(a) It is given that $f(x) = (x - a)^2 g(x)$, where $f(x)$ and $g(x)$ are polynomials. Show that $(x - a)$ is a factor of $f'(x)$. (b) It is given that $(x - 3)^2$ is a factor of $2x^3 - 4x^2 + px + q$, where $p$ and $q$ are constants. Find the values of $p$ and $q$.

(a) (x - a) is a factor of f'(x) · (b) p = -30, q = 72

(a) By sketching a suitable pair of graphs, show that the equation $\cot 2x = 2\sin 2x - 1$ has exactly one root in the interval $0 < x < \frac{1}{2}\pi$. (b) Show by calculation that the root is in the interval $0.4 < x < 0.6$. (c) Use the iterative formula $x_{n+1} = \frac{1}{2} \tan^{-1} \left( \frac{1}{2\sin 2x_n - 1} \right)$ to calculate the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

(a) The curves y = cot(2x) and y = 2sin(2x) - 1 intersect at exactly one point in 0 0 and f(0.6) ≈ -0.475 < 0, so there is a sign change and a root in 0.4 < x < 0.6. · (c) 0.49

The equation of a curve is $2y^3 - 3x^2y - x^3 = 16$. (a) Show that $$ \frac{dy}{dx} = \frac{x^2 + 2xy}{2y^2 - x^2}. $$ (b) Hence find the coordinates of the points on the curve at which the normal is parallel to the $y$-axis.

(a) dy/dx = (x^2 + 2xy)/(2y^2 - x^2) · (b) (0, 2) and (4, -2)

(a) Prove the identity $\sin 4x \equiv 4\sin x(2\cos^3 x - \cos x)$. (b) Hence find the exact value of $\int_0^{\frac{1}{4}\pi} \cos^3 x \sin 4x \, dx$.

(a) sin 4x = 4 sin x(2 cos^3 x - cos x) · (b) (sqrt(2) + 12)/35

Let $f(x) = \frac{x^2 + 4ax + 6a^2}{(x + 2a)(x + 3a)}$, where $a$ is a positive constant. (a) Express $f(x)$ in partial fractions. (b) Hence find the exact value of $\int_{-a}^a f(x) \, dx$. Give your answer in the form $a(p + \ln q)$, where $p$ and $q$ are rational.

(a) f(x) = 1 + 2a/(x + 2a) - 3a/(x + 3a) · (b) a(2 + ln(9/8))

9709/32 Oct/Nov 2025 — Mathematics Solved Past Paper

Q1MediumAlgebra

(a)

Sketch the graph of $y = |x + 3a|$ , where $a$ is a positive constant.

1M Similar

DifficultyEasy

Worked solution

Approach

The graph of $y = |x + 3a|$ is a V-shaped graph with its vertex on the x-axis where the expression inside the modulus is zero.

Working

The vertex occurs where $x + 3a = 0$ , so $x = -3a$ . The vertex is at $(-3a, 0)$ .

The y-intercept is found by setting $x = 0$ :

y = |0 + 3a| = 3a

The y-intercept is $(0, 3a)$ .

For $x \geq -3a$ , $y = x + 3a$ , which is a straight line with gradient 1.
For $x < -3a$ , $y = -(x + 3a) = -x - 3a$ , which is a straight line with gradient -1.

The graph is symmetrical about the line $x = -3a$ and consists of two solid line segments above the x-axis.

Answer

The graph is a V-shape with vertex at $(-3a, 0)$ and y-intercept at $(0, 3a)$ .

Final answer

V-shaped graph with vertex at (-3a, 0) and y-intercept at (0, 3a)

Detailed explanation

Walkthrough

First, identify the vertex of the modulus function $y = |x + 3a|$ . The vertex occurs where the expression inside the modulus equals zero, which is $x + 3a = 0$ , giving $x = -3a$ . Since the modulus function is always non-negative, the vertex is at $(-3a, 0)$ on the x-axis.

Next, find the y-intercept by substituting $x = 0$ : $y = |3a| = 3a$ (since $a$ is positive). This gives the point $(0, 3a)$ .

The graph consists of two straight line segments meeting at the vertex. For $x \geq -3a$ , the expression inside is positive, so $y = x + 3a$ (gradient 1). For $x < -3a$ , the expression is negative, so $y = -x - 3a$ (gradient -1). The graph is entirely above or on the x-axis, forming a V-shape.

Key Takeaways

The graph of $y = |x + c|$ is a V-shape with its vertex on the x-axis at $x = -c$ .
The y-intercept is found by evaluating the modulus at $x = 0$ .
The two arms of the V have gradients of $1$ and $-1$ .

Common Mistakes

Drawing the graph below the x-axis (modulus is always $\geq 0$ ).
Failing to mark the vertex at $-3a$ and the y-intercept at $3a$ .
Drawing curved lines instead of straight line segments.

Things to Be Careful About

Ensure the graph is symmetrical about the vertical line $x = -3a$ .
Use solid lines for the graph; ignore any dotted lines below the axis.
The constant $a$ is positive, so $-3a$ is on the negative x-axis and $3a$ is on the positive y-axis.

Techniques used

sketch modulus functionidentify vertex and y-intercept

(b)

Hence or otherwise solve the inequality $|x + 3a| > a - 2x$ .

2M Similar

DifficultyMedium

Worked solution

Approach

To solve $|x + 3a| > a - 2x$ , we find the critical value where the two expressions are equal, then use the graph from part (a) or algebraic reasoning to determine the region where the inequality holds. We must be careful to reject any extraneous solutions.

Working

Method 1: Using the graph
From part (a), the graph of $y = |x + 3a|$ is a V-shape with vertex at $(-3a, 0)$ and y-intercept at $(0, 3a)$ .
We want to find where this graph lies above the line $y = a - 2x$ .

The critical value occurs where the two expressions are equal:

x + 3a = a - 2x

3x = -2a

x = -\frac{2}{3}a

At this value, $y = -\frac{2}{3}a + 3a = \frac{7}{3}a$ .

Since $a > 0$ , the line $y = a - 2x$ has a negative gradient and intersects the right arm of the V-shape at $x = -\frac{2}{3}a$ .

We test a value in the region $x > -\frac{2}{3}a$ , for example $x = 0$ :
LHS: $|0 + 3a| = 3a$
RHS: $a - 2(0) = a$
Since $a > 0$ , $3a > a$ , so $x = 0$ satisfies the inequality.

Thus, the solution is $x > -\frac{2}{3}a$ .

Method 2: Algebraic (squaring both sides)
Note that squaring is only directly valid when both sides are non-negative, but we can solve $(x + 3a)^2 = (a - 2x)^2$ to find critical values:

x^2 + 6ax + 9a^2 = a^2 - 4ax + 4x^2

3x^2 - 10ax - 8a^2 = 0

(3x + 2a)(x - 4a) = 0

This gives $x = -\frac{2}{3}a$ or $x = 4a$ .

However, $x = 4a$ is an extraneous solution introduced by squaring. We must check the regions:

For $x > -\frac{2}{3}a$ , the inequality $|x + 3a| > a - 2x$ holds.
For $x \leq -\frac{2}{3}a$ , the inequality does not hold.

The clear conclusion is $x > -\frac{2}{3}a$ .

Answer

x > -\frac{2}{3}a

Final answer

x > -2/3 a

Detailed explanation

Walkthrough

To solve the inequality $|x + 3a| > a - 2x$ , we first find the critical value where the two sides are equal. Setting $x + 3a = a - 2x$ gives $3x = -2a$ , so $x = -\frac{2}{3}a$ .

This critical value divides the number line into two regions: $x > -\frac{2}{3}a$ and $x < -\frac{2}{3}a$ . We test a value in each region to see where the inequality holds.

Testing $x = 0$ (which is $> -\frac{2}{3}a$ since $a > 0$ ):
LHS: $|0 + 3a| = 3a$
RHS: $a - 0 = a$
Since $3a > a$ , the inequality is satisfied for $x > -\frac{2}{3}a$ .

Testing $x = -3a$ (which is $< -\frac{2}{3}a$ ):
LHS: $|-3a + 3a| = 0$
RHS: $a - 2(-3a) = 7a$
Since $0 < 7a$ , the inequality is not satisfied.

Thus, the solution is $x > -\frac{2}{3}a$ .

If using the squaring method, $(x + 3a)^2 = (a - 2x)^2$ gives $x = -\frac{2}{3}a$ or $x = 4a$ . The value $x = 4a$ is extraneous because squaring can introduce false solutions when one side is negative. The correct solution must be stated as $x > -\frac{2}{3}a$ with a clear implication that $x = 4a$ is rejected.

Key Takeaways

To solve modulus inequalities, find the critical value where the expressions are equal.
Test regions to determine where the inequality holds.
Be careful with algebraic methods like squaring, which can introduce extraneous solutions.

Common Mistakes

Forgetting to reject extraneous solutions when squaring both sides.
Using $\geq$ instead of $>$ in the final answer (the original inequality is strict).
Failing to test regions to confirm the solution.

Things to Be Careful About

The constant $a$ is positive, so $-\frac{2}{3}a$ is negative.
Ensure the final answer uses strict inequality $>$ as in the original question.
If using the squaring method, explicitly reject $x = 4a$ and justify why.

Techniques used

solve modulus inequality using critical valuecheck solution regions

Q24MMediumLogarithmic and Exponential Functions

Solve the equation $3 \times 2^{x+1} = 4 \times 3^{2x-3}$ . Give your answer correct to 3 significant figures.

DifficultyMedium

Worked solution

Approach

Take natural logarithms of both sides to convert the exponential equation into a linear equation in $x$ . Then solve the linear equation and collect the result.

