(a) Show that the equation $\log_4(2x + 1) = 2\log_4(3x - 1) - 2$ can be written as a quadratic equation in $x$. (b) Hence solve the equation $\log_4(2x + 1) = 2\log_4(3x - 1) - 2$.

(a) 9x^2 - 38x - 15 = 0 (or equivalently 16(2x+1) = (3x-1)^2) · (b) x = (19 + 4√31)/9 ≈ 4.59

(a) Express $3\sqrt{2}\sin(x + 45^\circ) + \cos x$ in the form $R\cos(x - \alpha)$, where $R > 0$ and $0^\circ < \alpha < 90^\circ$. (b) Hence solve the equation $3\sqrt{2}\sin(3\theta + 45^\circ) + \cos 3\theta = -4$ for $0^\circ < \theta < 180^\circ$.

(a) R = 5, alpha = 36.9°; expression = 5 cos(x - 36.9°) · (b) θ = 60.0° and θ = 84.6°

(a) By sketching a suitable pair of graphs, show that the equation $$ \sec 2x = -e^x $$ has only one root in the interval $0 < x < \frac{1}{2}\pi$. (b) Show by calculation that this root lies between 0.9 and 1. (c) Show that if a sequence of values given by the iterative formula $$ x_{n+1} = \frac{1}{2}\cos^{-1}(-e^{-x_n}) $$ converges, then it converges to the root of the equation in part (a). (d) Use the iterative formula given in part (c) to calculate $x$ correct to 3…

(a) The graphs intersect exactly once in the interval (pi/4, pi/2). · (b) The root lies between 0.9 and 1. · (c) The iterative formula converges to the root of sec 2x = -e^x. · (d) 0.978

Let $f(x) = \frac{x^3 + 2x - 11}{(3 + x)(2 + x^2)}$. (a) Express $f(x)$ in partial fractions. (b) Hence obtain the expansion of $f(x)$ in ascending powers of $x$, up to and including the term in $x^2$.

(a) f(x) = 1 - 4/(3+x) + (x-3)/(2+x^2) · (b) f(x) = -11/6 + (17/18)x + (65/108)x^2

9709/31 Oct/Nov 2025 — Mathematics Solved Past Paper

Q14MMedium-EasyIntegration

Find the exact value of $\int_1^2 \ln 3x \, dx$ . Give your answer in the form $a + \ln b$ , where $a$ and $b$ are integers.

DifficultyMedium-Easy

Worked solution

Approach

Use integration by parts with $u = \ln 3x$ and $\frac{dv}{dx} = 1$ , then evaluate the antiderivative between the limits 1 and 2, combining the logarithms at the end.

Working

Let $u = \ln 3x$ and $\frac{dv}{dx} = 1$ . Then:

\frac{du}{dx} = \frac{1}{x}, \quad v = x

Using integration by parts:

\int \ln 3x \, dx = x \ln 3x - \int x \cdot \frac{1}{x} \, dx

Simplify the integrand:

\int \ln 3x \, dx = x \ln 3x - \int 1 \, dx = x \ln 3x - x

Evaluate the definite integral from 1 to 2:

\left[ x \ln 3x - x \right]_1^2 = (2 \ln 6 - 2) - (1 \ln 3 - 1)

Simplify:

= 2 \ln 6 - 2 - \ln 3 + 1 = -1 + 2 \ln 6 - \ln 3

Combine the logarithms:

-1 + \ln 36 - \ln 3 = -1 + \ln\left(\frac{36}{3}\right) = -1 + \ln 12

Answer

-1 + \ln 12

So $a = -1$ and $b = 12$ .

Final answer

-1 + ln 12

Detailed explanation

Walkthrough

We need to find the exact value of $\int_1^2 \ln 3x \, dx$ . Since $\ln 3x$ is not a standard integrand we recognise directly, we use integration by parts:

\int u \, \frac{dv}{dx} \, dx = uv - \int v \, \frac{du}{dx} \, dx

We choose $u = \ln 3x$ (to be differentiated) and $\frac{dv}{dx} = 1$ (to be integrated). Differentiating $\ln 3x$ using the chain rule gives $\frac{1}{x}$ , and integrating 1 gives $x$ . Substituting into the formula:

\int \ln 3x \, dx = x \ln 3x - \int x \cdot \frac{1}{x} \, dx = x \ln 3x - \int 1 \, dx = x \ln 3x - x

We then evaluate this antiderivative at the limits. At $x = 2$ the value is $2 \ln 6 - 2$ ; at $x = 1$ the value is $\ln 3 - 1$ . Subtracting the lower from the upper:

(2 \ln 6 - 2) - (\ln 3 - 1) = 2 \ln 6 - 2 - \ln 3 + 1 = -1 + 2 \ln 6 - \ln 3

Finally we combine the logarithms. Since $2 \ln 6 = \ln 6^2 = \ln 36$ , we have $\ln 36 - \ln 3 = \ln(36/3) = \ln 12$ . So the final answer is $-1 + \ln 12$ , matching the required form with $a = -1$ and $b = 12$ .

Key Takeaways

This question tests the ability to apply integration by parts to a logarithmic function, to differentiate $\ln(ax)$ correctly using the chain rule, and to combine logarithms using the laws of logarithms. It also tests careful substitution of limits into a definite integral.

Common Mistakes

Forgetting that $\frac{d}{dx}(\ln 3x) = \frac{1}{x}$ (the 3 cancels), and instead writing $\frac{1}{3x}$ .
Incorrectly substituting the limits, e.g. mixing up which limit is subtracted from which.
Failing to combine $2 \ln 6 - \ln 3$ into $\ln 12$ , leaving the answer in an uncombined form that does not match the required structure.

Things to Be Careful About

The final answer must be in the form $a + \ln b$ with integers $a$ and $b$ , so the logarithm must be fully combined.
The mark scheme requires working to be shown; an unsupported answer gains no marks.
When substituting limits, be careful with the signs: the lower-limit contribution is subtracted.

Techniques used

apply integration by partsdifferentiate a logarithmic function using the chain ruleevaluate a definite integral at its limitscombine logarithms using the laws of logarithms

Similar questions

Q2MediumLogarithmic and Exponential Functions

(a)

Show that the equation $\log_4(2x + 1) = 2\log_4(3x - 1) - 2$ can be written as a quadratic equation in $x$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Express the constant 2 as a logarithm in base 4, apply the power law to move the coefficient inside the logarithm, combine the logarithms using the quotient law, then equate arguments and expand to obtain a quadratic equation.

Working

Start with the equation:

\log_4(2x + 1) = 2\log_4(3x - 1) - 2

Since $2 = \log_4(4^2) = \log_4 16$ , rewrite the equation:

\log_4(2x + 1) = 2\log_4(3x - 1) - \log_4 16

Apply the power law $n\log_b a = \log_b(a^n)$ :

2\log_4(3x - 1) = \log_4(3x - 1)^2

So:

\log_4(2x + 1) = \log_4(3x - 1)^2 - \log_4 16

Combine using the quotient law $\log_b a - \log_b c = \log_b\left(\frac{a}{c}\right)$ :

\log_4(2x + 1) = \log_4\left(\frac{(3x - 1)^2}{16}\right)

Since the logarithms are equal, their arguments are equal:

2x + 1 = \frac{(3x - 1)^2}{16}

Multiply through by 16:

16(2x + 1) = (3x - 1)^2

Expand and rearrange:

32x + 16 = 9x^2 - 6x + 1

9x^2 - 38x - 15 = 0

This is the required quadratic equation.

Answer

The equation can be written as:

9x^2 - 38x - 15 = 0

(or equivalently $16(2x + 1) = (3x - 1)^2$ ).

Final answer

9x^2 - 38x - 15 = 0 (or equivalently 16(2x+1) = (3x-1)^2)

Detailed explanation

Walkthrough

We start with the equation $\log_4(2x + 1) = 2\log_4(3x - 1) - 2$ . The goal is to eliminate the logarithms and obtain a quadratic equation in $x$ .

First, we express the constant 2 as a logarithm in base 4. Since $\log_4(4^2) = 2$ , we can write $2 = \log_4 16$ . This earns the B1 mark — stating or implying that $2 = \log_4 4^2$ .

Next, we apply the power law of logarithms to the term $2\log_4(3x - 1)$ . The power law says $n\log_b a = \log_b(a^n)$ , so $2\log_4(3x - 1) = \log_4(3x - 1)^2$ . This earns the M1 mark.

Now the equation becomes $\log_4(2x + 1) = \log_4(3x - 1)^2 - \log_4 16$ . We combine the two logarithms on the right using the quotient law: $\log_b a - \log_b c = \log_b(a/c)$ . So $\log_4(3x - 1)^2 - \log_4 16 = \log_4\left(\frac{(3x - 1)^2}{16}\right)$ .

Since both sides are logarithms with the same base 4, their arguments must be equal: $2x + 1 = \frac{(3x - 1)^2}{16}$ . Multiplying by 16 gives $16(2x + 1) = (3x - 1)^2$ . Expanding and rearranging gives $9x^2 - 38x - 15 = 0$ , which is the required quadratic. This earns the A1 mark.

Key Takeaways

A constant can be written as a logarithm in any base: $c = \log_b(b^c)$ .
The power law $n\log_b a = \log_b(a^n)$ moves a coefficient inside the logarithm.
The quotient law $\log_b a - \log_b c = \log_b(a/c)$ combines two logarithms.
If $\log_b A = \log_b B$ , then $A = B$ (provided the base is the same).
The final equation must be free of logarithms to be a "quadratic equation in $x$ ".

Common Mistakes

Forgetting to express the constant 2 as a logarithm — students sometimes try to equate arguments directly, which fails because the right-hand side is not a single logarithm.
Incorrectly applying the power law, e.g. writing $2\log_4(3x - 1) = \log_4(2(3x - 1))$ instead of $\log_4(3x - 1)^2$ .
Sign errors when combining logarithms with the quotient law (adding instead of subtracting).
Errors in expanding $(3x - 1)^2$ — forgetting the middle term $-6x$ .

Things to Be Careful About

The mark scheme requires showing the step $2 = \log_4 4^2$ (or equivalent) for the B1 mark.
The M1 mark requires using the power law (or product/quotient law) correctly.
The A1 mark requires a correct equation free of logs, in any form — e.g. $16(2x + 1) = (3x - 1)^2$ or $9x^2 - 38x - 15 = 0$ .
All logarithms in the original equation must have positive arguments; this becomes relevant in part (b).

Techniques used

express the constant 2 as a logarithm in base 4apply the power law of logarithmscombine logarithms using the quotient lawequate logarithmic arguments and expand to form a quadratic

(b)

Hence solve the equation $\log_4(2x + 1) = 2\log_4(3x - 1) - 2$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Solve the quadratic obtained in part (a) using the quadratic formula, then reject any root that makes the argument of a logarithm negative.

Working

From part (a):

9x^2 - 38x - 15 = 0

Apply the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 9$ , $b = -38$ , $c = -15$ :

x = \frac{38 \pm \sqrt{(-38)^2 - 4(9)(-15)}}{2(9)}

x = \frac{38 \pm \sqrt{1444 + 540}}{18}

x = \frac{38 \pm \sqrt{1984}}{18}

Since $1984 = 64 \times 31$ , we have $\sqrt{1984} = 8\sqrt{31}$ :

x = \frac{38 \pm 8\sqrt{31}}{18} = \frac{19 \pm 4\sqrt{31}}{9}

The two roots are:

x = \frac{19 + 4\sqrt{31}}{9} \approx 4.59

and

x = \frac{19 - 4\sqrt{31}}{9} \approx -0.363

The original equation contains $\log_4(2x + 1)$ and $\log_4(3x - 1)$ , which require $2x + 1 > 0$ and $3x - 1 > 0$ , i.e. $x > \frac{1}{3}$ . The root $\frac{19 - 4\sqrt{31}}{9} \approx -0.363$ is negative, so it is rejected.

