2\cos^2 x - 1$ :

4\cos x(3+2\cos^2 x-1) + 8\sin x(2\sin x\cos x) = 0

4\cos x(2+2\cos^2 x) + 16\sin^2 x\cos x = 0

8\cos x(1+\cos^2 x) + 16\sin^2 x\cos x = 0

8\cos x(1+\cos^2 x + 2\sin^2 x) = 0

Since $\sin^2 x = 1 - \cos^2 x$ ,

8\cos x(1+\cos^2 x + 2 - 2\cos^2 x) = 0

8\cos x(3 - \cos^2 x) = 0

i.e.

24\cos x - 8\cos^3 x = 0

Because $\cos^2 x \leq 1$ , the factor $3 - \cos^2 x$ is never zero, so

\cos x = 0

For $0 \leq x \leq 2\pi$ , this gives

x = \frac{\pi}{2}, \quad x = \frac{3\pi}{2}

Now evaluate $y$ :

y\left(\frac{\pi}{2}\right) = \frac{4\sin\frac{\pi}{2}}{3+\cos\pi} = \frac{4(1)}{3-1} = 2

y\left(\frac{3\pi}{2}\right) = \frac{4\sin\frac{3\pi}{2}}{3+\cos 3\pi} = \frac{4(-1)}{3-1} = -2

Answer

The stationary points are

\left(\frac{\pi}{2}, 2\right) \quad \text{and} \quad \left(\frac{3\pi}{2}, -2\right)

and there are no others.

Final answer

(pi/2, 2) and (3pi/2, -2)

Detailed explanation

Walkthrough

Stationary points are where the gradient is zero. The derivative is a fraction, and its denominator $(3+\cos 2x)^2$ is always positive (since $3+\cos 2x \ge 2$ ), so the derivative is zero exactly when the numerator is zero. The numerator from part (a) is $4\cos x(3+\cos 2x)+8\sin x\sin 2x$ . To solve this, use the double-angle identities to write everything in terms of $\cos x$ . Replace $\cos 2x$ by $2\cos^2 x - 1$ and $\sin 2x$ by $2\sin x\cos x$ . After simplifying, the equation becomes $8\cos x(3-\cos^2 x)=0$ . Since $\cos^2 x$ cannot exceed 1, $3-\cos^2 x$ is never zero, so the only possibility is $\cos x=0$ . On $0 \le x \le 2\pi$ , this gives $x=\pi/2$ and $x=3\pi/2$ . Finally substitute back into the original equation for $y$ to get the coordinates.

Key Takeaways

Stationary points occur where $dy/dx=0$ .
Double-angle identities allow a trigonometric equation with $\sin 2x$ and $\cos 2x$ to be reduced to a single variable.
Always check the domain when solving trigonometric equations.

Common Mistakes

Forgetting that the derivative is zero when the numerator is zero, and trying to solve the whole fraction equal to zero.
Incorrectly simplifying the numerator; sign errors in the double-angle substitution are common.
Dividing by $\cos x$ without considering $\cos x=0$ as a solution. Here $\cos x=0$ is exactly the solution, so dividing would lose it.
Forgetting to evaluate $y$ at the stationary points.

Things to Be Careful About

The factor $3-\cos^2 x$ is never zero, so the equation reduces cleanly to $\cos x=0$ . Do not try to solve $3-\cos^2 x=0$ .
The interval is $0 \leq x \leq 2\pi$ , so both $\pi/2$ and $3\pi/2$ are included; no other $x$ satisfies $\cos x=0$ in this interval.
The denominator is always positive, so no vertical asymptotes affect the stationary-point calculation.

Techniques used

set the derivative equal to zerouse double-angle identities to simplifysolve a trigonometric equation on a given intervalevaluate the original function at critical points

Q5MediumAlgebraLogarithmic and Exponential FunctionsNumerical Solution of Equations

(a)

Sketch on the same diagram the graphs of $y = |2x - 3|$ and $y = \ln(x + 1)$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Sketch $y=\ln(x+1)$ and $y=|2x-3|$ on the same axes, showing the key features and the two intersection points.

Working

The graph of $y=\ln(x+1)$ has domain $x>-1$ and a vertical asymptote at $x=-1$ . It passes through $(0,0)$ and increases, so it appears in the third and first quadrants.

The graph of $y=|2x-3|$ is V-shaped, with vertex at $(1.5,0)$ . For $x<1.5$ it is the line $y=3-2x$ ; for $x>1.5$ it is the line $y=2x-3$ .