Working

Take natural logarithms of both sides:

\ln\left(3 \times 2^{x+1}\right) = \ln\left(4 \times 3^{2x-3}\right)

Use $\ln(ab) = \ln a + \ln b$ and $\ln(a^k) = k\ln a$ :

\ln 3 + (x+1)\ln 2 = \ln 4 + (2x-3)\ln 3

Expand the brackets:

\ln 3 + x\ln 2 + \ln 2 = \ln 4 + 2x\ln 3 - 3\ln 3

Collect the $x$ -terms on one side:

x\ln 2 - 2x\ln 3 = \ln 4 - \ln 2 - 4\ln 3

Factor out $x$ and combine using logarithm laws:

x(\ln 2 - 2\ln 3) = \ln\left(\frac{2}{81}\right)

Since $\ln 2 - 2\ln 3 = \ln\left(\frac{2}{9}\right)$ , divide:

x = \frac{\ln\left(\frac{2}{81}\right)}{\ln\left(\frac{2}{9}\right)}

Evaluate:

x \approx 2.46

Answer

x = 2.46 \quad (3 \text{ s.f.})

Final answer

x = 2.46

Detailed explanation

Walkthrough

Since the variable is in the exponents, we cannot solve the equation with ordinary algebra. Taking logarithms of both sides is the standard technique because it turns powers into multipliers.

Both sides of the equation are positive, so taking natural logarithms is valid. Apply two laws:

$\ln(ab) = \ln a + \ln b$ for the products $3 \times 2^{x+1}$ and $4 \times 3^{2x-3}$ .
$\ln(a^k) = k\ln a$ for the powers $2^{x+1}$ and $3^{2x-3}$ .

This gives a linear equation in $x$ :

\ln 3 + (x+1)\ln 2 = \ln 4 + (2x-3)\ln 3

Next, expand the brackets so that all terms involving $x$ are explicit. Then move all $x$ -terms to one side and all constant terms to the other, as in a normal linear equation. Factor out $x$ and divide by its coefficient.

The coefficient simplifies to $\ln 2 - 2\ln 3 = \ln\left(\frac{2}{9}\right)$ , and the constant side simplifies to $\ln\left(\frac{2}{81}\right)$ . Therefore

x = \frac{\ln\left(\frac{2}{81}\right)}{\ln\left(\frac{2}{9}\right)}

Finally, evaluate this quotient with a calculator and round to 3 significant figures, obtaining $2.46$ .

Key Takeaways

When the unknown appears in an exponent, taking logarithms is the standard strategy.
The laws $\ln(ab) = \ln a + \ln b$ and $\ln(a^k) = k\ln a$ convert an exponential equation into a linear equation.
Careful collection of terms is required before solving. The final answer must be rounded to the requested degree of accuracy.

Common Mistakes

Incorrectly combining $3 \times 2^{x+1}$ as $6^{x+1}$ . This is invalid because the bases are different; the mark scheme explicitly says it is not allowed.
Forgetting to apply the power law to both $2^{x+1}$ and $3^{2x-3}$ .
Making sign errors when moving terms across the equation.
Leaving the answer as an unsimplified logarithm expression when the question asks for a value to 3 significant figures.

Things to Be Careful About

Both sides of the equation are positive, so taking natural logarithms is valid.
When simplifying, note that $\ln 2 - 2\ln 3 = \ln 2 - \ln 9 = \ln\left(\frac{2}{9}\right)$ .
The final answer must be $2.46$ , not an unrounded value such as $2.459...$ .

Techniques used

take logarithms of both sidesapply product and power laws of logarithmscollect terms to isolate xevaluate a logarithmic quotient

Approach

P is the lower intersection of the line $\text{Re } z = 2$ with the circle $|z - (3 + i)| = 2$ . Substitute $x = 2$ into the circle equation, solve for $y$ , and take the smaller root.

Working

The circle $|z - (3+i)| = 2$ corresponds to the equation

(x - 3)^2 + (y - 1)^2 = 4.

Set $x = 2$ (since $\text{Re } z = 2$ ):

(2 - 3)^2 + (y - 1)^2 = 4

1 + (y - 1)^2 = 4

(y - 1)^2 = 3

y - 1 = \pm\sqrt{3}.

The two intersection points have $y = 1 + \sqrt{3}$ (upper) and $y = 1 - \sqrt{3}$ (lower). From the diagram, $P$ is the lower intersection, so $y = 1 - \sqrt{3}$ .

Answer

P = 2 + i(1 - \sqrt{3})

Final answer

2 + i(1 − √3)

Detailed explanation

Walkthrough

The shaded region is bounded by the vertical line $\text{Re } z = 2$ and the circle $|z - (3 + i)| = 2$ (centre $(3, 1)$ , radius $2$ ). The point $P$ is the lower intersection of this line and the circle, so it lies on both loci.

The circle equation $(x-3)^2 + (y-1)^2 = 4$ is the expanded form of $|z - (3+i)| = 2$ . The line is just $x = 2$ , so we substitute $x = 2$ into the circle equation. This gives $(2-3)^2 + (y-1)^2 = 4$ , i.e. $(y-1)^2 = 3$ , so $y = 1 \pm \sqrt{3}$ . The two intersection points are $(2, 1+\sqrt{3})$ (upper) and $(2, 1-\sqrt{3})$ (lower). From the diagram, $P$ is the lower intersection, so $y = 1 - \sqrt{3}$ and $P = 2 + i(1 - \sqrt{3})$ .

Key Takeaways

The locus $|z - a| = r$ in the Argand diagram is a circle of centre $a$ and radius $r$ .
The line $\text{Re } z = k$ is the vertical line $x = k$ .
Intersections of a vertical line with a circle are found by substituting $x = k$ into the circle equation.

Common Mistakes

Choosing the upper intersection $2 + i(1 + \sqrt{3})$ instead of the lower one.
Forgetting to write the final answer in the form $x + iy$ (e.g. giving the coordinates $(2, 1-\sqrt{3})$ instead).

Things to Be Careful About

Make sure the answer is written as a single complex number, not as a coordinate pair.
Both intersection points satisfy the equation; read the diagram carefully to pick the correct one.

Techniques used

substitute the line equation x=2 into the circle equationsolve the resulting quadratic in yselect the lower intersection point from the diagram

(b)

Find the greatest value of $\arg z$ for points in the shaded region.

3M Similar

DifficultyMedium

Worked solution

Approach

The maximum $\arg z$ in the shaded region occurs at the point on the circle where a ray from the origin is tangent to it. The required angle is the angle from the positive real axis to this tangent, which equals the angle from the positive real axis to the line $OC$ (where $C = 3 + i$ is the centre) plus the half-angle between $OC$ and the tangent (computed from the right triangle with hypotenuse $|OC|$ and opposite side equal to the radius).

Working

The centre of the circle is $C = 3 + i$ , with modulus

|OC| = \sqrt{3^2 + 1^2} = \sqrt{10}.

Angle from the positive real axis to the line $OC$ :

\theta_1 = \tan^{-1}\left(\frac{1}{3}\right).

Half-angle of the tangent at the origin (right triangle with hypotenuse $|OC| = \sqrt{10}$ and side opposite the angle at $O$ equal to the radius $2$ ):

\theta_2 = \sin^{-1}\left(\frac{2}{\sqrt{10}}\right).

The maximum $\arg z$ in the shaded region is the sum of these two angles (the upper tangent gives the larger argument):

\arg z_{\max} = \tan^{-1}\left(\frac{1}{3}\right) + \sin^{-1}\left(\frac{2}{\sqrt{10}}\right).

Evaluating numerically:

\tan^{-1}\left(\frac{1}{3}\right) \approx 0.32175 \text{ rad}, \qquad \sin^{-1}\left(\frac{2}{\sqrt{10}}\right) \approx 0.68472 \text{ rad}.

\arg z_{\max} \approx 1.00647 \text{ rad} \approx 57.7^\circ.

(Sanity check: the top of the line $x = 2$ inside the circle is at $2 + i(1 + \sqrt{3})$ , with $\arg z = \tan^{-1}\!\left(\frac{1+\sqrt{3}}{2}\right) \approx 0.939$ rad, which is less than $1.01$ rad, so the maximum is indeed on the circle, not on the line.)

Answer

\arg z_{\max} \approx 1.01 \text{ rad} \; (\approx 57.7^\circ)

Final answer

1.01 radians (≈ 57.7°)

Detailed explanation

Walkthrough

The shaded region is the part of the disc $|z - (3+i)| \le 2$ with $\text{Re } z \le 2$ . The argument of $z$ is the angle that the line from the origin $O$ to $z$ makes with the positive real axis. As we rotate counterclockwise from the positive real axis, the ray first hits a boundary point; the largest such angle is the argument of the upper tangent line from $O$ to the circle.

The tangent point is the foot of the perpendicular from the centre $C = 3 + i$ to the tangent line. The triangle formed by $O$ , the tangent point and $C$ is right-angled at the tangent point, with hypotenuse $|OC| = \sqrt{3^2 + 1^2} = \sqrt{10}$ and the side opposite the angle at $O$ being the radius $2$ . If $\theta_2$ is the angle at $O$ in this triangle, $\sin \theta_2 = 2/\sqrt{10}$ , giving $\theta_2 = \sin^{-1}(2/\sqrt{10})$ .

The line $OC$ itself makes angle $\theta_1 = \tan^{-1}(1/3)$ with the positive real axis. The upper tangent is rotated $\theta_2$ further anticlockwise from $OC$ , so the maximum argument in the shaded region is

\arg z_{\max} = \theta_1 + \theta_2 = \tan^{-1}\left(\frac{1}{3}\right) + \sin^{-1}\left(\frac{2}{\sqrt{10}}\right).

This evaluates to about $0.3217 + 0.6847 \approx 1.0065$ rad, or about $57.7°$ .