Therefore:

x = \frac{19 + 4\sqrt{31}}{9} \approx 4.59

Answer

x = \frac{19 + 4\sqrt{31}}{9} \approx 4.59

Final answer

x = (19 + 4√31)/9 ≈ 4.59

Detailed explanation

Walkthrough

From part (a) we have the quadratic $9x^2 - 38x - 15 = 0$ . We solve it using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 9$ , $b = -38$ , $c = -15$ .

This gives $x = \frac{38 \pm \sqrt{1984}}{18} = \frac{19 \pm 4\sqrt{31}}{9}$ . The two roots are approximately 4.59 and -0.363.

Now we must check the roots against the domain of the original equation. The original equation contains $\log_4(2x + 1)$ and $\log_4(3x - 1)$ , which require $2x + 1 > 0$ and $3x - 1 > 0$ . The stricter condition is $3x - 1 > 0$ , i.e. $x > \frac{1}{3}$ . The root $\frac{19 - 4\sqrt{31}}{9} \approx -0.363$ is negative, so it is rejected. Only $\frac{19 + 4\sqrt{31}}{9} \approx 4.59$ is valid.

The M1 mark is for solving the 3-term quadratic; the A1 mark is for the correct answer "4.59 only" — the word "only" signals that the negative root must be rejected.

Key Takeaways

The quadratic formula solves any 3-term quadratic.
Solutions of a transformed equation must be checked against the domain of the original equation.
Logarithms are only defined for positive arguments, so any root making $\log_4(2x + 1)$ or $\log_4(3x - 1)$ undefined must be rejected.

Common Mistakes

Forgetting to reject the negative root — the mark scheme explicitly requires "4.59 only".
Arithmetic errors in the quadratic formula, especially with the discriminant $38^2 - 4(9)(-15)$ .
Not simplifying $\sqrt{1984}$ to $8\sqrt{31}$ (though this is not required for the marks, it gives the exact answer).

Things to Be Careful About

The mark scheme accepts AWRT 4.59 (anything within reasonable tolerance of 4.59).
The exact answer is $\frac{19 + 4\sqrt{31}}{9}$ .
The domain condition is $x > \frac{1}{3}$ , since $3x - 1 > 0$ is stricter than $2x + 1 > 0$ .
Do not round to 4.6 unless the question allows it; AWRT 4.59 is the accepted value.

Techniques used

solve the quadratic using the quadratic formulareject the root outside the domain of the logarithms

Q3MediumTrigonometry

(a)

Express $3\sqrt{2}\sin(x + 45^\circ) + \cos x$ in the form $R\cos(x - \alpha)$ , where $R > 0$ and $0^\circ < \alpha < 90^\circ$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Expand $\sin(x+45^\circ)$ using the compound angle formula, simplify to $3\sin x+4\cos x$ , then write this in the form $R\cos(x-\alpha)$ by comparing coefficients.

Working

Using $\sin(A+B)=\sin A\cos B+\cos A\sin B$ ,

\sin(x+45^\circ)=\sin x\cos45^\circ+\cos x\sin45^\circ

Since $\cos45^\circ=\sin45^\circ=\frac{1}{\sqrt2}$ , the given expression becomes

3\sqrt2\sin(x+45^\circ)+\cos x=3\sin x+3\cos x+\cos x=3\sin x+4\cos x.

Now write $R\cos(x-\alpha)=R\cos x\cos\alpha+R\sin x\sin\alpha$ . Comparing with $3\sin x+4\cos x$ gives

R\cos\alpha=4,\qquad R\sin\alpha=3.

Hence

R=\sqrt{3^2+4^2}=5

and

\tan\alpha=\frac{R\sin\alpha}{R\cos\alpha}=\frac34.

So

\alpha=\tan^{-1}\left(\frac34\right)=36.9^\circ.

Answer

3\sqrt2\sin(x+45^\circ)+\cos x=5\cos(x-36.9^\circ)

Final answer

R = 5, alpha = 36.9°; expression = 5 cos(x - 36.9°)

Detailed explanation

Walkthrough

Start with the given expression. The angle $x+45^\circ$ is a compound angle, so expand $\sin(x+45^\circ)$ using the addition formula. Substitute $\cos45^\circ=\sin45^\circ=\frac{1}{\sqrt2}$ and multiply by $3\sqrt2$ . This gives $3\sin x+3\cos x$ . Adding the extra $\cos x$ gives $3\sin x+4\cos x$ , which is exactly the $a\sin x+b\cos x$ form.

To rewrite this as $R\cos(x-\alpha)$ , expand the target form using the cosine subtraction formula and equate the coefficients of $\sin x$ and $\cos x$ . This gives $R\cos\alpha=4$ and $R\sin\alpha=3$ . Square and add these equations to eliminate $\alpha$ and find $R=5$ . Dividing the second equation by the first gives $\tan\alpha=\frac34$ , then taking inverse tangent in degrees gives $\alpha=36.9^\circ$ .

Be careful not to write $\cos\alpha=4$ and $\sin\alpha=3$ without the $R$ factors; because $R=5$ , the correct statements are $\cos\alpha=\frac45$ and $\sin\alpha=\frac35$ .

Key Takeaways

An expression of the form $a\sin x+b\cos x$ can be rewritten as a single sinusoidal expression $R\cos(x-\alpha)$ .
The compound angle formula lets us simplify angles such as $x+45^\circ$ .
Comparing coefficients is a powerful technique for matching two equivalent trigonometric forms.
For $R\cos(x-\alpha)=a\cos x+b\sin x$ , we have $R=\sqrt{a^2+b^2}$ and $\tan\alpha=\frac{b}{a}$ .

Common Mistakes

Expanding $\sin(x+45^\circ)$ incorrectly, especially swapping the sine and cosine terms.
Forgetting to add the remaining $\cos x$ after multiplying the expansion by $3\sqrt2$ .
Writing $\cos\alpha=4$ and $\sin\alpha=3$ instead of including the $R$ factors; the mark scheme gives M0A0 for this.
Mixing up the coefficients: the coefficient of $\cos x$ is $4$ and the coefficient of $\sin x$ is $3$ .
Working in radians when the question is in degrees.

Things to Be Careful About

Keep all angles in degrees throughout.
$R$ must be positive, so take the positive square root.
The required answer has $0^\circ<\alpha<90^\circ$ , which $36.9^\circ$ satisfies.
Show the intermediate step $\alpha=\tan^{-1}\left(\frac34\right)$ to earn the M1 mark.

Techniques used

expand a compound angle using the sine addition formulacombine like terms to convert to a sin x + b cos x formmatch coefficients with the R cos(x - α) formuse inverse tangent to determine the phase angle

(b)

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

4M Similar

DifficultyMedium

Worked solution

Approach

Use the result from part (a) to rewrite the left-hand side as $5\cos(3\theta-36.9^\circ)$ . Then solve $\cos(3\theta-36.9^\circ)=-\frac45$ by finding the principal angle and the two families of solutions.

Working

From part (a),

3\sqrt2\sin(3\theta+45^\circ)+\cos3\theta=5\cos(3\theta-36.9^\circ).

The equation is therefore

5\cos(3\theta-36.9^\circ)=-4

so

\cos(3\theta-36.9^\circ)=-\frac45.

Let $u=3\theta-36.9^\circ$ . The principal value is

\cos^{-1}\left(-\frac45\right)=143.13^\circ.

Thus the general solutions are

u=\pm143.13^\circ+360n,

where $n$ is an integer.

Case 1:

3\theta-36.9^\circ=143.13^\circ+360n

3\theta=180^\circ+360n

\theta=60^\circ+120n.

For $0^\circ<\theta<180^\circ$ , $n=0$ gives $\theta=60^\circ$ ; $n=1$ gives $180^\circ$ , which is not included.

Case 2:

3\theta-36.9^\circ=-143.13^\circ+360n

3\theta=-106.26^\circ+360n

\theta=-35.42^\circ+120n.

For $0^\circ<\theta<180^\circ$ , $n=1$ gives $\theta=84.6^\circ$ ; $n=2$ gives $204.6^\circ$ , which is outside the interval.

Answer

\theta=60^\circ\quad\text{or}\quad\theta=84.6^\circ

Final answer

θ = 60.0° and θ = 84.6°

Detailed explanation

Walkthrough

The word "Hence" tells us to use the result of part (a). Replace the left-hand side by $5\cos(3\theta-36.9^\circ)$ . Then set this equal to $-4$ . Dividing by $5$ isolates the cosine: $\cos(3\theta-36.9^\circ)=-\frac45$ .

Let $u=3\theta-36.9^\circ$ . The principal angle whose cosine is $-\frac45$ is $143.13^\circ$ , because cosine is negative in the second quadrant. For a cosine equation, the general solution is $u=\pm143.13^\circ+360n$ . This accounts for both the second-quadrant angle and its reflection in the x-axis, and for all full rotations.

Solve the two cases for $\theta$ :

$3\theta-36.9^\circ=143.13^\circ+360n$ gives $\theta=60^\circ+120n$ . Within $0<\theta<180$ , only $\theta=60^\circ$ is valid; $180^\circ$ is excluded by the strict inequality.
$3\theta-36.9^\circ=-143.13^\circ+360n$ gives $\theta=-35.42^\circ+120n$ . Within the interval, $n=1$ gives $84.6^\circ$ ; $n=2$ gives $204.6^\circ$ , which is outside.

So the two solutions are $60^\circ$ and $84.6^\circ$ .

Key Takeaways

A single sinusoidal form such as $R\cos(\dots)$ turns a more complicated equation into a standard $\cos u=k$ equation.
The general solution of $\cos u=c$ is $u=\pm\cos^{-1}c+360n$ .
After solving for $u$ , remember to convert to the required variable, here $\theta$ , by dividing by the coefficient.
Always test each integer $n$ to see whether the resulting value lies in the required interval.

Common Mistakes

Forgetting the $\pm$ sign in the general solution and losing one of the two families.
Not adding $360n$ , so missing repeated solutions for larger $n$ .
Solving for $u$ but forgetting to divide by $3$ to get $\theta$ .
Including $\theta=180^\circ$ , even though the interval is $0<\theta<180^\circ$ and therefore excludes the endpoint.
Stopping after finding only one solution.
If part (a) gives a different $R$ , keep using that value consistently; the mark scheme allows follow-through marks.

Things to Be Careful About

All angles are in degrees; set the calculator to degree mode.
The inequality is strict, so neither $0^\circ$ nor $180^\circ$ is allowed.
Acceptable answers are AWRT $60.0$ and AWRT $84.6$ .
Answers outside the given interval should be ignored, not counted as extra solutions.
Show the step $\cos^{-1}(-4/5)=143.13^\circ$ clearly, as this earns the B1 mark.

Techniques used

substitute the R cos form from part (a)divide by R to isolate a cosine equationtake inverse cosine and apply the general solution with ± and period 360°check all solutions in the given interval

Q45MMediumDifferentiationTrigonometry

Fig. 1

The diagram shows the graph of $y = e^{\sin 2x} \cos 4x$ for $0 \le x \le \frac{1}{4}\pi$ , and its maximum point $M$ .