The two curves intersect twice: once on the left arm of the V and once on the right arm.

Answer

The sketch shows the increasing logarithmic curve and the V-shaped modulus graph with two intersection points.

Final answer

Sketch of y = ln(x+1) and y = |2x-3| with two intersection points

Detailed explanation

Walkthrough

Start with $y=\ln(x+1)$ . Since $\ln(x+1)$ is only defined for $x>-1$ , the curve has a vertical asymptote at $x=-1$ . At $x=0$ , $y=\ln 1=0$ , so it passes through the origin. It increases slowly, so for $-1<x<0$ it is below the $x$ -axis and for $x>0$ it is above it.

Next sketch $y=|2x-3|$ . The expression $2x-3$ is zero at $x=1.5$ , so the vertex is at $(1.5,0)$ . To the left the graph is $y=3-2x$ ; to the right it is $y=2x-3$ . This gives the V shape.

The logarithmic curve crosses the left arm of the V once and the right arm once, so there are two intersections.

Key Takeaways

A modulus graph $y=|ax+b|$ is V-shaped with its vertex where $ax+b=0$ . The graph of $y=\ln(x+1)$ is a translation of $y=\ln x$ one unit left, with a vertical asymptote at $x=-1$ . The number of intersections of two graphs can be seen from their shapes.

Common Mistakes

Drawing $y=\ln(x+1)$ with an $x$ -intercept other than $0$ .
Forgetting the vertical asymptote at $x=-1$ .
Placing the modulus vertex at $x=3$ instead of $x=1.5$ .
Drawing only one intersection instead of two.

Things to Be Careful About

The modulus graph has two linear branches; the logarithmic graph is defined only for $x>-1$ . Both intersections must be shown for the 2 marks.

Techniques used

sketch modulus graphsketch logarithmic graphidentify intersection points

(b)

The $x$ -coordinates of the points where the graphs intersect are denoted by $\alpha$ and $\beta$ , where $\alpha < \beta$ .

Show that $\alpha = 1.5 - 0.5\ln(\alpha + 1)$ .

1M Similar

(c)

Use an iterative formula, based on the equation in part (b), to find the value of $\alpha$ correct to 3 significant figures. Give the result of each iteration to 5 significant figures.

3M Similar

DifficultyMedium

Worked solution

Approach

Use the rearrangement from part (b) as an iteration:

x_{n+1}=1.5-0.5\ln(x_n+1)

Start with $x_0=1.1$ and record each value to 5 significant figures.

Working

$n$	$x_n$ (5 s.f.)
0	1.1000
1	1.1290
2	1.1222
3	1.1238
4	1.1234
5	1.1235
6	1.1235

The values settle at $1.1235$ to 5 significant figures. To justify the rounding, let $g(x)=2x-3+\ln(x+1)$ . Then

g(1.115)\approx -0.0209<0,\qquad g(1.125)\approx 0.00377>0

so the root lies in $[1.115,1.125]$ , and $\alpha=1.12$ correct to 3 significant figures.

Answer

\alpha=1.12\quad(3\text{ s.f.})

Final answer

1.12 (3 significant figures)

Detailed explanation

Walkthrough

The equation from part (b) is already in the form $x=F(x)$ , so we can iterate $x_{n+1}=1.5-0.5\ln(x_n+1)$ . From the sketch, $\alpha$ is a little above 1, so start at $x_0=1.1$ . Apply the formula repeatedly, keeping 5 significant figures in the table. The iterates oscillate slightly around the root: 1.1290, 1.1222, 1.1238, 1.1234, 1.1235, 1.1235. Once two consecutive values agree to 5 significant figures, the root is $1.1235$ to 5 s.f., which is $1.12$ to 3 s.f. To justify the final rounding, evaluate $g(x)=2x-3+\ln(x+1)$ at $1.115$ and $1.125$ ; the sign change shows the root is in that interval.

Key Takeaways

An equation written as $x=F(x)$ can be solved by iteration. You must record enough iterations to be confident of the required accuracy, and a sign change over a small interval is a good way to justify the final rounded answer.

Common Mistakes

Stopping after one or two iterations, which is not enough to justify 3 significant figures.
Rounding intermediate values too early.
Giving the final answer as 1.123 instead of 1.12.
Not showing any justification for the rounded value.