Check: the top of the line segment on $x = 2$ inside the circle is at $2 + i(1 + \sqrt{3})$ , whose argument is $\tan^{-1}\!\left(\frac{1+\sqrt{3}}{2}\right) \approx 0.939$ rad, which is smaller, so the maximum is on the circle.

Key Takeaways

For a closed region, the maximum $\arg z$ is attained on its boundary.
For a circle not containing the origin, the largest $\arg z$ on the circle corresponds to the upper tangent from the origin.
The geometry of a tangent from an external point can be exploited using a right triangle whose hypotenuse is the distance from the origin to the centre and one leg is the radius.
The maximum $\arg z$ over the whole shaded region is the larger of the two candidate maxima (on the circle vs. on the line $x = 2$ ); here the circle wins.

Common Mistakes

Forgetting to include the angle $\theta_1 = \tan^{-1}(1/3)$ that accounts for the offset of the centre from the real axis; using only $\sin^{-1}(2/\sqrt{10})$ underestimates the maximum.
Using the lower tangent (giving a negative or smaller angle).
Confusing $\tan^{-1}(1/3)$ with $\tan^{-1}(3)$ .
Failing to verify that the tangent point lies inside the shaded region (it does, with $x \approx 1.31 < 2$ ).
Quoting the answer in degrees without the radian value (or vice versa).

Things to Be Careful About

The maximum $\arg z$ is taken over the shaded region, not the entire circle; the boundary line $x = 2$ must be considered, although here the circle gives a larger angle.
$\arg z$ is typically taken in $(-\pi, \pi]$ , so we look at the upper (positive) half of the figure.
The numerical value is $\approx 1.01$ rad; the mark scheme accepts $1.01$ rad or $57.7°$ (AWRT).

Techniques used

identify that the maximum argument occurs at the upper tangent from the origin to the circlecompute the angle from the positive real axis to the line from the origin to the centrecompute the half-angle of the tangent using the right triangle (radius / hypotenuse)add the two angles to obtain the maximum argument

Q46MMediumIntegration

Find the exact value of $\int_0^1 x \tan^{-1} x \, dx$ .

DifficultyMedium

Worked solution

Approach

Use integration by parts with $u = \tan^{-1} x$ and $\frac{dv}{dx} = x$ . Then split the resulting rational integrand and evaluate the definite integral.

Working

Let

u = \tan^{-1} x, \quad \frac{dv}{dx} = x.

Then

\frac{du}{dx} = \frac{1}{1+x^2}, \quad v = \frac{x^2}{2}.

Integration by parts gives

\int x \tan^{-1} x \, dx = \frac{x^2}{2}\tan^{-1} x - \int \frac{x^2}{2} \cdot \frac{1}{1+x^2} \, dx

= \frac{x^2}{2}\tan^{-1} x - \frac{1}{2}\int \frac{x^2}{1+x^2} \, dx.

Now split the fraction:

\frac{x^2}{1+x^2} = 1 - \frac{1}{1+x^2}.

Therefore

\int \frac{x^2}{1+x^2} \, dx = \int \left(1 - \frac{1}{1+x^2}\right) dx = x - \tan^{-1} x.

Hence an antiderivative is

F(x) = \frac{x^2}{2}\tan^{-1} x - \frac{1}{2}\left(x - \tan^{-1} x\right).

Evaluate from $0$ to $1$ :

\int_0^1 x \tan^{-1} x \, dx = \left[\frac{x^2}{2}\tan^{-1} x - \frac{1}{2}\left(x - \tan^{-1} x\right)\right]_0^1.

At $x = 1$ :

F(1) = \frac{1}{2} \cdot \frac{\pi}{4} - \frac{1}{2}\left(1 - \frac{\pi}{4}\right) = \frac{\pi}{8} - \frac{1}{2} + \frac{\pi}{8} = \frac{\pi}{4} - \frac{1}{2}.

At $x = 0$ :

F(0) = 0 - \frac{1}{2}(0 - 0) = 0.

Therefore

\int_0^1 x \tan^{-1} x \, dx = \frac{\pi}{4} - \frac{1}{2}.

Answer

\frac{\pi}{4} - \frac{1}{2}

Final answer

π/4 - 1/2

Detailed explanation

Walkthrough

We need to integrate a product $x \tan^{-1} x$ . Since $\tan^{-1} x$ is not a simple derivative, integration by parts is the natural method. Choose $u = \tan^{-1} x$ so that its derivative $\frac{1}{1+x^2}$ simplifies the remaining integral, and choose $\frac{dv}{dx} = x$ because its integral $\frac{x^2}{2}$ is easy. This gives

\int x \tan^{-1} x \, dx = \frac{x^2}{2}\tan^{-1} x - \frac{1}{2}\int \frac{x^2}{1+x^2} \, dx.

The new integral is not immediately standard, but we can rewrite the fraction:

\frac{x^2}{1+x^2} = 1 - \frac{1}{1+x^2}.

This splits it into a polynomial term and the standard derivative of $\tan^{-1} x$ . Integrating gives $x - \tan^{-1} x$ . Substituting back gives the antiderivative. Finally, evaluate at the limits $x=1$ and $x=0$ , using $\tan^{-1} 1 = \frac{\pi}{4}$ and $\tan^{-1} 0 = 0$ . The lower limit contributes zero, so the exact value is $\frac{\pi}{4} - \frac{1}{2}$ .

Key Takeaways

This question tests integration by parts, recognising the derivative of inverse tangent, rewriting a rational function, and evaluating definite integrals. A student should understand how to choose $u$ and $\frac{dv}{dx}$ so the new integral is simpler, and how to split $\frac{x^2}{1+x^2}$ into $1 - \frac{1}{1+x^2}$ .

Common Mistakes

Choosing $u = x$ and $\frac{dv}{dx} = \tan^{-1} x$ leads to a harder integral, since integrating $\tan^{-1} x$ is possible but less direct.
Forgetting the factor $\frac{1}{2}$ after integration by parts.
Incorrectly simplifying $\frac{x^2}{1+x^2}$ ; it is not $x^2$ or $1$ , but $1 - \frac{1}{1+x^2}$ .
Using degrees instead of radians when evaluating $\tan^{-1} 1$ ; the mark scheme requires radians.
Forgetting to subtract the lower-limit value, or substituting $x=0$ incorrectly.

Things to Be Careful About

The integration by parts formula has a minus sign; it is condoned in the mark scheme but must be handled correctly.
The lower limit at $x=0$ : $x^2 \tan^{-1} x = 0$ and $x - \tan^{-1} x = 0$ , so it contributes $0$ , but evidence of substitution at least once is needed.
Use exact values: $\tan^{-1} 1 = \frac{\pi}{4}$ , not a decimal.
The final answer must be exact; decimals are not accepted.

Techniques used

apply integration by partsdifferentiate inverse tangentsplit rational integrand into polynomial and standard formevaluate definite integral using exact values

Approach

Use the product rule to differentiate $f(x) = (x-a)^2g(x)$ , then factor out $(x-a)$ and apply the factor theorem.

Working

Let $u = (x-a)^2$ and $v = g(x)$ . Then

u' = 2(x-a), \quad v' = g'(x)

By the product rule,

f'(x) = u'v + uv' = 2(x-a)g(x) + (x-a)^2g'(x)

Factor out the common factor $(x-a)$ :

f'(x) = (x-a)(2g(x) + (x-a)g'(x))

This has the form $(x-a)h(x)$ , where $h(x) = 2g(x) + (x-a)g'(x)$ .

Answer

Since $f'(x) = (x-a)h(x)$ , $(x-a)$ is a factor of $f'(x)$ .

Final answer

(x - a) is a factor of f'(x)

Detailed explanation

Walkthrough

We need to show that the derivative contains $(x-a)$ as a factor. Since $f(x)$ is written as the product of two functions, the product rule is the natural tool. Let $u=(x-a)^2$ and $v=g(x)$ . The derivative of $u$ is $2(x-a)$ , because the derivative of the inside $x-a$ is $1$ . The derivative of $v$ is just $g'(x)$ . Applying the product rule gives two terms. Both terms contain one copy of $(x-a)$ , so factor it out. The remaining bracket is a new polynomial $h(x)$ , so $f'(x)=(x-a)h(x)$ . By the factor theorem, this means $(x-a)$ is a factor of $f'(x)$ .

Key Takeaways

The product rule applies whenever a function is a product of two different functions.
A common factor can be taken out of a derivative just as from any algebraic expression.
The factor theorem says: if a polynomial can be written as $(x-a)h(x)$ , then $(x-a)$ is a factor.

Common Mistakes

Applying the product rule as $u'v'$ instead of $u'v + uv'$ .
Forgetting the $g'(x)$ term in the second product.
Writing the derivative without factorising and therefore not giving a clear conclusion.

Things to Be Careful About

The derivative of $(x-a)^2$ is $2(x-a)$ , not $(x-a)^2$ , because the derivative of the inside $x-a$ is $1$ .
To score the final mark, make the factorisation explicit, or state that $f'(a)=0$ and invoke the factor theorem.

Techniques used

apply the product ruledifferentiate a power using the chain rulefactor out a common factorapply the factor theorem

(b)

It is given that $(x - 3)^2$ is a factor of $2x^3 - 4x^2 + px + q$ , where $p$ and $q$ are constants.

Find the values of $p$ and $q$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Because $(x-3)^2$ is a factor, $x=3$ is a repeated root of $f(x)$ . Hence $f(3)=0$ and, from part (a), $f'(3)=0$ . Use these two conditions to form equations in $p$ and $q$ and solve them.