Find the $x$ -coordinate of $M$ .

DifficultyMedium

Worked solution

Approach

Differentiate $y = e^{\sin 2x} \cos 4x$ using the product rule and chain rule, set $\frac{dy}{dx} = 0$ at the maximum point $M$ , and solve the resulting equation by expressing everything in terms of $\sin 2x$ using double-angle identities.

Working

Differentiate using the product rule:

\frac{dy}{dx} = e^{\sin 2x} \cdot \frac{d}{dx}(\cos 4x) + \cos 4x \cdot \frac{d}{dx}(e^{\sin 2x})

Compute each derivative using the chain rule:

\frac{d}{dx}(\cos 4x) = -4\sin 4x

\frac{d}{dx}(e^{\sin 2x}) = e^{\sin 2x} \cdot \cos 2x \cdot 2 = 2\cos 2x \, e^{\sin 2x}

Substitute back:

\frac{dy}{dx} = -4\sin 4x \, e^{\sin 2x} + 2\cos 2x \cos 4x \, e^{\sin 2x}

At the maximum point $M$ , $\frac{dy}{dx} = 0$ . Since $e^{\sin 2x} \neq 0$ , divide through:

-4\sin 4x + 2\cos 2x \cos 4x = 0

Use the double-angle identity $\sin 4x = 2\sin 2x \cos 2x$ :

-4(2\sin 2x \cos 2x) + 2\cos 2x \cos 4x = 0

-8\sin 2x \cos 2x + 2\cos 2x \cos 4x = 0

Factor out $2\cos 2x$ :

2\cos 2x (-4\sin 2x + \cos 4x) = 0

So either $\cos 2x = 0$ or $\cos 4x - 4\sin 2x = 0$ .

For $\cos 2x = 0$ with $0 \le x \le \frac{1}{4}\pi$ , we get $2x = \frac{\pi}{2}$ , so $x = \frac{\pi}{4}$ . At this point, $\cos 4x = \cos \pi = -1$ , giving $y = e^1 \cdot (-1) = -e$ , which is not a maximum.

Now solve $\cos 4x - 4\sin 2x = 0$ . Use $\cos 4x = 1 - 2\sin^2 2x$ :

1 - 2\sin^2 2x - 4\sin 2x = 0

2\sin^2 2x + 4\sin 2x - 1 = 0

Let $u = \sin 2x$ :

2u^2 + 4u - 1 = 0

Apply the quadratic formula:

u = \frac{-4 \pm \sqrt{16 - 4(2)(-1)}}{2(2)} = \frac{-4 \pm \sqrt{24}}{4} = \frac{-4 \pm 2\sqrt{6}}{4} = \frac{-2 \pm \sqrt{6}}{2}

The two solutions are:

u_1 = \frac{-2 + \sqrt{6}}{2} \approx 0.2247, \quad u_2 = \frac{-2 - \sqrt{6}}{2} \approx -2.225

Since $-1 \le \sin 2x \le 1$ , reject $u_2$ .

So $\sin 2x = \frac{-2 + \sqrt{6}}{2} \approx 0.2247$ .

For $0 \le x \le \frac{1}{4}\pi$ , we have $0 \le 2x \le \frac{\pi}{2}$ , so:

2x = \arcsin\left(\frac{-2 + \sqrt{6}}{2}\right) \approx 0.2266

x = \frac{0.2266}{2} \approx 0.1133

Rounding to 3 decimal places:

x = 0.113

Answer

x = 0.113

Final answer

x = 0.113

Detailed explanation

Walkthrough

Step 1: Differentiate using the product rule. The function $y = e^{\sin 2x} \cos 4x$ is a product of two functions: $u = e^{\sin 2x}$ and $v = \cos 4x$ . The product rule gives $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ . This earns the first mark (M1) for correctly applying the product rule.

Step 2: Apply the chain rule to each part. For $\frac{d}{dx}(\cos 4x)$ , the outer function is $\cos(\cdot)$ and the inner function is $4x$ , giving $-4\sin 4x$ . For $\frac{d}{dx}(e^{\sin 2x})$ , the outer function is $e^{(\cdot)}$ and the inner function is $\sin 2x$ , giving $e^{\sin 2x} \cdot \cos 2x \cdot 2 = 2\cos 2x \, e^{\sin 2x}$ . Substituting these back yields the full derivative $\frac{dy}{dx} = -4\sin 4x \, e^{\sin 2x} + 2\cos 2x \cos 4x \, e^{\sin 2x}$ , which earns the second mark (A1).

Step 3: Set the derivative to zero and simplify. At a maximum point, $\frac{dy}{dx} = 0$ . Since $e^{\sin 2x}$ is always positive and never zero, we can divide it out, leaving $-4\sin 4x + 2\cos 2x \cos 4x = 0$ . This earns the third mark (M1) for equating the derivative to zero and obtaining an equation in one trig function.

Step 4: Use double-angle identities to reduce to a single trig function. Substitute $\sin 4x = 2\sin 2x \cos 2x$ to get $-8\sin 2x \cos 2x + 2\cos 2x \cos 4x = 0$ . Factor out $2\cos 2x$ to get $2\cos 2x(-4\sin 2x + \cos 4x) = 0$ . The factor $\cos 2x = 0$ gives $x = \frac{\pi}{4}$ , but checking this in the original function shows it is not a maximum (it gives $y = -e$ ). For the other factor, substitute $\cos 4x = 1 - 2\sin^2 2x$ to get $1 - 2\sin^2 2x - 4\sin 2x = 0$ , which rearranges to $2\sin^2 2x + 4\sin 2x - 1 = 0$ . This earns the fourth mark (A1).

Step 5: Solve the quadratic in $\sin 2x$ . Let $u = \sin 2x$ , giving $2u^2 + 4u - 1 = 0$ . The quadratic formula gives $u = \frac{-2 \pm \sqrt{6}}{2}$ . Since $\sin 2x$ must lie in $[-1, 1]$ , only $u = \frac{-2 + \sqrt{6}}{2} \approx 0.2247$ is valid. Then $2x = \arcsin(0.2247) \approx 0.2266$ , so $x \approx 0.1133$ , which rounds to $x = 0.113$ . This earns the final mark (A1).

Key Takeaways

The product rule combined with the chain rule is essential for differentiating functions involving products of exponentials and trigonometric functions.
Setting $\frac{dy}{dx} = 0$ and factoring out common terms (especially $e^{\text{something}}$ which is never zero) simplifies the equation significantly.
Double-angle identities ( $\sin 2A = 2\sin A \cos A$ and $\cos 2A = 1 - 2\sin^2 A$ ) are powerful tools for reducing equations involving multiple angles to a single trig function.
Always check whether solutions from factored equations actually correspond to maxima or minima by evaluating the original function.

Common Mistakes

Forgetting the chain rule when differentiating $e^{\sin 2x}$ , leading to an incorrect derivative such as $e^{\sin 2x} \cos 4x$ instead of $2\cos 2x \, e^{\sin 2x} \cos 4x$ .
Not dividing out $e^{\sin 2x}$ when setting the derivative to zero, making the equation appear more complicated than it is.
Failing to check the extraneous solution $x = \frac{\pi}{4}$ from $\cos 2x = 0$ , which does not correspond to the maximum point $M$ .
Making sign errors when applying the double-angle identity for $\cos 4x$ , particularly confusing $\cos 4x = 1 - 2\sin^2 2x$ with $2\cos^2 2x - 1$ .
Rejecting the valid root $\frac{-2 + \sqrt{6}}{2}$ or incorrectly keeping $\frac{-2 - \sqrt{6}}{2}$ which is outside the range of $\sin 2x$ .

Things to Be Careful About

The interval $0 \le x \le \frac{1}{4}\pi$ means $0 \le 2x \le \frac{\pi}{2}$ , so $\sin 2x \ge 0$ and there is only one solution to $\sin 2x = k$ in this range. Always check the domain when solving trigonometric equations.
The mark scheme condones a sign error in the derivative and allows arithmetical errors, but requires showing the method clearly.
The answer $x = 0.113$ is given to 3 decimal places; more accuracy (e.g., $x = 0.1133$ ) is also acceptable.
When factoring $2\cos 2x(-4\sin 2x + \cos 4x) = 0$ , remember to check both factors in the original context — $\cos 2x = 0$ gives a point on the curve, but it is not the maximum $M$ .

Techniques used

apply product ruleapply chain rule for composite exponentialuse double angle identities to simplifysolve quadratic equation in trig functionsolve trigonometric equation in given interval

Similar questions

Q5MediumComplex Numbers

Fig. 2

The shaded region on the Argand diagram shows points representing complex numbers $z$ defined by two inequalities. The shaded region is bounded by a circle and a line parallel to the imaginary axis. The boundaries of the region are included in the shaded region.

(a)

Find two inequalities in terms of $z$ that define the shaded region.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The shaded region is bounded by a vertical line and a circle. We translate these geometric boundaries into inequalities involving $z$ .

Working

The vertical line is at $\operatorname{Re}(z) = 2$ . Since the shaded region is to the right of this line (including the boundary), the first inequality is:

\operatorname{Re}(z) \ge 2

The circle is centred at $1 - 2\mathrm{i}$ with radius 2. The region inside the circle (including the boundary) is described by the modulus inequality $|z - \text{centre}| \le \text{radius}$ :

|z - (1 - 2\mathrm{i})| \le 2

Simplifying the expression inside the modulus:

|z - 1 + 2\mathrm{i}| \le 2

Answer

\operatorname{Re}(z) \ge 2 \quad \text{and} \quad |z - 1 + 2\mathrm{i}| \le 2

Final answer

Re(z) >= 2 and |z - 1 + 2i| <= 2

Detailed explanation

Walkthrough

First, observe the vertical boundary of the shaded region. It is a straight line parallel to the imaginary axis passing through $\operatorname{Re}(z) = 2$ . The shaded area lies to the right of this line, which directly translates to the inequality $\operatorname{Re}(z) \ge 2$ .

Next, observe the circular boundary. The diagram shows a circle centred at the point $(1, -2)$ , which corresponds to the complex number $1 - 2\mathrm{i}$ . The radius of the circle is 2 units. The shaded region lies inside this circle. The set of points $z$ whose distance from a fixed point $a$ is less than or equal to a constant $r$ is given by the modulus inequality $|z - a| \le r$ . Substituting $a = 1 - 2\mathrm{i}$ and $r = 2$ , we get $|z - (1 - 2\mathrm{i})| \le 2$ , which simplifies to $|z - 1 + 2\mathrm{i}| \le 2$ .

Key Takeaways

A vertical line $x = c$ on the Argand diagram corresponds to $\operatorname{Re}(z) = c$ . The half-plane to the right is $\operatorname{Re}(z) \ge c$ .
A circle with centre $a$ and radius $r$ on the Argand diagram corresponds to the locus $|z - a| = r$ . The interior (including boundary) is $|z - a| \le r$ .

Common Mistakes

Forgetting to include the equality sign when the boundary is included in the shaded region. The question states "The boundaries of the region are included", so $\ge$ and $\le$ must be used, not $>$ and $<$ .
Writing the centre incorrectly, e.g., using $1 + 2\mathrm{i}$ instead of $1 - 2\mathrm{i}$ , which leads to $|z - 1 - 2\mathrm{i}| \le 2$ .