Things to Be Careful About

The question asks for each iteration to 5 significant figures. Use $\ln$ , not $\log_{10}$ . The final answer must be to 3 significant figures, so $\alpha=1.12$ , not $1.123$ .

Techniques used

set up iterative formulaiterate to convergenceround to required precisionjustify with sign change

(d)

Show by calculation that $2.055 < \beta < 2.065$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $\beta$ is the larger intersection, it lies on the right branch of the modulus graph, where $|2x-3|=2x-3$ . Define $h(x)=2x-3-\ln(x+1)$ and evaluate it at the two endpoints.

Working

At $x=2.055$ :

h(2.055)=2(2.055)-3-\ln(3.055)=1.11-\ln(3.055)\approx -0.00678

At $x=2.065$ :

h(2.065)=2(2.065)-3-\ln(3.065)=1.13-\ln(3.065)\approx 0.00995

The sign of $h$ changes from negative to positive between $2.055$ and $2.065$ . Since $h$ is continuous, its zero $\beta$ lies in this interval.

Answer

2.055<\beta<2.065

Final answer

2.055 < beta < 2.065

Detailed explanation

Walkthrough

For the right intersection, $x>1.5$ , so the modulus is positive: $|2x-3|=2x-3$ . The intersection condition is $2x-3=\ln(x+1)$ , or $h(x)=2x-3-\ln(x+1)=0$ . Evaluate $h$ at the endpoints. At $2.055$ , $h\approx -0.00678$ ; at $2.065$ , $h\approx 0.00995$ . A continuous function that changes sign between two points must have a root between them, so $\beta$ lies between $2.055$ and $2.065$ .

Key Takeaways

A sign change of a continuous function over an interval locates a root. For a modulus equation, choose the correct branch before forming the function.

Common Mistakes

Using the left branch $-(2x-3)$ for $\beta$ .
Using the equation from part (b), which applies only to $\alpha$ .
Not evaluating the function at both endpoints or not giving the approximate values.

Things to Be Careful About

The mark scheme requires the sign change to be based on $2x-3-\ln(x+1)$ or an equivalent right-branch function, not on the left-branch equation. The values do not need to be exact; approximate values with the correct signs are enough.

Techniques used

evaluate function at interval endpointsapply sign-change criterion

Q6MediumAlgebraIntegration

(a)

Find the quotient and remainder when $18x^3 - 6x^2 - 30x + 4$ is divided by $(3x - 1)$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use polynomial long division by $3x-1$ , matching the leading term of the dividend at each step. The final result can be written as

18x^3 - 6x^2 - 30x + 4 = (3x-1)(6x^2 - 10) - 6

so the quotient is $6x^2 - 10$ and the remainder is $-6$ .

Working

Divide the leading term $18x^3$ by $3x$ :

6x^2(3x-1) = 18x^3 - 6x^2

Subtract this from the dividend:

(18x^3 - 6x^2 - 30x + 4) - (18x^3 - 6x^2) = -30x + 4

Now divide $-30x$ by $3x$ :

-10(3x-1) = -30x + 10

Subtract:

(-30x + 4) - (-30x + 10) = -6

Since $-6$ has degree less than the divisor $3x-1$ , the division stops. Hence

18x^3 - 6x^2 - 30x + 4 = (3x-1)(6x^2 - 10) - 6

Answer

Quotient $= 6x^2 - 10$ ; remainder $= -6$ .

Final answer

Quotient = 6x^2 - 10; remainder = -6

Detailed explanation

Walkthrough

We need to divide $18x^3 - 6x^2 - 30x + 4$ by $3x-1$ . The divisor has leading term $3x$ , not $x$ , so each quotient term is found by dividing the current leading term by $3x$ .

First, $18x^3 \div 3x = 6x^2$ . Multiplying back gives $18x^3 - 6x^2$ , which cancels the first two terms of the dividend. The remaining expression is $-30x + 4$ .

Next, $(-30x) \div 3x = -10$ . Multiplying back gives $-30x + 10$ ; subtracting leaves $-6$ . Since $-6$ has degree lower than the divisor $3x-1$ , the division stops. Thus the quotient is $6x^2 - 10$ and the remainder is $-6$ .

Key Takeaways

This question tests polynomial long division by a linear divisor. The key relationship is

\text{dividend} = \text{divisor} \times \text{quotient} + \text{remainder}.

This form is essential for part (b), where the same relationship is used to rewrite the integrand.