Working

Let $f(x) = 2x^3 - 4x^2 + px + q$ . Since $x=3$ is a root,

f(3) = 0

2(3)^3 - 4(3)^2 + 3p + q = 0

54 - 36 + 3p + q = 0

So

3p + q = -18

Now differentiate $f$ :

f'(x) = 6x^2 - 8x + p

Since $x=3$ is a repeated root, $f'(3)=0$ :

6(3)^2 - 8(3) + p = 0

54 - 24 + p = 0

30 + p = 0

Hence

p = -30

Substitute $p=-30$ into $3p+q=-18$ :

3(-30) + q = -18

-90 + q = -18

q = 72

Answer

p = -30, \quad q = 72

Final answer

p = -30, q = 72

Detailed explanation

Walkthrough

A repeated factor $(x-3)^2$ means that $x=3$ is a root with multiplicity two. Therefore the value of the polynomial is zero at $x=3$ , and so is its derivative. This is exactly part (a) applied with $a=3$ .

First impose $f(3)=0$ :

2(3)^3 - 4(3)^2 + 3p + q = 0

After evaluating the powers, this becomes

18 + 3p + q = 0

so

3p + q = -18.

Next differentiate the polynomial:

f'(x) = 6x^2 - 8x + p.

Since $x=3$ is a repeated root, set $f'(3)=0$ :

6(3)^2 - 8(3) + p = 0.

This simplifies to $30 + p = 0$ , so $p = -30$ .

Finally, substitute $p=-30$ into $3p+q=-18$ :

-90 + q = -18

so $q = 72$ .

Key Takeaways

A repeated factor $(x-a)^2$ gives two conditions: $f(a)=0$ and $f'(a)=0$ .
Unknown coefficients in a polynomial can be found by using root conditions to form equations.
The derivative of a cubic is a quadratic, and evaluating it gives a separate linear equation in the coefficients.

Common Mistakes

Using only $f(3)=0$ , which leaves one equation with two unknowns and so $p$ and $q$ cannot be found.
Forgetting the derivative condition for a repeated root.
Making sign errors when substituting $x=3$ or when moving terms across the equation.
Differentiating incorrectly, for example writing the derivative of $-4x^2$ as $-4x$ instead of $-8x$ .

Things to Be Careful About

The condition $f'(3)=0$ is only valid because the factor is repeated; a single factor would not give a derivative condition.
Evaluate powers correctly: $3^3=27$ , $3^2=9$ , $f'(x)=6x^2-8x+p$ , not $6x^2-4x+p$ .
The final answers are $p=-30$ and $q=72$ ; check by substituting back into the cubic or by expanding if desired.

Techniques used

use the repeated-root conditions f(3)=0 and f'(3)=0differentiate a polynomialform and solve simultaneous linear equations

Q6MediumTrigonometryNumerical Solution of Equations

(a)

By sketching a suitable pair of graphs, show that the equation $\cot 2x = 2\sin 2x - 1$ has exactly one root in the interval $0 < x < \frac{1}{2}\pi$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

To show that the equation $\cot 2x = 2\sin 2x - 1$ has exactly one root in $0 < x < \frac{1}{2}\pi$ , we sketch the graphs of $y = \cot 2x$ and $y = 2\sin 2x - 1$ on the same axes and identify their points of intersection.

Working

Consider the two functions:

y_1 = \cot 2x

y_2 = 2\sin 2x - 1

For $y_1 = \cot 2x$ on $0 < x < \frac{\pi}{2}$ :

As $x \to 0^+$ , $\cot 2x \to +\infty$ (vertical asymptote at $x = 0$ ).
$y_1 = 0$ when $2x = \frac{\pi}{2}$ , so $x = \frac{\pi}{4}$ .
As $x \to \frac{\pi}{2}^-$ , $\cot 2x \to -\infty$ (vertical asymptote at $x = \frac{\pi}{2}$ ).
The curve is continuous and strictly decreasing from $+\infty$ to $-\infty$ .

For $y_2 = 2\sin 2x - 1$ on $0 < x < \frac{\pi}{2}$ :

At $x = 0$ , $y_2 = 2\sin 0 - 1 = -1$ .
Maximum value occurs when $\sin 2x = 1$ , i.e., $2x = \frac{\pi}{2} \Rightarrow x = \frac{\pi}{4}$ . The maximum is $y_2 = 2(1) - 1 = 1$ .
At $x = \frac{\pi}{2}$ , $y_2 = 2\sin \pi - 1 = -1$ .
The curve is a sine wave shifted downwards by 1 unit.

By sketching both curves:

$y_1$ starts at $+\infty$ , decreases through $\left(\frac{\pi}{4}, 0\right)$ , and goes to $-\infty$ .
$y_2$ starts at $(0, -1)$ , increases to a maximum at $\left(\frac{\pi}{4}, 1\right)$ , and decreases back to $\left(\frac{\pi}{2}, -1\right)$ .

The curves intersect exactly once in the interval $0 < x < \frac{\pi}{4}$ (marked with a red dot in the diagram). For $x > \frac{\pi}{4}$ , $y_1 < 0$ while $y_2 \leq 1$ , but $y_1$ drops to $-\infty$ much faster than $y_2$ reaches $-1$ , and they do not cross again before $x = \frac{\pi}{2}$ . Thus, there is exactly one point of intersection in $0 < x < \frac{1}{2}\pi$ .

Answer

The graphs intersect at exactly one point in the interval $0 < x < \frac{1}{2}\pi$ , confirming exactly one root.

Final answer

The curves y = cot(2x) and y = 2sin(2x) - 1 intersect at exactly one point in 0 < x < π/2.

Detailed explanation

Walkthrough

We are asked to show that the equation $\cot 2x = 2\sin 2x - 1$ has exactly one root in $0 < x < \frac{1}{2}\pi$ . The most direct method is to treat each side of the equation as a separate function and sketch them on the same set of axes. The roots of the equation correspond to the x-coordinates of the intersection points of the two graphs.

First, we analyze $y_1 = \cot 2x$ . The cotangent function $\cot \theta$ has vertical asymptotes where $\sin \theta = 0$ . For $\cot 2x$ , this occurs when $2x = 0, \pi, 2\pi, \dots$ , so $x = 0, \frac{\pi}{2}, \pi, \dots$ . In the interval $0 < x < \frac{\pi}{2}$ , the asymptotes are at the boundaries $x = 0$ and $x = \frac{\pi}{2}$ . The function crosses the x-axis when $\cos 2x = 0$ , which gives $2x = \frac{\pi}{2} \Rightarrow x = \frac{\pi}{4}$ . Since $\cot 2x$ is decreasing between its asymptotes, $y_1$ falls from $+\infty$ to $-\infty$ .

Next, we analyze $y_2 = 2\sin 2x - 1$ . This is a sine wave with period $\pi$ , amplitude 2, and a vertical shift of $-1$ . At $x = 0$ , $y_2 = -1$ . It reaches its maximum of $1$ when $\sin 2x = 1$ , i.e., at $x = \frac{\pi}{4}$ . At $x = \frac{\pi}{2}$ , $y_2 = -1$ . The curve rises from $(0, -1)$ to a peak at $\left(\frac{\pi}{4}, 1\right)$ and then falls back to $\left(\frac{\pi}{2}, -1\right)$ .

By drawing these two curves, we see that $y_1$ starts high above $y_2$ (since $y_1 \to +\infty$ and $y_2 = -1$ at $x \to 0^+$ ). At $x = \frac{\pi}{4}$ , $y_1 = 0$ and $y_2 = 1$ , so $y_1 < y_2$ . By continuity, they must cross at least once in $\left(0, \frac{\pi}{4}\right)$ . Since $y_1$ is strictly decreasing and $y_2$ is strictly increasing in this sub-interval, there is exactly one crossing. For $x > \frac{\pi}{4}$ , $y_1 < 0$ and decreasing to $-\infty$ , while $y_2$ is decreasing but bounded below by $-1$ , so they do not cross again. Thus, there is exactly one root.

Key Takeaways

Sketching two functions on the same axes is a powerful graphical method to locate roots of equations.
Understanding the key features of trigonometric graphs (asymptotes, intercepts, maxima, minima) is essential for accurate sketching.
The number of intersections between two graphs directly corresponds to the number of roots of the equation formed by equating them.

Common Mistakes

Sketching $\cot 2x$ with the wrong period or missing the vertical asymptotes at $x = 0$ and $x = \frac{\pi}{2}$ .
Forgetting to shift the sine curve down by 1 unit, which would place the maximum at $y = 2$ instead of $y = 1$ .
Not justifying why there is only one intersection (e.g., by noting the monotonic behavior in the relevant sub-interval).

Things to Be Careful About

Ensure all angles are in radians, as the interval is given in terms of $\pi$ .
The sketch must be clear enough to show that no second intersection occurs in the interval. Marking the intersection point explicitly helps.
Scales do not need to be perfectly accurate, but key points (asymptotes, x-intercept at $\frac{\pi}{4}$ , maximum at $\frac{\pi}{4}$ ) must be correctly positioned.

Techniques used

sketch cotangent curvesketch transformed sine curveidentify intersection points

(b)

Show by calculation that the root is in the interval $0.4 < x < 0.6$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

To show that the root lies in $0.4 < x < 0.6$ , we define a function $f(x) = \cot 2x - 2\sin 2x + 1$ and evaluate it at the endpoints $x = 0.4$ and $x = 0.6$ . If $f(0.4)$ and $f(0.6)$ have opposite signs, then by the Intermediate Value Theorem, there is a root in the interval.

Working

Let $f(x) = \cot 2x - 2\sin 2x + 1$ . We calculate $f(0.4)$ and $f(0.6)$ using radians.