Things to Be Careful About

Ensure the modulus inequality is written in the form $|z - a| \le b$ where $a$ is complex and $b$ is real. The expression $|z - 1 + 2\mathrm{i}|$ is correct because $-(1 - 2\mathrm{i}) = -1 + 2\mathrm{i}$ .
The argument of $z$ is typically defined in $(-\pi, \pi]$ or $[0, 2\pi)$ . Part (a) does not require argument, but part (b) will.

Techniques used

interpret half-plane as real part inequalityidentify circle locus as modulus inequality

(b)

Calculate the least value of $\arg z$ for points in this region.

3M Similar

DifficultyMedium

Worked solution

Approach

The least value of $\arg z$ occurs at the point in the shaded region that makes the most negative angle with the positive real axis. This point lies on the boundary of the region, specifically where the vertical line $\operatorname{Re}(z) = 2$ intersects the lower part of the circle.

Working

The boundaries are given by $\operatorname{Re}(z) = 2$ and $|z - 1 + 2\mathrm{i}| = 2$ . Let $z = x + y\mathrm{i}$ . The circle equation is:

(x - 1)^2 + (y + 2)^2 = 4

Substitute $x = 2$ into the circle equation to find the intersection points:

(2 - 1)^2 + (y + 2)^2 = 4

1 + (y + 2)^2 = 4

(y + 2)^2 = 3

y + 2 = \pm\sqrt{3}

y = -2 \pm \sqrt{3}

The two intersection points are $(2, -2 + \sqrt{3})$ and $(2, -2 - \sqrt{3})$ . Since $\sqrt{3} \approx 1.732$ , the $y$ -coordinates are approximately $-0.268$ and $-3.732$ .

The point $(2, -2 - \sqrt{3})$ is in the fourth quadrant and has the most negative $y$ -value for $x = 2$ . For points in the region $x \ge 2$ and inside the circle, the argument $\theta = \tan^{-1}(y/x)$ is minimized when $y$ is as negative as possible and $x$ is as small as possible. The point $(2, -2 - \sqrt{3})$ satisfies this.

Calculate the argument of $z = 2 - (2 + \sqrt{3})\mathrm{i}$ :

\tan \theta = \frac{-(2 + \sqrt{3})}{2} = -1 - \frac{\sqrt{3}}{2} \approx -1.866

\theta = \tan^{-1}(-1.866) \approx -61.8^\circ

In radians:

\theta \approx -1.08 \text{ rad}

Alternatively, using the positive angle measure:

360^\circ - 61.8^\circ = 298.2^\circ \quad \text{or} \quad 2\pi - 1.08 \approx 5.20 \text{ rad}

The least value is $-61.8^\circ$ or $-1.08$ rad.

Answer

-61.8^\circ \quad \text{or} \quad -1.08 \text{ rad}

Final answer

-61.8 degrees or -1.08 rad

Detailed explanation

Walkthrough

To find the least value of $\arg z$ , we need to identify the point in the shaded region that has the smallest angle measured from the positive real axis. Since the region is in the fourth quadrant (mostly), the least argument will be the most negative angle.

The boundary of the region consists of the line $x = 2$ and the circle $(x-1)^2 + (y+2)^2 = 4$ . The extreme points for the argument will lie on these boundaries. We find the intersection of the line $x = 2$ and the circle by substituting $x = 2$ into the circle equation:

$(2-1)^2 + (y+2)^2 = 4 \implies 1 + (y+2)^2 = 4 \implies (y+2)^2 = 3 \implies y = -2 \pm \sqrt{3}$ .

The intersection points are $(2, -2+\sqrt{3})$ and $(2, -2-\sqrt{3})$ . The point $(2, -2-\sqrt{3})$ is lower down, meaning it has a more negative $y$ -coordinate. For a fixed $x > 0$ , a more negative $y$ gives a more negative argument. We also check if any other point in the region could give a smaller argument. The circle extends to $x=3$ at $y=-2$ , giving $\arg(3-2i) = \tan^{-1}(-2/3) \approx -33.7^\circ$ , which is greater than $-61.8^\circ$ . Thus, the minimum argument occurs at $(2, -2-\sqrt{3})$ .

Calculating the argument: $\tan \theta = \frac{-2-\sqrt{3}}{2}$ . Using a calculator, $\theta \approx -61.8^\circ$ or $-1.08$ rad.

Key Takeaways

The extreme values of $\arg z$ in a bounded region often occur at boundary points, particularly intersections of boundaries.
When finding the least argument, consider the quadrant and whether you want the most negative value (principal value) or the smallest positive value.

Common Mistakes

Finding the wrong intersection point (e.g., using $y = -2 + \sqrt{3}$ instead of $y = -2 - \sqrt{3}$ ).
Forgetting that $\arg z$ can be negative and giving $298.2^\circ$ as the answer when $-61.8^\circ$ is the least value (though both may be accepted depending on the convention, the question asks for the least value, which is $-61.8^\circ$ ).
Rounding errors in the final answer.

Things to Be Careful About

Ensure the argument is calculated using the correct quadrant. The point $(2, -2-\sqrt{3})$ is in the fourth quadrant, so the argument is negative (or between $270^\circ$ and $360^\circ$ ).
The mark scheme accepts $-61.8^\circ$ , $298.2^\circ$ , $-1.08$ rad, or $5.20$ rad. Be consistent with the required format.

Techniques used

find intersection of circle and linecalculate argument from Cartesian coordinates

Q65MMediumComplex Numbers

Solve the quadratic equation $(2 + i)w^2 + 4w + 2 - i = 0$ . Give your answers in the form $x + iy$ , where $x$ and $y$ are real.

DifficultyMedium

Worked solution

Approach

This is a quadratic equation with complex coefficients, so use the quadratic formula

w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

with $a = 2 + \mathrm{i}$ , $b = 4$ , $c = 2 - \mathrm{i}$ . Simplify the discriminant using $\mathrm{i}^2 = -1$ , then write each root in the form $x + \mathrm{i}y$ by multiplying numerator and denominator by the conjugate of the denominator.

Working

First compute the discriminant:

\begin{aligned} b^2 - 4ac &= 4^2 - 4(2 + \mathrm{i})(2 - \mathrm{i}) \\ &= 16 - 4(5) \\ &= -4 \end{aligned}

so $\sqrt{b^2 - 4ac} = \pm 2\mathrm{i}$ . Therefore

w = \frac{-4 \pm 2\mathrm{i}}{2(2 + \mathrm{i})} = \frac{-4 \pm 2\mathrm{i}}{4 + 2\mathrm{i}}

For the first root:

\begin{aligned} w_1 &= \frac{-4 + 2\mathrm{i}}{4 + 2\mathrm{i}} \\ &= \frac{(-4 + 2\mathrm{i})(4 - 2\mathrm{i})}{(4 + 2\mathrm{i})(4 - 2\mathrm{i})} \\ &= \frac{-12 + 16\mathrm{i}}{20} \\ &= -\frac{3}{5} + \frac{4}{5}\mathrm{i} \end{aligned}

For the second root:

\begin{aligned} w_2 &= \frac{-4 - 2\mathrm{i}}{4 + 2\mathrm{i}} \\ &= \frac{(-4 - 2\mathrm{i})(4 - 2\mathrm{i})}{(4 + 2\mathrm{i})(4 - 2\mathrm{i})} \\ &= \frac{-20}{20} \\ &= -1 \end{aligned}

Answer

w = -\frac{3}{5} + \frac{4}{5}\mathrm{i} \quad \text{or} \quad w = -1

Final answer

w = -3/5 + 4/5 i or w = -1 + 0i

Detailed explanation

Walkthrough

We are asked to solve a quadratic equation, but now the coefficients are complex numbers. The quadratic formula still applies: for $aw^2 + bw + c = 0$ ,

w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Here $a = 2 + \mathrm{i}$ , $b = 4$ and $c = 2 - \mathrm{i}$ . First calculate the discriminant. The product $(2 + \mathrm{i})(2 - \mathrm{i})$ is a difference of squares: it equals $4 - \mathrm{i}^2 = 4 - (-1) = 5$ . Hence $b^2 - 4ac = 16 - 4(5) = -4$ . Since $\sqrt{-4} = \pm 2\mathrm{i}$ , the two roots are $(-4 \pm 2\mathrm{i})/(4 + 2\mathrm{i})$ .

Now we need each answer in the form $x + \mathrm{i}y$ . The denominators are complex, so multiply numerator and denominator by the conjugate $4 - 2\mathrm{i}$ . This makes the denominator real: $(4 + 2\mathrm{i})(4 - 2\mathrm{i}) = 20$ . For the first numerator, $(-4 + 2\mathrm{i})(4 - 2\mathrm{i}) = -12 + 16\mathrm{i}$ , so the first root is $-\frac{3}{5} + \frac{4}{5}\mathrm{i}$ . For the second numerator, $(-4 - 2\mathrm{i})(4 - 2\mathrm{i}) = -20$ , so the second root is $-1$ , which can be written as $-1 + 0\mathrm{i}$ .

An alternative approach is to multiply the whole equation by $2 - \mathrm{i}$ first. This gives $5w^2 + 4(2 - \mathrm{i})w + (2 - \mathrm{i})^2 = 0$ , which can then be solved by factorising or the quadratic formula. Both methods lead to the same two roots.

Key Takeaways

The quadratic formula can be used with complex coefficients as well as real coefficients.
Use $\mathrm{i}^2 = -1$ whenever simplifying products of complex numbers.
To write a quotient of complex numbers in Cartesian form, multiply numerator and denominator by the conjugate of the denominator.
A real answer such as $-1$ is also a complex number, written as $-1 + 0\mathrm{i}$ .

Common Mistakes

Forgetting that $\mathrm{i}^2 = -1$ when multiplying conjugates, giving wrong real parts.
Dropping the $\pm$ sign when taking the square root of the discriminant.
Leaving the answer as a quotient of complex numbers instead of simplifying to $x + \mathrm{i}y$ .
Sign errors when expanding $(-4 \pm 2\mathrm{i})(4 - 2\mathrm{i})$ .
Not showing evidence of multiplying by the conjugate; the mark scheme requires some evidence of this step.

Things to Be Careful About

The discriminant is negative, so the square root introduces $\mathrm{i}$ .
The coefficient $a$ is complex, so $2a = 4 + 2\mathrm{i}$ , not $4$ .
When rationalising, multiply both numerator and denominator by the same conjugate.
Give the final answers in the required form $x + \mathrm{i}y$ ; write $-1$ as $-1 + 0\mathrm{i}$ if needed.

Techniques used

apply the quadratic formula to complex coefficientssimplify a complex discriminant using i^2 = -1rationalise a complex denominator using its conjugateexpress roots in Cartesian form x + iy

Similar questions

Q75MMediumDifferentiation

The parametric equations of a curve are

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

Obtain a simplified expression for $\frac{dy}{dx}$ in terms of $t$ .

DifficultyMedium

Worked solution

Approach

Since $x$ and $y$ are given parametrically in terms of $t$ , differentiate each expression with respect to $t$ , then use

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

Working

Differentiate $x = t^2 - \ln(2t + 1)$ with respect to $t$ . The derivative of $t^2$ is $2t$ , and by the chain rule the derivative of $\ln(2t + 1)$ is $\frac{2}{2t + 1}$ :

\frac{dx}{dt} = 2t - \frac{2}{2t + 1}.