Common Mistakes

Stopping after finding only the first quotient term.
Writing the remainder as $+6$ instead of $-6$ .
Dividing by $x$ instead of by $3x$ when finding each quotient term.

Things to Be Careful About

The remainder must have degree less than the divisor. Check the result by expanding:

(3x-1)(6x^2 - 10) - 6 = 18x^3 - 6x^2 - 30x + 10 - 6 = 18x^3 - 6x^2 - 30x + 4.

Techniques used

divide a polynomial by a linear divisormatch leading terms at each division stepexpress the dividend as divisor times quotient plus remainder

(b)

Hence find

\int_1^5 \frac{18x^3 - 6x^2 - 30x + 4}{3x - 1} \,dx

Give your answer in the form $a - \ln b$ , where $a$ and $b$ are integers.

5M Similar

DifficultyMedium

Worked solution

Approach

Use the result from part (a) to rewrite the integrand as a polynomial plus a term of the form $\frac{k}{3x-1}$ . Then integrate each term and apply the limits.

Working

From part (a),

18x^3 - 6x^2 - 30x + 4 = (3x-1)(6x^2 - 10) - 6

so dividing by $3x-1$ gives

\frac{18x^3 - 6x^2 - 30x + 4}{3x-1} = 6x^2 - 10 - \frac{6}{3x-1}

Integrate term by term:

\int \left(6x^2 - 10 - \frac{6}{3x-1}\right)\,dx = 2x^3 - 10x - 2\ln(3x-1)

because

\int \frac{6}{3x-1}\,dx = 2\ln(3x-1)

so the negative term gives $-2\ln(3x-1)$ . Now apply the limits:

\left[2x^3 - 10x - 2\ln(3x-1)\right]_1^5

At $x=5$ :

2(5)^3 - 10(5) - 2\ln(14) = 200 - 2\ln 14

At $x=1$ :

2(1)^3 - 10(1) - 2\ln(2) = -8 - 2\ln 2

Subtract the lower-limit value from the upper-limit value:

(200 - 2\ln 14) - (-8 - 2\ln 2) = 208 - 2\ln 14 + 2\ln 2

Use logarithm laws:

208 - 2(\ln 14 - \ln 2) = 208 - 2\ln 7 = 208 - \ln 49

Answer

208 - \ln 49

Final answer

208 - ln 49

Detailed explanation

Walkthrough

The word "Hence" tells us to use the result from part (a). Since

18x^3 - 6x^2 - 30x + 4 = (3x-1)(6x^2 - 10) - 6,

dividing by $3x-1$ gives

6x^2 - 10 - \frac{6}{3x-1}.

Now integrate each term separately. The polynomial part integrates to $2x^3 - 10x$ . For the reciprocal term, recall that

\frac{d}{dx}\ln(3x-1) = \frac{3}{3x-1},

so to obtain $-\frac{6}{3x-1}$ we multiply by $-2$ , giving $-2\ln(3x-1)$ .

Then evaluate the antiderivative at the limits. At $x=5$ the value is $200 - 2\ln 14$ ; at $x=1$ it is $-8 - 2\ln 2$ . Subtracting the lower value from the upper value gives

208 - 2\ln 14 + 2\ln 2 = 208 - 2\ln 7 = 208 - \ln 49.

This is in the required form $a - \ln b$ with $a = 208$ and $b = 49$ .

Key Takeaways

This question connects algebraic division with integration. The quotient-remainder form lets us split a complicated rational integrand into a polynomial and a simple reciprocal term. It also reinforces the integral

\int \frac{1}{ax+b}\,dx = \frac{1}{a}\ln|ax+b| + C

and the logarithm law $\ln p - \ln q = \ln\left(\frac{p}{q}\right)$ .

Common Mistakes

Forgetting that the remainder is negative, so the reciprocal term is $-\frac{6}{3x-1}$ , not $+\frac{6}{3x-1}$ .
Writing $\int \frac{6}{3x-1}\,dx = 6\ln(3x-1)$ instead of $2\ln(3x-1)$ , because the derivative of $3x-1$ contributes a factor of $3$ .
Forgetting to subtract the lower-limit value when evaluating the definite integral.
Stopping at $208 - 2\ln 7$ instead of simplifying to $208 - \ln 49$ .

Things to Be Careful About

On the interval $1 \leq x \leq 5$ , the expression $3x-1$ is positive, so no absolute value signs are needed in the logarithm. When applying limits, be careful with the signs of the lower-limit terms: the entire lower value, including $-2\ln 2$ , must be subtracted. Finally, use

2\ln 7 = \ln 7^2 = \ln 49

to match the requested form $a - \ln b$ .