For $x = 0.4$ :

2x = 0.8

\cot(0.8) = \frac{\cos(0.8)}{\sin(0.8)} \approx \frac{0.6967}{0.7174} \approx 0.9712

2\sin(0.8) \approx 2 \times 0.7174 = 1.4348

f(0.4) = 0.9712 - 1.4348 + 1 = 0.5364

f(0.4) > 0

For $x = 0.6$ :

2x = 1.2

\cot(1.2) = \frac{\cos(1.2)}{\sin(1.2)} \approx \frac{0.3624}{0.9320} \approx 0.3888

2\sin(1.2) \approx 2 \times 0.9320 = 1.8640

f(0.6) = 0.3888 - 1.8640 + 1 = -0.4752

f(0.6) < 0

Since $f(0.4) \approx 0.537 > 0$ and $f(0.6) \approx -0.475 < 0$ , there is a sign change in the interval $0.4 < x < 0.6$ . Because $f(x)$ is continuous on this interval, there must be at least one root in $0.4 < x < 0.6$ .

Answer

$f(0.4) \approx 0.537 > 0$ and $f(0.6) \approx -0.475 < 0$ , so the root lies in $0.4 < x < 0.6$ .

Final answer

f(0.4) ≈ 0.537 > 0 and f(0.6) ≈ -0.475 < 0, so there is a sign change and a root in 0.4 < x < 0.6.

Detailed explanation

Walkthrough

We need to prove that the root of $\cot 2x = 2\sin 2x - 1$ lies between $x = 0.4$ and $x = 0.6$ . The standard method for this is to rearrange the equation into the form $f(x) = 0$ and check the signs of $f(x)$ at the endpoints of the interval.

Rearranging gives $f(x) = \cot 2x - 2\sin 2x + 1 = 0$ . We must ensure our calculator is in radian mode, as the interval is given in radians.

At $x = 0.4$ , we compute $2x = 0.8$ . Then $\cot(0.8) \approx 0.9712$ and $2\sin(0.8) \approx 1.4348$ . Substituting these into $f(x)$ gives $f(0.4) = 0.9712 - 1.4348 + 1 = 0.5364$ , which is positive.

At $x = 0.6$ , we compute $2x = 1.2$ . Then $\cot(1.2) \approx 0.3888$ and $2\sin(1.2) \approx 1.8640$ . Substituting gives $f(0.6) = 0.3888 - 1.8640 + 1 = -0.4752$ , which is negative.

Since $f(x)$ is continuous on $(0.4, 0.6)$ and changes sign from positive to negative, the Intermediate Value Theorem guarantees that there is at least one root in this interval. The question asks to "show by calculation", so presenting these values with a clear statement of the sign change is sufficient.

Key Takeaways

The sign-change method is a reliable way to locate roots within a given interval.
Always ensure the calculator is in the correct mode (radians for trigonometric equations involving $\pi$ ).
Values should be given to at least 2 significant figures to justify the sign change.

Common Mistakes

Using degrees instead of radians when evaluating trigonometric functions. This will give completely wrong values and no sign change.
Making arithmetic errors when calculating $\cot 2x$ (remembering that $\cot \theta = \frac{\cos \theta}{\sin \theta}$ , not $\tan \theta$ ).
Not explicitly stating that a sign change implies a root (though often implied, it is good practice to mention continuity).

Things to Be Careful About

The mark scheme awards M0 if working in degrees. Always use radians.
Values should be accurate to at least 2 significant figures. Using more decimal places is fine and helps avoid rounding errors.
Ensure the function $f(x)$ is defined and continuous on the entire interval $(0.4, 0.6)$ . Since $\cot 2x$ has asymptotes at $x = 0$ and $x = \frac{\pi}{2} \approx 1.57$ , it is continuous on $(0.4, 0.6)$ .

Techniques used

define function from equationevaluate at interval endpointsapply intermediate value theorem

(c)

Use the iterative formula $x_{n+1} = \frac{1}{2} \tan^{-1} \left( \frac{1}{2\sin 2x_n - 1} \right)$ to calculate the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

3M Similar

DifficultyMedium

Worked solution

Approach

We are given the iterative formula:

x_{n+1} = \frac{1}{2} \tan^{-1} \left( \frac{1}{2\sin 2x_n - 1} \right)

We start with an initial value $x_1$ in the interval $0.4 < x < 0.6$ (e.g., $x_1 = 0.5$ ) and iterate until the result is correct to 2 decimal places. Each iteration must be given to 4 decimal places.

Working

Let $x_1 = 0.5$ .

Iteration 1:

x_2 = \frac{1}{2} \tan^{-1} \left( \frac{1}{2\sin(2 \times 0.5) - 1} \right) = \frac{1}{2} \tan^{-1} \left( \frac{1}{2\sin(1.0) - 1} \right)

2\sin(1.0) - 1 \approx 2(0.8415) - 1 = 0.6830

x_2 = \frac{1}{2} \tan^{-1} \left( \frac{1}{0.6830} \right) = \frac{1}{2} \tan^{-1}(1.4641) \approx \frac{1}{2}(0.9719) = 0.4860

Iteration 2:

x_3 = \frac{1}{2} \tan^{-1} \left( \frac{1}{2\sin(2 \times 0.4860) - 1} \right) = \frac{1}{2} \tan^{-1} \left( \frac{1}{2\sin(0.9720) - 1} \right)

2\sin(0.9720) - 1 \approx 2(0.8253) - 1 = 0.6506

x_3 = \frac{1}{2} \tan^{-1} \left( \frac{1}{0.6506} \right) = \frac{1}{2} \tan^{-1}(1.5370) \approx \frac{1}{2}(0.9961) = 0.4981

Iteration 3:

x_4 = \frac{1}{2} \tan^{-1} \left( \frac{1}{2\sin(2 \times 0.4981) - 1} \right) = \frac{1}{2} \tan^{-1} \left( \frac{1}{2\sin(0.9962) - 1} \right)

2\sin(0.9962) - 1 \approx 2(0.8390) - 1 = 0.6780

x_4 = \frac{1}{2} \tan^{-1} \left( \frac{1}{0.6780} \right) = \frac{1}{2} \tan^{-1}(1.4749) \approx \frac{1}{2}(0.9746) = 0.4873

Iteration 4:

x_5 = \frac{1}{2} \tan^{-1} \left( \frac{1}{2\sin(2 \times 0.4873) - 1} \right) = \frac{1}{2} \tan^{-1} \left( \frac{1}{2\sin(0.9746) - 1} \right)

2\sin(0.9746) - 1 \approx 2(0.8270) - 1 = 0.6540

x_5 = \frac{1}{2} \tan^{-1} \left( \frac{1}{0.6540} \right) = \frac{1}{2} \tan^{-1}(1.5290) \approx \frac{1}{2}(0.9934) = 0.4967

Iteration 5:

x_6 = \frac{1}{2} \tan^{-1} \left( \frac{1}{2\sin(2 \times 0.4967) - 1} \right) = \frac{1}{2} \tan^{-1} \left( \frac{1}{2\sin(0.9934) - 1} \right)

2\sin(0.9934) - 1 \approx 2(0.8378) - 1 = 0.6756

x_6 = \frac{1}{2} \tan^{-1} \left( \frac{1}{0.6756} \right) = \frac{1}{2} \tan^{-1}(1.4802) \approx \frac{1}{2}(0.9758) = 0.4879

Iteration 6:

x_7 = \frac{1}{2} \tan^{-1} \left( \frac{1}{2\sin(2 \times 0.4879) - 1} \right) = \frac{1}{2} \tan^{-1} \left( \frac{1}{2\sin(0.9758) - 1} \right)

2\sin(0.9758) - 1 \approx 2(0.8279) - 1 = 0.6558

x_7 = \frac{1}{2} \tan^{-1} \left( \frac{1}{0.6558} \right) = \frac{1}{2} \tan^{-1}(1.5247) \approx \frac{1}{2}(0.9921) = 0.4961

Continuing this process, the values oscillate and converge:

x_8 \approx 0.4895, \quad x_9 \approx 0.4947, \quad x_{10} \approx 0.4915, \quad x_{11} \approx 0.4933, \quad x_{12} \approx 0.4923

To confirm the root is $0.49$ to 2 decimal places, we can check for a sign change in $f(x) = \cot 2x - 2\sin 2x + 1$ around $0.485$ and $0.495$ :

f(0.485) = \cot(0.970) - 2\sin(0.970) + 1 \approx 0.9755 - 1.6513 + 1 = -0.0758 < 0

f(0.495) = \cot(0.990) - 2\sin(0.990) + 1 \approx 0.9500 - 1.6345 + 1 = 0.3155 > 0

Since $f(0.485) < 0$ and $f(0.495) > 0$ , the root lies in $(0.485, 0.495)$ , which means it is $0.49$ to 2 decimal places.

Answer

The root correct to 2 decimal places is $0.49$ .

Final answer

0.49

Detailed explanation

Walkthrough

We are given an iterative formula to find the root of the equation. The formula is:

x_{n+1} = \frac{1}{2} \tan^{-1} \left( \frac{1}{2\sin 2x_n - 1} \right)

We choose an initial guess $x_1 = 0.5$ , which is within the interval $(0.4, 0.6)$ established in part (b). We then substitute this into the right-hand side to find $x_2$ , and repeat the process.

Iteration 1: $x_1 = 0.5$
$2x_1 = 1.0$ . $2\sin(1.0) - 1 \approx 0.6830$ . $\frac{1}{0.6830} \approx 1.4641$ . $\tan^{-1}(1.4641) \approx 0.9719$ . $x_2 = \frac{0.9719}{2} = 0.4860$ .

Iteration 2: $x_2 = 0.4860$
$2x_2 = 0.9720$ . $2\sin(0.9720) - 1 \approx 0.6506$ . $\frac{1}{0.6506} \approx 1.5370$ . $\tan^{-1}(1.5370) \approx 0.9961$ . $x_3 = \frac{0.9961}{2} = 0.4981$ .