Differentiate $y = \frac{t}{2t + 1}$ using the quotient rule:

\begin{aligned} \frac{dy}{dt} &= \frac{(2t + 1)(1) - t(2)}{(2t + 1)^2} \\ &= \frac{2t + 1 - 2t}{(2t + 1)^2} \\ &= \frac{1}{(2t + 1)^2}. \end{aligned}

Now use the parametric differentiation formula:

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

Simplify the denominator first:

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

Therefore

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

Answer

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

Final answer

dy/dx = 1 / [2(2t + 1)(2t - 1)(t + 1)]

Detailed explanation

Walkthrough

We are given $x$ and $y$ separately in terms of a parameter $t$ , so to find $\frac{dy}{dx}$ we differentiate both $x$ and $y$ with respect to $t$ , then use the parametric formula $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ .

For $x = t^2 - \ln(2t + 1)$ , the derivative of $t^2$ is $2t$ . For the logarithmic term, the chain rule says that the derivative of $\ln(2t + 1)$ is $\frac{2}{2t + 1}$ , because the derivative of the inner function $2t + 1$ is $2$ . This gives $\frac{dx}{dt} = 2t - \frac{2}{2t + 1}$ .

For $y = \frac{t}{2t + 1}$ , use the quotient rule:

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

Here $u = t$ and $v = 2t + 1$ , so $\frac{du}{dt} = 1$ and $\frac{dv}{dt} = 2$ . Substituting gives $\frac{(2t + 1)(1) - t(2)}{(2t + 1)^2}$ , which simplifies to $\frac{1}{(2t + 1)^2}$ .

The final step is to divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ . Write $\frac{dx}{dt}$ as a single fraction and factor the numerator:

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

Then dividing gives the simplified result.

Key Takeaways

This question tests the standard parametric differentiation technique: differentiate each parametric equation with respect to the parameter and divide the two derivatives. It also requires the chain rule for differentiating $\ln(2t + 1)$ and the quotient rule for differentiating a rational function of $t$ . Finally, it tests algebraic fluency in rewriting a compound fraction as a single simplified fraction.

Common Mistakes

A common error is to forget the factor $2$ when differentiating $\ln(2t + 1)$ , writing the derivative as $\frac{1}{2t + 1}$ instead of $\frac{2}{2t + 1}$ . Another common error is applying the quotient rule incorrectly, especially getting the order of numerator terms wrong. Students also sometimes divide $\frac{dx}{dt}$ by $\frac{dy}{dt}$ instead of the other way round. Finally, marks can be lost by not simplifying the compound fraction fully; the answer should be written as a single fraction with numerator $1$ .

Things to Be Careful About

The expression $\ln(2t + 1)$ is only defined for $2t + 1 > 0$ , so $t > -\frac{1}{2}$ . Within this domain, $2t + 1 \neq 0$ . Also note that $\frac{dx}{dt} = 0$ when $t = \frac{1}{2}$ or $t = -1$ , but $t = -1$ is outside the domain; at $t = \frac{1}{2}$ the tangent is vertical and the formula for $\frac{dy}{dx}$ is not valid. In the final answer, ensure the denominator is left fully factorised and that the numerator is exactly $1$ .

Techniques used

differentiate parametric equations with respect to the parameterapply the chain rule to the logarithmic termapply the quotient ruledivide the two parametric derivatives to find dy/dxsimplify the resulting rational expression

Similar questions

Q88MMedium-HardDifferential EquationsIntegration

The variables $x$ and $y$ satisfy the differential equation

(x^2 + 1)\frac{dy}{dx} = kxe^{2y}

where $k$ is a constant. It is given that $y = 0$ when $x = 0$ and that $y = -\frac{1}{2}$ when $x = 1$ .

Solve the differential equation and find the exact value of $y$ when $x = \sqrt{3}$ .

DifficultyMedium-Hard

Worked solution

Approach

Separate the variables so that all terms in $y$ are on one side and all terms in $x$ are on the other. Integrate both sides, using $\int \mathrm{e}^{-2y}\,dy = -\frac{1}{2}\mathrm{e}^{-2y}$ and $\int \frac{kx}{x^2+1}\,dx = \frac{k}{2}\ln(x^2+1)$ . Then use the two given pairs of values to determine the constants $c$ and $k$ , and finally substitute $x=\sqrt{3}$ .

Working

Separate variables:

\mathrm{e}^{-2y}\,dy = \frac{kx}{x^2+1}\,dx

Integrate both sides:

\int \mathrm{e}^{-2y}\,dy = \int \frac{kx}{x^2+1}\,dx

-\frac{1}{2}\mathrm{e}^{-2y} = \frac{k}{2}\ln(x^2+1) + c

Use $y=0$ when $x=0$ :

-\frac{1}{2} = 0 + c \quad \Rightarrow \quad c = -\frac{1}{2}

Use $y=-\frac{1}{2}$ when $x=1$ :

-\frac{1}{2}\mathrm{e} = \frac{k}{2}\ln 2 - \frac{1}{2}

-\mathrm{e} = k\ln 2 - 1

k = \frac{1-\mathrm{e}}{\ln 2}

Thus the particular solution is

-\frac{1}{2}\mathrm{e}^{-2y} = \frac{1-\mathrm{e}}{2\ln 2}\ln(x^2+1) - \frac{1}{2}

When $x=\sqrt{3}$ , $x^2+1=4$ , so

-\frac{1}{2}\mathrm{e}^{-2y} = \frac{1-\mathrm{e}}{2\ln 2}\ln 4 - \frac{1}{2} = \frac{1-\mathrm{e}}{2\ln 2}(2\ln 2) - \frac{1}{2} = 1 - \mathrm{e} - \frac{1}{2} = \frac{1}{2} - \mathrm{e}

\mathrm{e}^{-2y} = 2\mathrm{e} - 1

-2y = \ln(2\mathrm{e} - 1)

Answer

y = -\frac{1}{2}\ln(2\mathrm{e} - 1)

Final answer

y = -1/2 ln(2e - 1)

Detailed explanation

Walkthrough

We start with the differential equation $(x^2+1)\frac{dy}{dx}=kxe^{2y}$ . The first step is to separate the variables: put all $y$ terms with $dy$ and all $x$ terms with $dx$ . Dividing by $(x^2+1)e^{2y}$ gives

\mathrm{e}^{-2y}\,dy = \frac{kx}{x^2+1}\,dx.

Now integrate each side. The left side is $\int \mathrm{e}^{-2y}\,dy = -\frac{1}{2}\mathrm{e}^{-2y}$ . On the right, the numerator $kx$ is a constant multiple of the derivative of $x^2+1$ (since the derivative is $2x$ ), so $\int \frac{kx}{x^2+1}\,dx = \frac{k}{2}\ln(x^2+1)$ . Including the constant of integration gives

-\frac{1}{2}\mathrm{e}^{-2y} = \frac{k}{2}\ln(x^2+1)+c.

Use $y=0$ when $x=0$ . Since $\ln 1=0$ , this gives $-\frac{1}{2}=c$ , so $c=-\frac{1}{2}$ .

Now use $y=-\frac{1}{2}$ when $x=1$ . Substituting gives

-\frac{1}{2}\mathrm{e} = \frac{k}{2}\ln 2 - \frac{1}{2}.

Multiply by 2 and rearrange: $-\mathrm{e}=k\ln 2 - 1$ , so $k\ln 2 = 1-\mathrm{e}$ and $k=\frac{1-\mathrm{e}}{\ln 2}$ .

Finally substitute $x=\sqrt{3}$ . Since $x^2+1=4$ and $\ln 4=2\ln 2$ , the logarithm term becomes

\frac{1-\mathrm{e}}{2\ln 2}\cdot 2\ln 2 = 1-\mathrm{e}.

Thus

-\frac{1}{2}\mathrm{e}^{-2y} = 1-\mathrm{e} - \frac{1}{2} = \frac{1}{2}-\mathrm{e}.

Multiplying by $-2$ gives $\mathrm{e}^{-2y}=2\mathrm{e}-1$ , so $y=-\frac{1}{2}\ln(2\mathrm{e}-1)$ .

Key Takeaways

This question tests separation of variables for a first-order differential equation, integration of an exponential and of the form $kf'(x)/f(x)$ , and the use of two initial conditions to determine two constants. It also requires exact manipulation with logarithms, especially simplifying $\ln 4=2\ln 2$ .

Common Mistakes

Forgetting the constant of integration after integrating both sides. The mark scheme requires the constant $c$ to be included before using the initial conditions.
Sign error when integrating $\mathrm{e}^{-2y}$ : the integral is $-\frac{1}{2}\mathrm{e}^{-2y}$ , not $\frac{1}{2}\mathrm{e}^{-2y}$ .
Missing the factor $\frac{1}{2}$ on the exponential term or on the logarithm term.
Using only one of the two given conditions; both are needed, one to find $c$ and one to find $k$ .
Algebra error in the final substitution: remember $\ln 4=2\ln 2$ , so the logarithm term simplifies before solving for $y$ .
The mark scheme notes that when separating variables, missing integral signs or missing $dx$ , $dy$ is condoned, but not both missing.

Things to Be Careful About

The constant of integration may be written on either side; if written on the right, it must be carried through correctly.
When substituting $y=-\frac{1}{2}$ , the term $\mathrm{e}^{-2y}$ becomes $\mathrm{e}^{1}$ , not $\mathrm{e}^{-1}$ . This is a common sign mistake.
Since $x^2+1>0$ for all real $x$ , there is no domain restriction from the logarithm in this problem.
Keep the answer exact: do not replace $\mathrm{e}$ or $\ln(2\mathrm{e}-1)$ by a decimal approximation.
Check that $2\mathrm{e}-1>0$ , so the logarithm is defined.

Techniques used

separate variables in a first-order differential equationintegrate an exponential functionintegrate using the kf'(x)/f(x) formapply initial conditions to find constantsevaluate the particular solution at a given x

Similar questions

Q9MediumTrigonometryNumerical Solution of Equations

(a)

By sketching a suitable pair of graphs, show that the equation

\sec 2x = -e^x

has only one root in the interval $0 < x < \frac{1}{2}\pi$ .

2M Similar

DifficultyMedium

Worked solution

Approach

To show that $\sec 2x = -e^x$ has only one root in $0 < x < \frac{\pi}{2}$ , we sketch the graphs of $y = \sec 2x$ and $y = -e^x$ on the same axes and observe their intersection.

Working

Consider the function $y = \sec 2x = \frac{1}{\cos 2x}$ .

At $x = 0$ , $y = \sec 0 = 1$ .
As $x \to \frac{\pi}{4}^-$ , $\cos 2x \to 0^+$ , so $y \to +\infty$ .
As $x \to \frac{\pi}{4}^+$ , $\cos 2x \to 0^-$ , so $y \to -\infty$ .
At $x = \frac{\pi}{2}$ , $y = \sec \pi = -1$ .
There is a vertical asymptote at $x = \frac{\pi}{4}$ .

Consider the function $y = -e^x$ .

At $x = 0$ , $y = -1$ .
The function is strictly decreasing and negative for all $x > 0$ .
At $x = \frac{\pi}{4} \approx 0.785$ , $y = -e^{\pi/4} \approx -2.19$ .
At $x = \frac{\pi}{2} \approx 1.571$ , $y = -e^{\pi/2} \approx -4.81$ .

In the interval $0 < x < \frac{\pi}{4}$ :

$\sec 2x > 0$ and $-e^x < 0$ , so there is no intersection.

In the interval $\frac{\pi}{4} < x < \frac{\pi}{2}$ :

$\sec 2x$ increases from $-\infty$ to $-1$ .
$-e^x$ decreases from $-e^{\pi/4} \approx -2.19$ to $-e^{\pi/2} \approx -4.81$ .
Since $\sec 2x$ goes from $-\infty$ (below $-e^x$ ) to $-1$ (above $-e^x$ ), and both functions are continuous in this interval, they must intersect exactly once.