Techniques used

rewrite the integrand using the quotient and remainderintegrate polynomial and reciprocal termsapply the limits of a definite integralsimplify using logarithm laws

Q7MediumTrigonometry

(a)

Express $6\sin\theta - 4\cos\theta$ in the form $R\sin(\theta - \alpha)$ , where $R > 0$ and $0^\circ < \alpha < 90^\circ$ . Give the exact value of $R$ and the value of $\alpha$ correct to 2 decimal places.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the compound angle identity $R\sin(\theta-\alpha)=R\sin\theta\cos\alpha - R\cos\theta\sin\alpha$ . Compare coefficients of $\sin\theta$ and $\cos\theta$ to find $R$ and $\alpha$ .

Working

Let

6\sin\theta - 4\cos\theta = R\sin(\theta-\alpha)

Expanding:

R\sin(\theta-\alpha) = R\sin\theta\cos\alpha - R\cos\theta\sin\alpha

Comparing coefficients:

R\cos\alpha = 6, \qquad R\sin\alpha = 4

Therefore

R = \sqrt{6^2 + 4^2} = \sqrt{52} = 2\sqrt{13}

and

\tan\alpha = \frac{R\sin\alpha}{R\cos\alpha} = \frac{4}{6} = \frac{2}{3}

Since $0^\circ < \alpha < 90^\circ$ ,

\alpha = \tan^{-1}\left(\frac{2}{3}\right) = 33.690\ldots^\circ = 33.69^\circ

Answer

6\sin\theta - 4\cos\theta = 2\sqrt{13}\sin(\theta - 33.69^\circ)

Final answer

R = 2√13, α = 33.69°

Detailed explanation

Walkthrough

We want to write $6\sin\theta - 4\cos\theta$ as a single sine function. The identity $R\sin(\theta-\alpha)=R\sin\theta\cos\alpha - R\cos\theta\sin\alpha$ lets us match the coefficients of $\sin\theta$ and $\cos\theta$ . This gives $R\cos\alpha=6$ and $R\sin\alpha=4$ . Squaring and adding gives $R^2=6^2+4^2$ , so $R=\sqrt{52}=2\sqrt{13}$ . Dividing the two equations gives $\tan\alpha=4/6=2/3$ , and since both coefficients are positive, $\alpha$ is acute, so $\alpha=\tan^{-1}(2/3)=33.69^\circ$ .

Key Takeaways

This question tests the standard technique of expressing $a\sin\theta + b\cos\theta$ in the form $R\sin(\theta-\alpha)$ . The key steps are matching coefficients, finding $R$ by Pythagoras, and finding $\alpha$ using $\tan\alpha = b/a$ when the expression is $a\sin\theta - b\cos\theta$ .

Common Mistakes

Forgetting that $R$ is found from $R^2=a^2+b^2$ , not from just one coefficient.
Using $\tan\alpha = 6/4$ instead of $\tan\alpha = 4/6$ .
Quoting $\alpha$ in radians when the question asks for degrees.
Not giving $R$ in exact form.

Things to Be Careful About

Since $R>0$ and $0^\circ<\alpha<90^\circ$ , the signs of the coefficients tell us that $\alpha$ is acute. Here both $R\cos\alpha$ and $R\sin\alpha$ are positive, so no adjustment to the angle is needed. Round $\alpha$ to 2 decimal places as requested.

Techniques used

express a sin θ + b cos θ in R sin(θ - α) formcompare coefficients to identify R and αuse inverse tangent to determine α

(b)

Hence solve the equation $6\sin\theta - 4\cos\theta + 5 = 0$ for $0^\circ < \theta < 360^\circ$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use part (a) to replace $6\sin\theta - 4\cos\theta$ with $2\sqrt{13}\sin(\theta - 33.69^\circ)$ . Then isolate the sine term, find the acute reference angle, and use the sine graph to find both solutions in the required interval.