Iteration 3: $x_3 = 0.4981$
$2x_3 = 0.9962$ . $2\sin(0.9962) - 1 \approx 0.6780$ . $\frac{1}{0.6780} \approx 1.4749$ . $\tan^{-1}(1.4749) \approx 0.9746$ . $x_4 = \frac{0.9746}{2} = 0.4873$ .

The values oscillate: $0.4860, 0.4981, 0.4873, 0.4967, 0.4879, 0.4961, \dots$ . This oscillation indicates that the iterative formula is converging to a root, but slowly. We continue iterating until the values agree to 2 decimal places.

After several iterations, the values settle around $0.49$ . To rigorously justify that the root is $0.49$ to 2 decimal places, we can check the sign of $f(x) = \cot 2x - 2\sin 2x + 1$ at $x = 0.485$ and $x = 0.495$ . If $f(0.485) < 0$ and $f(0.495) > 0$ , then the root must lie in $(0.485, 0.495)$ , which means it rounds to $0.49$ .

Calculating:
$f(0.485) = \cot(0.970) - 2\sin(0.970) + 1 \approx 0.9755 - 1.6513 + 1 = -0.0758 < 0$
$f(0.495) = \cot(0.990) - 2\sin(0.990) + 1 \approx 0.9500 - 1.6345 + 1 = 0.3155 > 0$

Since there is a sign change in $(0.485, 0.495)$ , the root is indeed $0.49$ to 2 decimal places.

Key Takeaways

Iterative formulas can be used to find roots of equations to a specified degree of accuracy.
The sequence of values may oscillate around the root before converging; this is normal and does not mean the formula is incorrect.
To confirm the final answer to a given number of decimal places, it is best to check for a sign change in the function at the boundaries of the rounding interval (e.g., $\pm 0.005$ ).

Common Mistakes

Using degrees instead of radians. This will cause the iteration to diverge or give completely wrong values.
Rounding intermediate values too early. Always keep at least 4 decimal places during iterations to avoid accumulation of rounding errors.
Stopping the iteration too early. The values must agree to 2 decimal places, and even then, a sign change check is the most robust way to confirm the answer.
Not showing enough iterations. The mark scheme requires at least a second iteration to be completed, and sufficient iterations to justify the final answer.

Things to Be Careful About

The iterative formula may oscillate. Do not assume it is converging monotonically.
Always use radians in calculator mode.
When checking for the final answer, use the original function $f(x) = \cot 2x - 2\sin 2x + 1$ rather than relying solely on the iterative values, as the iteration itself may not have fully converged to the exact root.
The mark scheme accepts various starting values (e.g., $0.4, 0.6$ ) and will condone missing iterations in the middle as long as the start and end are correct.

Techniques used

apply iterative formulacompute successive approximationscheck for convergence to required precision

Q7MediumDifferentiation

The equation of a curve is $2y^3 - 3x^2y - x^3 = 16$ .

(a)

Show that

\frac{dy}{dx} = \frac{x^2 + 2xy}{2y^2 - x^2}.

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

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

Working

Differentiate each term:

\frac{d}{dx}(2y^3) = 6y^2 \frac{dy}{dx}

For $3x^2y$ , use the product rule:

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

Also:

\frac{d}{dx}(x^3) = 3x^2, \qquad \frac{d}{dx}(16) = 0

So differentiating the whole equation gives:

6y^2 \frac{dy}{dx} - \left(6xy + 3x^2 \frac{dy}{dx}\right) - 3x^2 = 0

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

(6y^2 - 3x^2)\frac{dy}{dx} = 6xy + 3x^2

Factor $3$ from the numerator and denominator and cancel the common factor:

\frac{dy}{dx} = \frac{3(x^2 + 2xy)}{3(2y^2 - x^2)} = \frac{x^2 + 2xy}{2y^2 - x^2}

as required.

Answer

\frac{dy}{dx} = \frac{x^2 + 2xy}{2y^2 - x^2}

Final answer

dy/dx = (x^2 + 2xy)/(2y^2 - x^2)

Detailed explanation

Walkthrough

We are given an equation that defines $y$ implicitly as a function of $x$ . To find $\frac{dy}{dx}$ , differentiate every term with respect to $x$ .

For $2y^3$ , since $y$ is a function of $x$ , the derivative is $6y^2 \frac{dy}{dx}$ .

For $3x^2y$ , use the product rule: differentiate $x^2$ to get $2x$ , and keep $y$ ; then keep $x^2$ and differentiate $y$ to get $\frac{dy}{dx}$ . This gives $3\left(2xy + x^2\frac{dy}{dx}\right)$ .

The derivative of $x^3$ is $3x^2$ , and the derivative of the constant $16$ is $0$ . Putting these together gives an equation involving $\frac{dy}{dx}$ . Collect the $\frac{dy}{dx}$ terms on one side, then factor $3$ from the numerator and denominator and cancel it. This is exactly the required expression.

Key Takeaways

This question tests implicit differentiation, the product rule, and algebraic rearrangement. When a variable $y$ is defined implicitly in terms of $x$ , every $y$ -term must be differentiated with an extra factor $\frac{dy}{dx}$ . The product rule is needed when a term contains both $x$ and $y$ .

Common Mistakes

Forgetting to multiply by $\frac{dy}{dx}$ when differentiating $2y^3$ .
Applying the product rule incorrectly to $3x^2y$ .
Losing the minus sign in front of $3x^2y$ .
Failing to collect the $\frac{dy}{dx}$ terms before solving.
Cancelling incorrectly; factor first, then cancel the common factor $3$ .

Things to Be Careful About

The derivative of $2y^3$ is $6y^2 \frac{dy}{dx}$ , not $6y^2$ . The right-hand side is a constant, so its derivative is $0$ . When simplifying, write the numerator and denominator with a common factor $3$ before cancelling. The mark scheme requires clear working showing how $\frac{dy}{dx}$ was obtained.

Techniques used

differentiate implicitly with respect to xapply the product rule to x^2 ycollect terms involving dy/dxsimplify by cancelling a common factor

(b)

Hence find the coordinates of the points on the curve at which the normal is parallel to the $y$ -axis.

4M Similar

DifficultyMedium

Worked solution

Approach

For the normal to be parallel to the $y$ -axis, the tangent must be horizontal, so $\frac{dy}{dx} = 0$ . Since the derivative is a fraction, set its numerator equal to zero. This gives two possible relationships between $x$ and $y$ . Substitute each into the curve equation to find the coordinates.

Working

The normal is parallel to the $y$ -axis when the tangent is horizontal, so:

\frac{dy}{dx} = 0

Using the derivative from part (a):

\frac{x^2 + 2xy}{2y^2 - x^2} = 0

So the numerator must be zero:

x^2 + 2xy = 0

Factorise:

x(x + 2y) = 0

Hence:

x = 0 \quad \text{or} \quad x = -2y

Case 1: $x = 0$ . Substitute into the curve:

2y^3 - 0 - 0 = 16

y^3 = 8

y = 2

So one point is $(0, 2)$ .

Case 2: $x = -2y$ . Substitute into the curve:

2y^3 - 3(-2y)^2 y - (-2y)^3 = 16

Simplify:

2y^3 - 3(4y^2)y + 8y^3 = 16

2y^3 - 12y^3 + 8y^3 = 16

-2y^3 = 16

y^3 = -8

y = -2

Then:

x = -2y = -2(-2) = 4

So the other point is $(4, -2)$ .

Answer

(0, 2) \quad \text{and} \quad (4, -2)

Final answer

(0, 2) and (4, -2)

Detailed explanation

Walkthrough

A normal line perpendicular to the curve is parallel to the $y$ -axis exactly when it is vertical. That happens when the tangent is horizontal, i.e. when $\frac{dy}{dx}=0$ . Since the derivative is a fraction, a fraction is zero only when its numerator is zero (provided the denominator is not also zero). Set $x^2+2xy=0$ and factorise to get $x=0$ or $x=-2y$ .

Each of these gives a possible family of points on the curve. Substitute each relation into the original curve equation to find the actual points.

For $x=0$ , the curve becomes $2y^3=16$ , so $y=2$ , giving $(0,2)$ .

For $x=-2y$ , substitute carefully: $x^2=4y^2$ and $x^3=-8y^3$ . The curve becomes $2y^3 - 3(4y^2)y - (-8y^3)=16$ , which simplifies to $-2y^3=16$ , so $y=-2$ and $x=4$ , giving $(4,-2)$ .

Both points have nonzero denominator in the derivative, so the condition is valid.

Key Takeaways

This question connects the geometry of tangents and normals to the value of the derivative. A vertical normal corresponds to a horizontal tangent, so $\frac{dy}{dx}=0$ . Solving an implicit derivative condition often leads to two cases that must each be substituted back into the original equation.

Common Mistakes

Setting the denominator equal to zero instead of the numerator.
Forgetting the case $x=0$ after factorising.
Making sign errors when substituting $x=-2y$ , especially with $(-2y)^3$ .
Stopping after finding $y=-2$ and not finding $x=4$ .

Things to Be Careful About

The normal is parallel to the $y$ -axis, so the tangent is horizontal: $\frac{dy}{dx}=0$ , not infinite. Use the numerator of the derivative. When substituting $x=-2y$ , remember $(-2y)^3=-8y^3$ . The mark scheme accepts coordinates stated separately, but both coordinates of each point are needed.

Techniques used

set the derivative equal to zerofactorise and solve for x in terms of ysubstitute each case into the original curve equationsolve the resulting equations for coordinates

Q8MediumTrigonometryIntegration

(a)

Prove the identity $\sin 4x \equiv 4\sin x(2\cos^3 x - \cos x)$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Start with $\sin 4x$ and use the double angle formula for sine, then express $\cos 2x$ in terms of $\sin x$ and $\cos x$ .