Thus, the equation has only one root in $0 < x < \frac{\pi}{2}$ .

Answer

The graphs intersect exactly once in the interval $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ , proving there is only one root.

Final answer

The graphs intersect exactly once in the interval (pi/4, pi/2).

Detailed explanation

Walkthrough

First, we analyze the behavior of $y = \sec 2x$ in the interval $0 < x < \frac{\pi}{2}$ . The function has a vertical asymptote at $x = \frac{\pi}{4}$ because $\cos(2 \cdot \frac{\pi}{4}) = \cos(\frac{\pi}{2}) = 0$ . For $0 < x < \frac{\pi}{4}$ , $\cos 2x$ is positive, so $\sec 2x > 0$ . For $\frac{\pi}{4} < x < \frac{\pi}{2}$ , $\cos 2x$ is negative, so $\sec 2x < 0$ . Specifically, as $x$ approaches $\frac{\pi}{4}$ from the right, $\sec 2x \to -\infty$ , and at $x = \frac{\pi}{2}$ , $\sec 2x = -1$ .

Next, we consider $y = -e^x$ . This is always negative and strictly decreasing. At $x = \frac{\pi}{4}$ , $y \approx -2.19$ . At $x = \frac{\pi}{2}$ , $y \approx -4.81$ .

Comparing the two graphs:

In $0 < x < \frac{\pi}{4}$ , $\sec 2x > 0$ while $-e^x < 0$ , so no intersection is possible.
In $\frac{\pi}{4} < x < \frac{\pi}{2}$ , $\sec 2x$ rises from $-\infty$ to $-1$ , while $-e^x$ falls from $-2.19$ to $-4.81$ . Since one starts below the other and ends above it, they must cross exactly once.

Key Takeaways

Sketching $\sec 2x$ requires identifying asymptotes at $x = \frac{\pi}{4} + \frac{n\pi}{2}$ and key points like $(0, 1)$ and $(\frac{\pi}{2}, -1)$ .
Graphical methods can prove the existence and uniqueness of roots by analyzing the monotonic behavior and bounds of the functions involved.

Common Mistakes

Forgetting the vertical asymptote at $x = \frac{\pi}{4}$ when sketching $y = \sec 2x$ .
Assuming the graphs intersect in $0 < x < \frac{\pi}{4}$ without checking the signs of the functions.
Not justifying why there is only one intersection (e.g., by noting the monotonic nature of both curves in the relevant interval).

Things to Be Careful About

Ensure the sketch clearly shows the asymptote and the correct branches of $\sec 2x$ .
The mark scheme also accepts sketching $y = \cos 2x$ and $y = -e^{-x}$ , which is obtained by rearranging $\sec 2x = -e^x$ to $\cos 2x = -e^{-x}$ . If using this alternative, note that $\cos 2x$ goes from $0$ to $-1$ and $-e^{-x}$ goes from $-e^{-\pi/4}$ to $-e^{-\pi/2}$ in the interval $(\frac{\pi}{4}, \frac{\pi}{2})$ .

Techniques used

sketch trigonometric functionsketch exponential functionidentify intersection points

(b)

Show by calculation that this root lies between 0.9 and 1.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Let $f(x) = \sec 2x + e^x$ . We evaluate $f(x)$ at $x = 0.9$ and $x = 1$ to show a sign change, indicating a root lies between these values.

Working

Calculate $f(0.9)$ :

\begin{aligned} f(0.9) &= \sec(1.8) + e^{0.9} \\ &= \frac{1}{\cos(1.8)} + e^{0.9} \\ &\approx \frac{1}{-0.2272} + 2.4596 \\ &\approx -4.4014 + 2.4596 \\ &\approx -1.9418 \end{aligned}

Calculate $f(1)$ :

\begin{aligned} f(1) &= \sec(2) + e^1 \\ &= \frac{1}{\cos(2)} + e \\ &\approx \frac{1}{-0.4161} + 2.7183 \\ &\approx -2.4033 + 2.7183 \\ &\approx 0.3150 \end{aligned}

Since $f(0.9) \approx -1.94 < 0$ and $f(1) \approx 0.315 > 0$ , there is a sign change. As $f(x)$ is continuous in this interval, the root lies between $0.9$ and $1$ .

Answer

The root lies between $0.9$ and $1$ .

Final answer

The root lies between 0.9 and 1.

Detailed explanation

Walkthrough

We define a function $f(x) = \sec 2x + e^x$ whose root corresponds to the solution of $\sec 2x = -e^x$ . To show the root is between 0.9 and 1, we calculate $f(0.9)$ and $f(1)$ .

At $x = 0.9$ : $\cos(1.8) \approx -0.2272$ , so $\sec(1.8) \approx -4.4014$ . Adding $e^{0.9} \approx 2.4596$ gives $f(0.9) \approx -1.9418$ .

At $x = 1$ : $\cos(2) \approx -0.4161$ , so $\sec(2) \approx -2.4033$ . Adding $e^1 \approx 2.7183$ gives $f(1) \approx 0.3150$ .

Because $f(0.9)$ is negative and $f(1)$ is positive, and the function is continuous between these points, the Intermediate Value Theorem guarantees a root exists in the interval $(0.9, 1)$ .

Key Takeaways

To locate a root between two values, evaluate a rearranged function $f(x) = 0$ at both endpoints.
A sign change in $f(x)$ confirms the presence of a root in the interval.

Common Mistakes

Using degrees instead of radians when calculating trigonometric values.
Forgetting to add the $e^x$ term or making sign errors when evaluating $\sec 2x + e^x$ .

Things to Be Careful About

Ensure your calculator is in radian mode.
The mark scheme also accepts evaluating the original expressions $\sec 2x$ and $-e^x$ separately and comparing them (e.g., $\sec(1.8) \approx -4.401 < -2.460 \approx -e^{0.9}$ and $\sec(2) \approx -2.403 > -2.718 \approx -e^1$ ).

Techniques used

evaluate function at endpointscheck for sign change

(c)

Show that if a sequence of values given by the iterative formula

x_{n+1} = \frac{1}{2}\cos^{-1}(-e^{-x_n})

converges, then it converges to the root of the equation in part (a).

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

If the sequence converges, let the limit be $\alpha$ . Then as $n \to \infty$ , $x_n \to \alpha$ and $x_{n+1} \to \alpha$ . Substitute $x = \alpha$ into the iterative formula and show it is equivalent to the original equation.

Working

Given the iterative formula:

x_{n+1} = \frac{1}{2}\cos^{-1}(-e^{-x_n})

If the sequence converges to a root $\alpha$ , we set $x_{n+1} = x_n = \alpha$ :

\alpha = \frac{1}{2}\cos^{-1}(-e^{-\alpha})

Rearrange to isolate the inverse cosine:

2\alpha = \cos^{-1}(-e^{-\alpha})

Take the cosine of both sides:

\cos(2\alpha) = -e^{-\alpha}

Multiply both sides by $e^\alpha$ :

e^\alpha \cos(2\alpha) = -1

Divide by $\cos(2\alpha)$ :

e^\alpha = \frac{-1}{\cos(2\alpha)}

Recognize that $\frac{1}{\cos(2\alpha)} = \sec(2\alpha)$ :

e^\alpha = -\sec(2\alpha)

Rearrange to obtain the original equation:

\sec(2\alpha) = -e^\alpha

Thus, if the sequence converges, it converges to the root of $\sec 2x = -e^x$ .

Answer

The iterative formula converges to the root of $\sec 2x = -e^x$ .

Final answer

The iterative formula converges to the root of sec 2x = -e^x.

Detailed explanation

Walkthrough

When an iterative sequence $x_{n+1} = F(x_n)$ converges, the limit $\alpha$ satisfies $\alpha = F(\alpha)$ . We substitute $x = \alpha$ into the given formula $x = \frac{1}{2}\cos^{-1}(-e^{-x})$ and rearrange to show it is equivalent to the original equation $\sec 2x = -e^x$ .

Multiplying by 2 gives $2x = \cos^{-1}(-e^{-x})$ . Taking the cosine of both sides yields $\cos 2x = -e^{-x}$ . Multiplying by $e^x$ gives $e^x \cos 2x = -1$ , which rearranges to $\frac{1}{\cos 2x} = -e^x$ , or $\sec 2x = -e^x$ . This proves the fixed point of the iteration is the root of the original equation.

Key Takeaways

The limit of a convergent iterative sequence is a fixed point of the iteration function.
Rearranging the fixed-point equation can verify it matches the original equation whose root is being found.

Common Mistakes

Forgetting to state that $x_{n+1} = x_n = \alpha$ at convergence.
Making algebraic errors when rearranging $\cos 2x = -e^{-x}$ back to $\sec 2x = -e^x$ .

Things to Be Careful About

The question only asks to show that IF it converges, it converges to the root. You do not need to prove convergence itself.
Ensure all algebraic steps are reversible or clearly stated.

Techniques used

rearrange equationsubstitute limit

(d)

Use the iterative formula given in part (c) to calculate $x$ correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

3M Similar

DifficultyMedium

Worked solution

Approach

Use the iterative formula $x_{n+1} = \frac{1}{2}\cos^{-1}(-e^{-x_n})$ starting with an initial value $x_1 = 1$ (or another value in the interval). Calculate successive values to 5 decimal places until the result is stable to 3 decimal places.

Working

Let $x_1 = 1$ .

Iteration 1:

x_2 = \frac{1}{2}\cos^{-1}(-e^{-1}) = \frac{1}{2}\cos^{-1}(-0.36788) \approx 0.97376

Iteration 2:

x_3 = \frac{1}{2}\cos^{-1}(-e^{-0.97376}) = \frac{1}{2}\cos^{-1}(-0.37768) \approx 0.97903

Iteration 3:

x_4 = \frac{1}{2}\cos^{-1}(-e^{-0.97903}) = \frac{1}{2}\cos^{-1}(-0.37568) \approx 0.97796

Iteration 4:

x_5 = \frac{1}{2}\cos^{-1}(-e^{-0.97796}) = \frac{1}{2}\cos^{-1}(-0.37608) \approx 0.97813

Iteration 5:

x_6 = \frac{1}{2}\cos^{-1}(-e^{-0.97813}) = \frac{1}{2}\cos^{-1}(-0.37602) \approx 0.97810

The values are stabilizing. To 3 decimal places:

$x_4 = 0.978$
$x_5 = 0.978$
$x_6 = 0.978$

Thus, the root is $0.978$ correct to 3 decimal places.

Answer

$x = 0.978$

Final answer

0.978

Detailed explanation

Walkthrough

We start with an initial guess $x_1 = 1$ (which is in the interval $(0.9, 1)$ identified in part (b)). We apply the formula $x_{n+1} = \frac{1}{2}\cos^{-1}(-e^{-x_n})$ repeatedly, keeping 5 decimal places for accuracy.

$x_1 = 1.00000$
$x_2 = 0.5 \times \cos^{-1}(-e^{-1}) = 0.5 \times \cos^{-1}(-0.36788) = 0.97376$
$x_3 = 0.5 \times \cos^{-1}(-e^{-0.97376}) = 0.5 \times \cos^{-1}(-0.37768) = 0.97903$
$x_4 = 0.5 \times \cos^{-1}(-e^{-0.97903}) = 0.5 \times \cos^{-1}(-0.37568) = 0.97796$
$x_5 = 0.5 \times \cos^{-1}(-e^{-0.97796}) = 0.5 \times \cos^{-1}(-0.37608) = 0.97813$
$x_6 = 0.5 \times \cos^{-1}(-e^{-0.97813}) = 0.5 \times \cos^{-1}(-0.37602) = 0.97810$

Since $x_5$ and $x_6$ both round to $0.978$ to 3 decimal places, the root is $0.978$ .