Working

6\sin\theta - 4\cos\theta + 5 = 0

Using part (a):

2\sqrt{13}\sin(\theta - 33.69^\circ) + 5 = 0

\sin(\theta - 33.69^\circ) = -\frac{5}{2\sqrt{13}}

The acute angle whose sine is $\frac{5}{2\sqrt{13}}$ is

\phi = \sin^{-1}\left(\frac{5}{2\sqrt{13}}\right) = 43.898^\circ

Since the sine is negative, the angles $\theta - 33.69^\circ$ lie in the third and fourth quadrants:

\theta - 33.69^\circ = 180^\circ + \phi \quad \text{or} \quad 360^\circ - \phi

First solution:

\theta = 33.69^\circ + 180^\circ + 43.898^\circ = 257.588^\circ = 257.6^\circ

Second solution:

\theta = 33.69^\circ + 360^\circ - 43.898^\circ = 349.792^\circ = 349.8^\circ

Both values lie in $0^\circ < \theta < 360^\circ$ , and there are no other solutions in this interval.

Answer

\theta = 257.6^\circ \quad \text{or} \quad \theta = 349.8^\circ

Final answer

θ = 257.6° or θ = 349.8°

Detailed explanation

Walkthrough

Start from the equation and substitute the $R\sin(\theta-\alpha)$ form found in part (a). This gives $2\sqrt{13}\sin(\theta-33.69^\circ)+5=0$ , so $\sin(\theta-33.69^\circ)=-5/(2\sqrt{13})$ . Let $\phi$ be the acute angle with $\sin\phi=5/(2\sqrt{13})$ . Because the required sine is negative, the angle $\theta-33.69^\circ$ must be in the third or fourth quadrant, so it is $180^\circ+\phi$ or $360^\circ-\phi$ . Adding $33.69^\circ$ gives the two values of $\theta$ : $257.6^\circ$ and $349.8^\circ$ .

Key Takeaways

This question combines the $R\sin(\theta-\alpha)$ form with solving a trigonometric equation. The important skill is using the reference angle and quadrant rules to find all solutions in a given interval, and checking that the final answers lie in the required range.

Common Mistakes

Forgetting to add $\alpha=33.69^\circ$ back to the angle $\theta-\alpha$ .
Using only one quadrant and missing the other solution.
Writing solutions in radians when the question specifies degrees.
Giving answers outside $0^\circ<\theta<360^\circ$ .

Things to Be Careful About

The reference angle $\phi$ is positive and acute. Since $\sin(\theta-\alpha)$ is negative, the angle must be in the third or fourth quadrant, not the first or second. Also, the final answers must be rounded to the accuracy shown; the mark scheme accepts greater accuracy such as $257.587^\circ$ and $349.792^\circ$ .

Techniques used

rewrite equation using R sin(θ - α) formisolate sine term and find reference angleapply quadrant rules to find both solutions in range

(c)

As the value of $\beta$ varies, find the greatest possible value of

(3\sin 4\beta - 2\cos 4\beta)^2 + 15

and determine the smallest positive value of $\beta$ , in degrees, for which this greatest value occurs.

3M Similar

DifficultyMedium

Worked solution

Approach

Write $3\sin 4\beta - 2\cos 4\beta$ in the form $R\sin(4\beta - \alpha)$ . Then $(3\sin 4\beta - 2\cos 4\beta)^2 = R^2\sin^2(4\beta-\alpha)$ , whose greatest value is $R^2$ . For the smallest positive $\beta$ , use $\sin(4\beta-\alpha)=1$ .

Working

For $3\sin 4\beta - 2\cos 4\beta$ ,

R = \sqrt{3^2 + 2^2} = \sqrt{13}, \qquad \tan\alpha = \frac{2}{3}, \quad \alpha = 33.69^\circ

Therefore

3\sin 4\beta - 2\cos 4\beta = \sqrt{13}\sin(4\beta - 33.69^\circ)

So

(3\sin 4\beta - 2\cos 4\beta)^2 + 15 = 13\sin^2(4\beta - 33.69^\circ) + 15

The greatest value of $\sin^2$ is $1$ , so the greatest value of the expression is

13(1) + 15 = 28

For this greatest value, $\sin(4\beta - 33.69^\circ)=\pm 1$ . The smallest positive $\beta$ is obtained from $\sin(4\beta - 33.69^\circ)=1$ :

4\beta - 33.69^\circ = 90^\circ

\beta = \frac{90^\circ + 33.69^\circ}{4} = \frac{123.69^\circ}{4} = 30.9225^\circ = 30.9^\circ

Answer

Greatest value is $28$ ; smallest positive $\beta = 30.9^\circ$ .