Working

\begin{aligned} \sin 4x &= 2\sin 2x \cos 2x \\ &= 2(2\sin x \cos x)(2\cos^2 x - 1) \\ &= 4\sin x \cos x(2\cos^2 x - 1) \\ &= 4\sin x(2\cos^3 x - \cos x). \end{aligned}

This proves the identity.

Answer

\sin 4x \equiv 4\sin x(2\cos^3 x - \cos x)

Final answer

sin 4x = 4 sin x(2 cos^3 x - cos x)

Detailed explanation

Walkthrough

We start with the left-hand side, $\sin 4x$ . The double angle formula for sine says $\sin 2\theta = 2\sin\theta\cos\theta$ . Taking $\theta = 2x$ gives $\sin 4x = 2\sin 2x \cos 2x$ . Next, expand $\sin 2x$ as $2\sin x\cos x$ , and expand $\cos 2x$ as $2\cos^2 x - 1$ . Substituting gives $2(2\sin x\cos x)(2\cos^2 x - 1) = 4\sin x\cos x(2\cos^2 x - 1)$ . Finally, multiply the $\cos x$ into the bracket to obtain $4\sin x(2\cos^3 x - \cos x)$ , which is exactly the right-hand side. Since every step is reversible and valid for all $x$ , the identity is proved.

Key Takeaways

This question tests the double angle formulae for sine and cosine. It also shows how to rewrite a trigonometric expression entirely in terms of $\sin x$ and $\cos x$ by expanding compound angles. Recognising which double angle form of $\cos 2x$ to use is an important skill.

Common Mistakes

Using $\sin 4x = 4\sin x\cos x$ directly, which is false.
Using the wrong form of $\cos 2x$ , such as $\cos^2 x + \sin^2 x$ .
Losing the factor $2$ when expanding $\sin 2x$ or $\cos 2x$ .
Stopping at $4\sin x\cos x(2\cos^2 x - 1)$ without distributing the $\cos x$ .

Things to Be Careful About

Because the statement is an identity, the working must be valid for all $x$ , not just particular values. Ensure the final expression has the factor $\cos x$ distributed correctly: $\cos x(2\cos^2 x - 1) = 2\cos^3 x - \cos x$ . The mark scheme requires full working to earn the final A1; an unsupported statement of the identity is not enough.

Techniques used

expand sin 4x using the double angle formulaexpand cos 2x in terms of sin x and cos xsimplify to the required identity

(b)

Hence find the exact value of $\int_0^{\frac{1}{4}\pi} \cos^3 x \sin 4x \, dx$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Use the identity from part (a) to rewrite the integrand as a sum of terms of the form $\sin x \cos^n x$ , integrate each by the reverse chain rule, then evaluate between the limits.

Working

Using $\sin 4x = 4\sin x(2\cos^3 x - \cos x)$ ,

\cos^3 x \sin 4x = 4\sin x \cos^3 x(2\cos^3 x - \cos x) = 8\sin x \cos^6 x - 4\sin x \cos^4 x.

Therefore

\int_0^{\frac{1}{4}\pi} \cos^3 x \sin 4x \, dx = 8\int_0^{\frac{1}{4}\pi} \sin x \cos^6 x \, dx - 4\int_0^{\frac{1}{4}\pi} \sin x \cos^4 x \, dx.

Since $\frac{d}{dx}(\cos x) = -\sin x$ ,

\int \sin x \cos^6 x \, dx = -\frac{1}{7}\cos^7 x + C, \qquad \int \sin x \cos^4 x \, dx = -\frac{1}{5}\cos^5 x + C.

So an antiderivative is

F(x) = -\frac{8}{7}\cos^7 x + \frac{4}{5}\cos^5 x.

Evaluate at the limits:

F\left(\frac{\pi}{4}\right) = -\frac{8}{7}\left(\frac{\sqrt{2}}{2}\right)^7 + \frac{4}{5}\left(\frac{\sqrt{2}}{2}\right)^5 = -\frac{1}{7\sqrt{2}} + \frac{1}{5\sqrt{2}},

F(0) = -\frac{8}{7} + \frac{4}{5}.

Hence

\int_0^{\frac{1}{4}\pi} \cos^3 x \sin 4x \, dx = \left(-\frac{1}{7\sqrt{2}} + \frac{1}{5\sqrt{2}}\right) - \left(-\frac{8}{7} + \frac{4}{5}\right) = \frac{\sqrt{2} + 12}{35}.

Answer

\frac{1}{35}(\sqrt{2} + 12)

Final answer

(sqrt(2) + 12)/35

Detailed explanation

Walkthrough

Part (b) says 'Hence', so we must use the identity from part (a). Substitute $\sin 4x = 4\sin x(2\cos^3 x - \cos x)$ into the integrand:

\cos^3 x \sin 4x = 4\sin x\cos^3 x(2\cos^3 x - \cos x) = 8\sin x\cos^6 x - 4\sin x\cos^4 x.

This splits the integral into two simpler integrals. For each, notice that the derivative of $\cos x$ is $-\sin x$ , so $\sin x\cos^n x$ is almost the derivative of $\cos^{n+1}x$ . Hence

\int \sin x\cos^n x\,dx = -\frac{\cos^{n+1}x}{n+1} + C.

Applying this with $n=6$ and $n=4$ gives the antiderivative $F(x) = -\frac{8}{7}\cos^7 x + \frac{4}{5}\cos^5 x$ . Then evaluate at $x = \frac{\pi}{4}$ and $x=0$ . At $x = \frac{\pi}{4}$ , $\cos x = \frac{\sqrt{2}}{2}$ , so $\cos^7 x = \frac{1}{8\sqrt{2}}$ and $\cos^5 x = \frac{1}{4\sqrt{2}}$ . At $x=0$ , both powers are $1$ . Subtracting $F(0)$ from $F(\pi/4)$ and simplifying gives $\frac{\sqrt{2}+12}{35}$ .

Key Takeaways

This question combines a trigonometric identity with integration by recognition (reverse chain rule). It shows how an identity can transform an apparently difficult integrand into terms of the form $\sin x\cos^n x$ , which integrate immediately. Exact trigonometric values at special angles are then used to obtain an exact final answer.

Common Mistakes

Forgetting to use the identity from part (a) and trying to integrate the original expression directly.
Incorrectly multiplying: $\cos^3 x(2\cos^3 x - \cos x) = 2\cos^6 x - \cos^4 x$ , then forgetting the factor $4$ from the identity.
Sign errors: the derivative of $\cos x$ is $-\sin x$ , so the antiderivative must have a negative sign.
Evaluating the upper limit incorrectly, especially the powers of $\frac{\sqrt{2}}{2}$ .
Forgetting to subtract $F(0)$ .

Things to Be Careful About

The mark scheme first awards a B1 for writing the integral as $8\int\sin x\cos^6 x\,dx - 4\int\sin x\cos^4 x\,dx$ , so show this line clearly. The antiderivative must be stated before substituting limits. When simplifying, $\frac{2}{35\sqrt{2}} = \frac{\sqrt{2}}{35}$ , and the final exact value is $\frac{1}{35}(\sqrt{2}+12)$ . Use radians throughout.

Techniques used

use the identity to rewrite the integrandintegrate powers of cos x times sin x by reverse chain ruleevaluate the definite integral at the limitssimplify the exact result

Q9MediumAlgebraIntegration

Let $f(x) = \frac{x^2 + 4ax + 6a^2}{(x + 2a)(x + 3a)}$ , where $a$ is a positive constant.

(a)

Express $f(x)$ in partial fractions.

5M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since the numerator and denominator have the same degree, the decomposition must include a constant term:

f(x) = A + \frac{B}{x + 2a} + \frac{C}{x + 3a}.

Find $A$ , $B$ and $C$ by multiplying both sides by $(x+2a)(x+3a)$ and comparing coefficients.

Working

x^2 + 4ax + 6a^2 = A(x+2a)(x+3a) + B(x+3a) + C(x+2a).

Comparing the coefficient of $x^2$ gives $A = 1$ .

Comparing the coefficient of $x$ and the constant term respectively gives

5aA + B + C = 4a,

6a^2A + 3aB + 2aC = 6a^2.

With $A = 1$ , these become

B + C = -a, \qquad 3B + 2C = 0.

From $3B + 2C = 0$ , $C = -\frac{3}{2}B$ . Substitute into $B + C = -a$ :

B - \frac{3}{2}B = -a \;\Rightarrow\; B = 2a.

Hence $C = -3a$ . Therefore

f(x) = 1 + \frac{2a}{x + 2a} - \frac{3a}{x + 3a}.

Final answer

f(x) = 1 + 2a/(x + 2a) - 3a/(x + 3a)

Detailed explanation

Walkthrough

The first thing to notice is that the numerator and denominator are both quadratic, so the fraction is not proper. If we try to write it as only two partial fractions, we cannot match the leading term. We therefore include a constant term:

f(x) = A + \frac{B}{x + 2a} + \frac{C}{x + 3a}.

Multiplying both sides by the denominator converts this into a polynomial identity. Comparing the coefficient of $x^2$ immediately gives $A = 1$ , because $A$ is multiplied by $(x+2a)(x+3a)$ , whose leading term is $x^2$ .

Next, comparing the coefficient of $x$ gives $5aA + B + C = 4a$ , and comparing the constant terms gives $6a^2A + 3aB + 2aC = 6a^2$ . Substituting $A = 1$ turns these into the two linear equations

B + C = -a, \qquad 3B + 2C = 0.

Solving this simple system gives $B = 2a$ and $C = -3a$ . The positive parameter $a$ is carried through the algebra and appears naturally in the coefficients.

Key Takeaways

This question tests the decomposition of an improper rational function. When the degrees of the numerator and denominator are equal, the partial fractions must begin with a non-constant term, here the constant $A$ . It also reinforces comparing coefficients as a systematic way of finding unknown constants, and solving a small linear system.