Key Takeaways

Iterative methods require sufficient iterations to ensure the result has stabilized to the required precision.
Always keep extra decimal places during intermediate steps to avoid rounding errors.

Common Mistakes

Rounding too early (e.g., to 3 decimal places) at each iteration, which can cause the sequence to oscillate or converge to the wrong value.
Using the wrong mode (degrees instead of radians) on the calculator.
Stopping the iteration too early before the value has stabilized.

Things to Be Careful About

The mark scheme requires showing sufficient iterations to 5 d.p. to justify the 3 d.p. answer. Showing a sign change in $(0.9775, 0.9785)$ is also acceptable.
Different starting values (e.g., $x_1 = 0.9$ or $x_1 = 0.95$ ) will produce different intermediate sequences but should converge to the same final answer.

Techniques used

apply iterative formulacheck convergence to required precision

Q10MediumAlgebra

Let $f(x) = \frac{x^3 + 2x - 11}{(3 + x)(2 + x^2)}$ .

(a)

Express $f(x)$ in partial fractions.

6M Similar

DifficultyMedium

Worked solution

Approach

The numerator and denominator have the same degree, so first perform polynomial long division to obtain a constant $A$ plus a proper fraction. Then express the proper fraction in the form $\frac{B}{3+x} + \frac{Cx + D}{2 + x^2}$ and solve for the constants.

Working

Step 1 — Long division.

\frac{x^3 + 2x - 11}{(3+x)(2+x^2)} = \frac{x^3 + 2x - 11}{x^3 + 3x^2 + 2x + 6}

Dividing $x^3 + 2x - 11$ by $x^3 + 3x^2 + 2x + 6$ :

\begin{aligned} x^3 + 2x - 11 &= 1 \cdot (x^3 + 3x^2 + 2x + 6) + (-3x^2 - 17) \\ &= (x^3 + 3x^2 + 2x + 6) + (-3x^2 - 17) \end{aligned}

Hence

f(x) = 1 + \frac{-3x^2 - 17}{(3+x)(2+x^2)} = A + \frac{B}{3+x} + \frac{Cx + D}{2+x^2}

Step 2 — Set up the identity.

-3x^2 - 17 = B(2 + x^2) + (Cx + D)(3 + x)

Step 3 — Find $B$ by substituting $x = -3$ .

-3(9) - 17 = B(2 + 9)

-44 = 11B \quad \Rightarrow \quad B = -4

Step 4 — Compare coefficients to find $C$ and $D$ .

Expanding the right side:

B(2+x^2) + (Cx+D)(3+x) = (B + C)x^2 + (3C + D)x + (2B + 3D)

Coefficient of $x^2$ : $\;B + C = -3 \;\Rightarrow\; C = -3 - (-4) = 1$ .

Coefficient of $x$ : $\;3C + D = 0 \;\Rightarrow\; D = -3C = -3(1) = -3$ .

Check the constant: $\;2B + 3D = 2(-4) + 3(-3) = -8 - 9 = -17$ ✓

So $A = 1$ , $B = -4$ , $C = 1$ , $D = -3$ .

Answer

f(x) = 1 - \frac{4}{3+x} + \frac{x - 3}{2 + x^2}

Final answer

f(x) = 1 - 4/(3+x) + (x-3)/(2+x^2)

Detailed explanation

Walkthrough

The denominator $(3+x)(2+x^2)$ has degree 3, the same as the numerator $x^3 + 2x - 11$ . Whenever the numerator and denominator have equal degree, we must do polynomial long division first to get a constant $A$ plus a proper fraction. Dividing gives the quotient $1$ and remainder $-3x^2 - 17$ , so

\frac{x^3 + 2x - 11}{(3+x)(2+x^2)} = 1 + \frac{-3x^2 - 17}{(3+x)(2+x^2)}.

The proper fraction has degree of numerator (2) less than degree of denominator (3), so it can be split into partial fractions. Because $2 + x^2$ is an irreducible quadratic, the second term must be linear in $x$ in its numerator: $\frac{Cx+D}{2+x^2}$ . The first term $\frac{B}{3+x}$ is the standard form for a distinct linear factor.

Clearing denominators gives the identity $-3x^2 - 17 = B(2+x^2) + (Cx+D)(3+x)$ . Setting $x = -3$ is the easiest way to find $B$ , because the $(Cx+D)$ factor vanishes, leaving $11B = -44$ , i.e. $B = -4$ .

With $B$ known, we expand the right side and compare coefficients of $x^2$ , $x$ and the constant with the left side:

$x^2$ : $B + C = -3 \Rightarrow C = 1$
$x$ : $3C + D = 0 \Rightarrow D = -3$
constant: $2B + 3D = -17$ — this is a useful check.

Key Takeaways

Always check degrees before attempting partial fractions: equal-degree case requires long division first.
When the denominator contains an irreducible quadratic, the corresponding partial fraction must have a linear numerator $Cx + D$ , not just a constant.
Substituting the root of a linear factor into the cleared identity is the fastest way to isolate one constant.
After finding constants, the constant-term check is a free way to confirm the work.

Common Mistakes

Trying to apply the partial fraction form directly without first doing long division (the $A$ term is needed because the fractions are not proper).
Writing the second term as $\frac{C}{2+x^2}$ instead of $\frac{Cx+D}{2+x^2}$ when the quadratic is irreducible.
Forgetting to expand the right side correctly and dropping a term when comparing coefficients.

Things to Be Careful About

The constant $A$ is not "free" — it is determined by the long division, and is $A = 1$ in this problem.
Sign errors when applying the substitution $x = -3$ (remember to substitute into the polynomial factors, not just leave the $(3+x)$ as $3+x$ ).
The form $A + \frac{B}{3+x} + \frac{Cx+D}{2+x^2}$ has four unknowns — make sure all four are obtained.

Techniques used

perform polynomial long division to extract the constant termset up partial fraction form with proper fractionuse substitution to find one partial fraction constantcompare coefficients to find remaining constants

(b)

Hence obtain the expansion of $f(x)$ in ascending powers of $x$ , up to and including the term in $x^2$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Rewrite each partial-fraction term so that its denominator is a constant times $(1 + u)$ with $u$ a multiple of $x$ or $x^2$ , then use the binomial expansion $(1+u)^{-1} = 1 - u + u^2 - \cdots$ up to the $x^2$ term. Finally multiply out the linear numerator $(Cx + D)$ in the second term and collect coefficients.

Working

Step 1 — Expand $\dfrac{4}{3+x}$ .

Factor the $3$ out of the denominator:

\frac{4}{3+x} = \frac{4}{3}\cdot\frac{1}{1 + x/3} = \frac{4}{3}\left(1 + \frac{x}{3}\right)^{-1}

Using the binomial expansion, up to the $x^2$ term:

\left(1 + \frac{x}{3}\right)^{-1} \approx 1 - \frac{x}{3} + \frac{x^2}{9}

Therefore

\frac{4}{3+x} \approx \frac{4}{3} - \frac{4x}{9} + \frac{4x^2}{27}

Step 2 — Expand $\dfrac{x-3}{2+x^2}$ .

Factor the $2$ out of the denominator:

\frac{x-3}{2+x^2} = \frac{x-3}{2}\cdot\frac{1}{1 + x^2/2} = \frac{x-3}{2}\left(1 + \frac{x^2}{2}\right)^{-1}

Using the binomial expansion, up to the $x^2$ term:

\left(1 + \frac{x^2}{2}\right)^{-1} \approx 1 - \frac{x^2}{2}

Multiplying out:

\frac{x-3}{2+x^2} \approx \frac{x-3}{2} - \frac{(x-3)x^2}{4} = \frac{x}{2} - \frac{3}{2} - \frac{x^3}{4} + \frac{3x^2}{4}

Keeping terms up to $x^2$ :

\frac{x-3}{2+x^2} \approx -\frac{3}{2} + \frac{x}{2} + \frac{3x^2}{4}

Step 3 — Combine the three pieces.

f(x) = 1 - \left(\frac{4}{3} - \frac{4x}{9} + \frac{4x^2}{27}\right) + \left(-\frac{3}{2} + \frac{x}{2} + \frac{3x^2}{4}\right)

Constant term:

1 - \frac{4}{3} - \frac{3}{2} = \frac{6 - 8 - 9}{6} = -\frac{11}{6}

Coefficient of $x$ :

\frac{4}{9} + \frac{1}{2} = \frac{8 + 9}{18} = \frac{17}{18}

Coefficient of $x^2$ :

-\frac{4}{27} + \frac{3}{4} = \frac{-16 + 81}{108} = \frac{65}{108}

Answer

f(x) = -\frac{11}{6} + \frac{17}{18}x + \frac{65}{108}x^2

Final answer

f(x) = -11/6 + (17/18)x + (65/108)x^2

Detailed explanation

Walkthrough

The idea is to write each denominator in the form $k(1+u)$ where $u$ is small when $|x|$ is small, then use the binomial series $(1+u)^{-1} = 1 - u + u^2 - u^3 + \cdots$ .

For the term $\frac{4}{3+x}$ , factor out $3$ to get $\frac{4}{3}(1 + x/3)^{-1}$ . The expansion up to $x^2$ uses $u = x/3$ :

(1 + x/3)^{-1} = 1 - \frac{x}{3} + \frac{x^2}{9} - \cdots

Multiplying by $\frac{4}{3}$ gives $\frac{4}{3} - \frac{4x}{9} + \frac{4x^2}{27}$ .

For the term $\frac{x-3}{2+x^2}$ , factor out $2$ to get $\frac{x-3}{2}(1 + x^2/2)^{-1}$ . Here $u = x^2/2$ , so the expansion up to $x^2$ is just $1 - x^2/2$ (the $u$ term is already an $x^2$ term, and there is no $x$ contribution). The key step is to multiply out the linear numerator $(x-3)$ by this expansion. The product

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

gives $-3 + x$ from the first part and $\frac{3x^2}{4} - \frac{x^3}{4}$ from the second; we only keep terms up to $x^2$ . After dividing by $2$ , we get

\frac{x-3}{2+x^2} \approx -\frac{3}{2} + \frac{x}{2} + \frac{3x^2}{4}.

Finally, the function $f(x) = 1 - \frac{4}{3+x} + \frac{x-3}{2+x^2}$ , so the constant $1$ adds to $-4/3 - 3/2 = -17/6$ , giving $-11/6$ . The $x$ coefficient is $+4/9 + 1/2 = 17/18$ (note both terms contribute with a positive sign because of the leading minus in front of the $4/(3+x)$ term). The $x^2$ coefficient is $-4/27 + 3/4 = 65/108$ .

A quick sanity check: setting $x = 0$ in the original $f$ gives $f(0) = -11/6$ , matching the constant in the expansion. ✓

Key Takeaways

Always rewrite a denominator as $k(1+u)$ with $u \to 0$ before applying the binomial expansion; never try to expand $(a + bx)^n$ directly without factoring.
For $(1 + x^2/2)^{-1}$ , there is no linear term in $x$ at all (only $1$ and $x^2$ ) — this is what makes the $x^2$ contribution come from the multiplication by $(x - 3)$ .
When collecting terms, pay attention to signs: the partial fraction decomposition has a minus in front of $\frac{4}{3+x}$ , so each of its contributions is subtracted.
A quick numerical check using $f(0)$ confirms the constant term.