Final answer

Greatest value = 28; β = 30.9°

Detailed explanation

Walkthrough

This part asks for the greatest value of a squared expression. First rewrite $3\sin 4\beta - 2\cos 4\beta$ using the same $R\sin$ technique: $R=\sqrt{13}$ and $\alpha=33.69^\circ$ . Then the expression becomes $13\sin^2(4\beta-33.69^\circ)+15$ . Since $\sin^2$ is at most $1$ , the greatest value is $13+15=28$ . To find the smallest positive $\beta$ , set $\sin(4\beta-33.69^\circ)=1$ , because this gives $4\beta-33.69^\circ=90^\circ$ , hence $\beta=(90+33.69)/4=30.9^\circ$ . Using $\sin=-1$ would give a larger positive $\beta$ .

Key Takeaways

The maximum of a squared sine term is $1$ , not $0$ or $-1$ . This question also shows how the $R\sin(\theta-\alpha)$ form can be used to analyse the maximum or minimum of an expression involving sine and cosine.

Common Mistakes

Forgetting that the square of the sine has maximum $1$ , so the greatest value is $R^2+15$ .
Setting $\sin(4\beta-\alpha)=-1$ first and obtaining a larger value of $\beta$ .
Using radians instead of degrees.
Forgetting to divide by $4$ when solving for $\beta$ .

Things to Be Careful About

The smallest positive $\beta$ comes from the $+90^\circ$ branch, not the $+270^\circ$ branch, because $\frac{90+33.69}{4}<\frac{270+33.69}{4}$ . The answer should be rounded to 1 decimal place as requested.

Techniques used

rewrite expression in R sin formidentify maximum of squared sine expressionsolve the sine equation for the smallest positive angle

Q8Medium-EasyDifferentiation

A curve has equation $3e^{2x}y + 4e^{3x} + y^3 = 18$ .

(a)

Show that $\frac{dy}{dx} = \frac{-2e^{2x}y - 4e^{3x}}{e^{2x} + y^2}$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Differentiate the equation implicitly with respect to $x$ , using the product rule on the term $3e^{2x}y$ . Then collect the terms containing $\frac{dy}{dx}$ and rearrange to obtain the given expression.

Working

Differentiating each term with respect to $x$ :

\frac{d}{dx}\left(3e^{2x}y\right) = 6e^{2x}y + 3e^{2x}\frac{dy}{dx}

\frac{d}{dx}\left(4e^{3x}\right) = 12e^{3x}

\frac{d}{dx}\left(y^3\right) = 3y^2\frac{dy}{dx}

So:

6e^{2x}y + 3e^{2x}\frac{dy}{dx} + 12e^{3x} + 3y^2\frac{dy}{dx} = 0

Collect the terms containing $\frac{dy}{dx}$ :

3e^{2x}\frac{dy}{dx} + 3y^2\frac{dy}{dx} = -6e^{2x}y - 12e^{3x}

\left(e^{2x} + y^2\right)\frac{dy}{dx} = -2e^{2x}y - 4e^{3x}

Hence:

\frac{dy}{dx} = \frac{-2e^{2x}y - 4e^{3x}}{e^{2x} + y^2}

Answer

\frac{dy}{dx} = \frac{-2e^{2x}y - 4e^{3x}}{e^{2x} + y^2}

Final answer

dy/dx = (-2e^(2x)y - 4e^(3x))/(e^(2x) + y^2)

Detailed explanation

Walkthrough

We are told to show the given derivative formula, so we differentiate the curve equation implicitly. The left-hand side has three terms. The first term, $3e^{2x}y$ , is a product of $3e^{2x}$ and $y$ , so we use the product rule. This gives $6e^{2x}y+3e^{2x}\frac{dy}{dx}$ . The second term, $4e^{3x}$ , is an ordinary exponential in $x$ , so its derivative is $12e^{3x}$ . The third term, $y^3$ , is a function of $y$ , and because $y$ depends on $x$ , its derivative is $3y^2\frac{dy}{dx}$ . The derivative of the right-hand side, $18$ , is zero.

Putting these together gives

6e^{2x}y+3e^{2x}\frac{dy}{dx}+12e^{3x}+3y^2\frac{dy}{dx}=0.

Then collect the terms containing $\frac{dy}{dx}$ on one side and the remaining terms on the other. Factor out $\frac{dy}{dx}$ and divide by the coefficient $e^{2x}+y^2$ to obtain the required expression.