Common Mistakes

Forgetting the leading $1$ (i.e. ignoring the constant $A$ from part (a)) and writing $f(x) = -\frac{4}{3+x} + \frac{x-3}{2+x^2}$ when combining.
Treating $1/(1+x^2/2)$ as having an $x$ term — it does not, because $u = x^2/2$ has no linear part.
Dropping the contribution from the $D$ part of the linear numerator when multiplying by the $(1+x^2/2)^{-1}$ expansion. In this problem $D = -3$ , which contributes an $x^2$ term: $-\frac{3}{2}\cdot\left(-\frac{x^2}{2}\right) = +\frac{3x^2}{4}$ .
Errors collecting the $x^2$ coefficient — there are two contributions: $-4/27$ from the first term and $+3/4$ from the second.

Things to Be Careful About

The validity of the expansion requires $|x/3| < 1$ and $|x^2/2| < 1$ , but for the algebraic manipulation this is not an issue — we are simply truncating a formal series.
Fractions should be combined over a common denominator; using a common denominator of $108$ for the $x^2$ coefficient avoids arithmetic slips.
The expansion of $(1+u)^{-1}$ up to $u^2$ is $1 - u + u^2$ ; with $u = x/3$ this is $1 - x/3 + x^2/9$ . The signs alternate, and this is easy to slip on.

Techniques used

write each fractional term in the form k*(1 + u)^-1apply binomial expansion up to x^2multiply out (Cx + D) by the expansion of the quadratic denominatorcollect coefficients of 1, x and x^2

Q11MediumVectors

With respect to the origin $O$ , the points $A, B, C$ and $D$ have position vectors given by

\vec{OA} = \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix}, \quad \vec{OB} = \begin{pmatrix} 0 \\ 4 \\ 1 \end{pmatrix}, \quad \vec{OC} = \begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix} \quad \text{and} \quad \vec{OD} = \begin{pmatrix} 3 \\ -5 \\ 4 \end{pmatrix}

The line $m$ passes through the points $A$ and $B$ .

(a)

Find a vector equation for $m$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The direction vector of the line is the displacement from one point to the other. Then write the equation in the form $\mathbf{r} = \mathbf{a} + t\mathbf{b}$ .

Working

The direction vector from $A$ to $B$ is

\overrightarrow{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 0 \\ 4 \\ 1 \end{pmatrix} - \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} = \begin{pmatrix} -1 \\ -1 \\ -2 \end{pmatrix}

Using $\vec{BA} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ as the direction vector and point $A$ , a vector equation for $m$ is

\mathbf{r} = \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}

Answer

\mathbf{r} = \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}

Final answer

r = (1, 5, 3) + λ(1, 1, 2)

Detailed explanation

Walkthrough

The vector equation of a line needs two pieces of information: a point on the line (given by its position vector) and a direction vector. Here we have two points on the line, $A$ and $B$ , so the direction vector is simply the displacement from one to the other: $\overrightarrow{AB}$ or $\overrightarrow{BA}$ . Either works — using $\overrightarrow{BA}$ just reverses the sign of the parameter. We choose $\overrightarrow{BA} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ and use $A$ 's position vector as the base point, giving $\mathbf{r} = \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ .

Key Takeaways

A line in 3D is written as $\mathbf{r} = \mathbf{a} + t\mathbf{b}$ , where $\mathbf{a}$ is a position vector of a point on the line and $\mathbf{b}$ is a direction vector.
The direction vector is any scalar multiple of the displacement between two points on the line.
The parameter $\lambda$ (or $t$ ) can be any real number.

Common Mistakes

Using a position vector as the direction vector (e.g., writing $\mathbf{r} = \lambda \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix}$ instead of adding it to a base point).
Forgetting the base point, writing only $\mathbf{r} = \lambda \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ .
Sign errors when computing $\overrightarrow{AB}$ .

Things to Be Careful About

Any scalar multiple of the direction vector is acceptable; the mark scheme accepts either $\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ or $\begin{pmatrix} -1 \\ -1 \\ -2 \end{pmatrix}$ .
The answer must be written in the form $\mathbf{r} = \dots$ — the mark scheme explicitly requires this.

Techniques used

compute direction vector between two pointswrite vector equation of a line

(b)

Find the position vector of the point of intersection of $m$ and the line passing through the points $C$ and $D$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Parametrise the line through $C$ and $D$ with a parameter $\mu$ , equate corresponding components of the two lines, solve for the parameters, then substitute back to find the intersection point.

Working

Line $m$ :

\mathbf{r} = \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 + \lambda \\ 5 + \lambda \\ 3 + 2\lambda \end{pmatrix}

Direction of line $CD$ :

\overrightarrow{CD} = \vec{OD} - \vec{OC} = \begin{pmatrix} 3 \\ -5 \\ 4 \end{pmatrix} - \begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ 3 \end{pmatrix}

Line through $C$ and $D$ :

\mathbf{r} = \begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ -2 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 + 2\mu \\ -3 - 2\mu \\ 1 + 3\mu \end{pmatrix}

Equate components:

1 + \lambda = 1 + 2\mu \quad \Rightarrow \quad \lambda = 2\mu

5 + \lambda = -3 - 2\mu \quad \Rightarrow \quad 5 + 2\mu = -3 - 2\mu \quad \Rightarrow \quad 4\mu = -8 \quad \Rightarrow \quad \mu = -2

Check third component:

3 + 2\lambda = 1 + 3\mu \quad \Rightarrow \quad 3 + 4\mu = 1 + 3\mu \quad \Rightarrow \quad \mu = -2 \quad \checkmark

So $\lambda = 2\mu = -4$ . The intersection point is on $m$ with $\lambda = -4$ :

\begin{pmatrix} 1 - 4 \\ 5 - 4 \\ 3 - 8 \end{pmatrix} = \begin{pmatrix} -3 \\ 1 \\ -5 \end{pmatrix}

Answer

\begin{pmatrix} -3 \\ 1 \\ -5 \end{pmatrix}

Final answer

(-3, 1, -5)

Detailed explanation

Walkthrough

To find where two lines intersect, we write both in parametric form and equate their components. The line through $C$ and $D$ has direction $\overrightarrow{CD} = \begin{pmatrix} 2 \\ -2 \\ 3 \end{pmatrix}$ , giving $\mathbf{r} = \begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ -2 \\ 3 \end{pmatrix}$ . Equating the first two components of this with line $m$ gives two equations in $\lambda$ and $\mu$ . Solving gives $\mu = -2$ and $\lambda = -4$ . We must verify the third component is consistent (it is), otherwise the lines would be skew. Substituting $\lambda = -4$ into $m$ (or $\mu = -2$ into the $CD$ line) gives the intersection point $\begin{pmatrix} -3 \\ 1 \\ -5 \end{pmatrix}$ .

Key Takeaways

Two lines in 3D intersect only if a consistent solution exists for all three component equations.
If two equations give a solution but the third does not match, the lines are skew.
The intersection point can be found by substituting the parameter back into either line.

Common Mistakes

Only equating two components and not checking the third — this can give a false intersection for skew lines.
Sign errors when writing the components of the $CD$ line (e.g., writing $-3 + 2\mu$ instead of $-3 - 2\mu$ ).
Confusing the parameters $\lambda$ and $\mu$ when substituting back.

Things to Be Careful About

The mark scheme notes that if the direction of $m$ is taken as $\begin{pmatrix} -1 \\ -1 \\ -2 \end{pmatrix}$ , then $\lambda = 4$ ; if the direction of $CD$ is $\begin{pmatrix} -2 \\ 2 \\ -3 \end{pmatrix}$ , then $\mu = 2$ . The final point is the same regardless.
The answer can be given as coordinates $(-3, 1, -5)$ or as a position vector.

Techniques used

parametrise both linesequate corresponding componentssolve simultaneous equations for parameterssubstitute parameter to find intersection point

(c)

Find the position vector of the foot of the perpendicular from $C$ to $m$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Let $P$ be a general point on $m$ . Form the vector $\overrightarrow{CP}$ . For the foot of the perpendicular, $\overrightarrow{CP}$ is perpendicular to the direction of $m$ , so their scalar product is zero. Solve for $\lambda$ and substitute.

Working

A general point $P$ on $m$ :

P = \begin{pmatrix} 1 + \lambda \\ 5 + \lambda \\ 3 + 2\lambda \end{pmatrix}

Vector from $C$ to $P$ :

\overrightarrow{CP} = P - C = \begin{pmatrix} 1 + \lambda - 1 \\ 5 + \lambda - (-3) \\ 3 + 2\lambda - 1 \end{pmatrix} = \begin{pmatrix} \lambda \\ 8 + \lambda \\ 2 + 2\lambda \end{pmatrix}

Perpendicular condition with direction $\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ :

\overrightarrow{CP} \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = \lambda(1) + (8 + \lambda)(1) + (2 + 2\lambda)(2) = \lambda + 8 + \lambda + 4 + 4\lambda = 6\lambda + 12 = 0

So $\lambda = -2$ . The foot of the perpendicular is

P = \begin{pmatrix} 1 - 2 \\ 5 - 2 \\ 3 - 4 \end{pmatrix} = \begin{pmatrix} -1 \\ 3 \\ -1 \end{pmatrix}

Answer

\begin{pmatrix} -1 \\ 3 \\ -1 \end{pmatrix}

Final answer

(-1, 3, -1)

Detailed explanation

Walkthrough

The foot of the perpendicular from $C$ to $m$ is the point $P$ on $m$ such that $CP$ is perpendicular to $m$ . We write a general point on $m$ as $P = \begin{pmatrix} 1 + \lambda \\ 5 + \lambda \\ 3 + 2\lambda \end{pmatrix}$ . Then $\overrightarrow{CP} = P - C = \begin{pmatrix} \lambda \\ 8 + \lambda \\ 2 + 2\lambda \end{pmatrix}$ . Perpendicularity means the scalar product of $\overrightarrow{CP}$ with the direction vector of $m$ , $\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ , is zero. Computing the scalar product gives $6\lambda + 12 = 0$ , so $\lambda = -2$ . Substituting back gives the foot of the perpendicular $\begin{pmatrix} -1 \\ 3 \\ -1 \end{pmatrix}$ .

Key Takeaways

The foot of the perpendicular is found by imposing the condition that the connecting vector is perpendicular to the line.
Perpendicularity in 3D is expressed through the scalar (dot) product being zero.
The scalar product of $\begin{pmatrix} a \\ b \\ c \end{pmatrix}$ and $\begin{pmatrix} d \\ e \\ f \end{pmatrix}$ is $ad + be + cf$ .

Common Mistakes

Using $\overrightarrow{PC}$ instead of $\overrightarrow{CP}$ — the sign flips but the scalar product being zero is unaffected, so either works, but be consistent.
Forgetting to subtract $C$ 's coordinates correctly (e.g., writing $5 - \lambda$ instead of $5 + \lambda - (-3) = 8 + \lambda$ ).
Arithmetic errors in expanding the scalar product.

Things to Be Careful About

The mark scheme accepts alternative forms of $\overrightarrow{CP}$ , e.g. $\begin{pmatrix} -1 \\ 7 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ .
The answer can be given as coordinates $(-1, 3, -1)$ or as a position vector.

Techniques used

form general point on the lineconstruct vector from C to general pointset scalar product with direction vector to zerosolve for parameter and substitute

Mathematics 9709/31 — October/November 2025

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