Key Takeaways

This part tests implicit differentiation. Whenever a term contains $y$ , we must multiply by $\frac{dy}{dx}$ when differentiating with respect to $x$ . When a term is a product of a function of $x$ and a function of $y$ , the product rule is needed. The final rearrangement should be done carefully so that the given form is reached exactly.

Common Mistakes

Forgetting to multiply the derivative of a $y$ -term by $\frac{dy}{dx}$ .
Differentiating $3e^{2x}y$ as if $y$ were constant.
Losing a factor of $3$ when differentiating $3e^{2x}y$ .
Sign errors when moving terms across the equation.
Since the answer is given (AG), the working must show all necessary detail; an unsupported answer is not enough.

Things to Be Careful About

The derivative of $e^{2x}$ is $2e^{2x}$ , not $e^{2x}$ .
The term $y^3$ differentiates to $3y^2\frac{dy}{dx}$ , not $3y^2$ .
When factorising, divide every term by the same factor; here dividing by $3$ gives the coefficient $e^{2x}+y^2$ .

Techniques used

differentiate implicitly with respect to xapply the product rule to products involving exponential functionscollect terms containing the derivativerearrange to isolate the derivative

(b)

Show that the curve has no stationary points.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

A stationary point occurs when $\frac{dy}{dx}=0$ . Set the numerator of the derivative equal to zero, solve for $y$ in terms of $x$ , substitute into the original curve equation, and show that the resulting equation has no real solution.

Working

Stationary points require:

\frac{-2e^{2x}y - 4e^{3x}}{e^{2x} + y^2} = 0

Since $e^{2x}+y^2>0$ , the numerator must be zero:

-2e^{2x}y - 4e^{3x} = 0

Divide by $-2e^{2x}$ :

y = -2e^x

Substitute $y=-2e^x$ into $3e^{2x}y+4e^{3x}+y^3=18$ :

3e^{2x}(-2e^x)+4e^{3x}+(-2e^x)^3=18

-6e^{3x}+4e^{3x}-8e^{3x}=18

-10e^{3x}=18

But $e^{3x}>0$ for all real $x$ , so $-10e^{3x}<0$ and cannot equal $18$ . Therefore no real $x$ satisfies the stationary condition.

Answer

The curve has no stationary points.

Final answer

No stationary points, since -10e^(3x) = 18 is impossible for real x.

Detailed explanation

Walkthrough

A stationary point is a point where $\frac{dy}{dx}=0$ . From the derivative, this means the numerator must be zero, because the denominator is always positive. So solve

-2e^{2x}y-4e^{3x}=0.

Dividing by $-2e^{2x}$ gives $y=-2e^x$ . This is the only possible $y$ -coordinate for a stationary point. Substitute this into the original curve equation. The left-hand side becomes

3e^{2x}(-2e^x)+4e^{3x}+(-2e^x)^3=-6e^{3x}+4e^{3x}-8e^{3x}=-10e^{3x}.

So a stationary point would require $-10e^{3x}=18$ . But $e^{3x}>0$ for every real $x$ , so $-10e^{3x}$ is always negative. It cannot equal $18$ . Hence no real $x$ satisfies the condition, and the curve has no stationary points.

Key Takeaways

This part connects differentiation with curve behaviour. To find stationary points, set the derivative to zero. When the derivative is a fraction, only the numerator matters, provided the denominator is not zero. Substituting back into the original equation is often needed to check whether the stationary condition can actually occur. The positivity of the exponential function is a useful property for ruling out solutions.

Common Mistakes

Forgetting to substitute $y=-2e^x$ back into the original equation.
Solving $-2e^{2x}y-4e^{3x}=0$ incorrectly and obtaining the wrong relation between $y$ and $x$ .
Ignoring the denominator when setting the derivative to zero.
Concluding that $-10e^{3x}=18$ has a solution by taking logarithms of a negative number; this is invalid.
Since the result is given (AG), the substitution and simplification must be shown.

Things to Be Careful About

$e^{2x}$ is never zero, so dividing by $e^{2x}$ is valid.
$e^{3x}$ is always positive, so $-10e^{3x}$ is always negative.
The original equation is used after substitution; do not use the differentiated equation for this step.
Check all signs when substituting $y=-2e^x$ , especially $(-2e^x)^3=-8e^{3x}$ .

Techniques used

set the derivative equal to zerosolve for y in terms of xsubstitute into the original equationuse the positivity of the exponential function to rule out solutions

Mathematics 9709/22 — February/March 2025

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