Common Mistakes

Omitting the constant term $A$ and trying to write the expression as only two fractions. The mark scheme only awards full credit if the form includes the constant term.
Substituting a numerical value such as $a = 1$ to make the algebra easier. The question requires an answer in terms of $a$ , and the mark scheme restricts credit if $a$ is replaced by a number.
Making sign errors when expanding $(x+2a)(x+3a)$ , especially in the constant term $6a^2$ .

Things to Be Careful About

The positive constant $a$ must be treated as a single unknown throughout. It is also important to expand correctly:

(x+2a)(x+3a) = x^2 + 5ax + 6a^2.

When solving $B + C = -a$ and $3B + 2C = 0$ , keep the signs in the simultaneous equations consistent; a single sign error will give incorrect coefficients.

Techniques used

write an improper rational function as a constant plus proper partial fractionsmultiply through by the denominator to form a polynomial identitycompare coefficients to obtain a linear systemsolve the linear system for the partial fraction coefficients

(b)

Hence find the exact value of $\int_{-a}^a f(x) \, dx$ . Give your answer in the form $a(p + \ln q)$ , where $p$ and $q$ are rational.

4M Similar

DifficultyMedium

Worked solution

Approach

Use the partial-fraction form from part (a):

f(x) = 1 + \frac{2a}{x + 2a} - \frac{3a}{x + 3a}.

Integrate term by term from $x = -a$ to $x = a$ . The reciprocal terms integrate to logarithms of the form $\ln(x + \text{constant})$ .

Working

\int_{-a}^{a} f(x)\,dx = \int_{-a}^{a} \left(1 + \frac{2a}{x+2a} - \frac{3a}{x+3a}\right)\,dx.

Since $a > 0$ , both $x+2a$ and $x+3a$ are positive on $[-a,a]$ , so we can write the antiderivative as

F(x) = x + 2a\ln(x+2a) - 3a\ln(x+3a).

At the upper limit $x = a$ :

F(a) = a + 2a\ln(3a) - 3a\ln(4a).

At the lower limit $x = -a$ :

F(-a) = -a + 2a\ln(a) - 3a\ln(2a).

Therefore

F(a) - F(-a) = \left(a - (-a)\right) + 2a\left[\ln(3a) - \ln a\right] - 3a\left[\ln(4a) - \ln(2a)\right].

= 2a + 2a\ln 3 - 3a\ln 2.

Now use logarithm laws:

2\ln 3 - 3\ln 2 = \ln\left(\frac{3^2}{2^3}\right) = \ln\left(\frac{9}{8}\right).

\int_{-a}^{a} f(x)\,dx = 2a + a\ln\left(\frac{9}{8}\right) = a\left(2 + \ln\frac{9}{8}\right).

So $p = 2$ and $q = \frac{9}{8}$ , both rational.

Answer

\int_{-a}^{a} f(x)\,dx = a\left(2 + \ln\frac{9}{8}\right).

Final answer

a(2 + ln(9/8))

Detailed explanation

Walkthrough

The function is already split into partial fractions from part (a). To integrate it, recall that

\int \frac{1}{x + c}\,dx = \ln(x + c) + K

when the argument is positive. Since the interval is $[-a,a]$ and $a > 0$ , $x + 2a$ and $x + 3a$ are positive throughout, so no absolute-value complications arise.

The antiderivative is therefore

F(x) = x + 2a\ln(x+2a) - 3a\ln(x+3a).

Substitute the upper limit $x = a$ and the lower limit $x = -a$ . A common mistake is to subtract the lower antiderivative incorrectly. In the log terms, the $a$ factors inside the arguments cancel:

\ln(3a) - \ln a = \ln 3,

\ln(4a) - \ln(2a) = \ln 2.

The $x$ -term contributes

a - (-a) = 2a.

Combine everything:

2a + 2a\ln 3 - 3a\ln 2 = 2a + a\left(2\ln 3 - 3\ln 2\right).

Finally,

2\ln 3 - 3\ln 2 = \ln 9 - \ln 8 = \ln\left(\frac{9}{8}\right),

so the final answer is $a(2 + \ln(9/8))$ .

Key Takeaways

This question requires:

using the result of a partial-fraction decomposition in integration,
integrating $\frac{k}{x+c}$ as $k \ln(x + c)$ ,
carefully substituting definite limits,
combining logarithms using laws, especially $x\ln p - y\ln q$ and $\ln p - \ln q$ .

Common Mistakes

Forgetting the constant term $1$ when integrating, i.e. missing the $x$ in the antiderivative.
Incorrectly substituting $-a$ into $\ln(x+2a)$ ; be careful not to write $\ln(-a)$ or anything similar. At $x = -a$ , we have $x+2a = a$ , not $-a$ .
Subtracting the lower antiderivative with the wrong signs. The final difference is $F(a) - F(-a)$ , so the $-a$ term contributes $a$ .
Losing the factor $a$ when combining the logarithmic parts. The expression is $2a\ln 3 - 3a\ln 2$ , not $2\ln 3 - 3\ln 2$ .

Things to Be Careful About

The mark scheme requires a completed partial-fraction form; if part (a) is wrong, the final answer may not be accepted even if the integration is correct.

Because $a$ is given as positive, no division by zero or negative argument to a logarithm occurs in the integration interval. Also remember that the final result must be stated in the requested form $a(p + \ln q)$ with rational $p$ and $q$ ; here $p=2$ and $q=\frac{9}{8}$ .

Techniques used

integrate a sum of a constant term and reciprocal fractionsuse logarithm integration for terms of the form k/(x+c)substitute upper and lower limits into the antiderivativesimplify logarithmic expressions using logarithm laws

Q10Medium-HardDifferential Equations

Fig. 2

The diagram shows a tank for holding water. The tank is in the shape of a cube of side $50\text{ cm}$ . At time $t$ seconds, the depth of water in the tank is $h\text{ cm}$ . Water is poured into the tank at a rate of $5000\text{ cm}^3\text{ s}^{-1}$ . Water pours out of the tank through a hole in the bottom at a rate proportional to $h^2$ .

When $h = 20$ , the depth of the water is increasing at a rate of $0.4\text{ cm s}^{-1}$ .

(a)

Show that

\frac{dh}{dt} = \frac{500 - h^2}{250}.

4M Similar

DifficultyMedium

Worked solution

Approach

Relate the volume of water $V$ to the depth $h$ . Differentiate with respect to $t$ using the chain rule to find $\frac{dh}{dt}$ in terms of $h$ . Use the given condition to find the constant of proportionality $k$ , then substitute back to obtain the required differential equation.

Working

The volume of water in the tank is:

V = 50 \times 50 \times h = 2500h

The rate of change of volume is the rate in minus the rate out:

\frac{dV}{dt} = 5000 - kh^2

Using the chain rule, $\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt} = 2500 \frac{dh}{dt}$ . Equating the two expressions:

2500 \frac{dh}{dt} = 5000 - kh^2

\frac{dh}{dt} = \frac{5000 - kh^2}{2500}

When $h = 20$ , $\frac{dh}{dt} = 0.4$ . Substitute these values to find $k$ :

0.4 = \frac{5000 - k(20)^2}{2500}

1000 = 5000 - 400k

400k = 4000 \implies k = 10

Substitute $k = 10$ into the expression for $\frac{dh}{dt}$ :

\frac{dh}{dt} = \frac{5000 - 10h^2}{2500} = \frac{500 - h^2}{250}

Answer

\frac{dh}{dt} = \frac{500 - h^2}{250}

Final answer

dh/dt = (500 - h^2) / 250

Detailed explanation

Walkthrough

First, we express the volume of water $V$ in terms of the depth $h$ . Since the tank has a square base of $50\text{ cm} \times 50\text{ cm}$ , the volume is $V = 2500h$ . The rate at which the volume changes, $\frac{dV}{dt}$ , is the rate at which water enters ( $5000$ ) minus the rate at which it leaves ( $kh^2$ ). This gives $\frac{dV}{dt} = 5000 - kh^2$ .

Next, we use the chain rule to connect $\frac{dV}{dt}$ to $\frac{dh}{dt}$ . Since $\frac{dV}{dh} = 2500$ , we have $\frac{dV}{dt} = 2500\frac{dh}{dt}$ . Equating the two expressions for $\frac{dV}{dt}$ gives $\frac{dh}{dt} = \frac{5000 - kh^2}{2500}$ .

To find the constant $k$ , we use the given condition that when $h = 20$ , $\frac{dh}{dt} = 0.4$ . Substituting these values yields $0.4 = \frac{5000 - 400k}{2500}$ , which simplifies to $k = 10$ . Finally, substituting $k = 10$ and dividing numerator and denominator by $10$ gives the required result $\frac{dh}{dt} = \frac{500 - h^2}{250}$ .

Key Takeaways

Relating physical quantities like volume and depth to set up differential equations.
Using the chain rule to change the variable of differentiation from volume to depth.
Determining unknown constants in a differential equation using provided initial or conditional data.

Common Mistakes

Forgetting to multiply the base area ( $50 \times 50$ ) when finding $\frac{dV}{dh}$ .
Failing to show the method for finding $k$ ; simply stating $k=10$ without working will not earn the mark.
Not simplifying the final fraction correctly to match the required form.

Things to Be Careful About

Ensure the final statement is complete and clearly shows $\frac{dh}{dt} = \frac{500 - h^2}{250}$ to secure the final mark.
Pay attention to units; all given values are in cm and seconds, so no unit conversion is needed, but consistency must be maintained.

Techniques used

relate volume to depthapply chain ruledetermine constant of proportionalitysubstitute into differential equation

(b)

Given that $h = 0$ when $t = 0$ , find the time taken for the depth of the water in the tank to reach $20\text{ cm}$ .

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

Separate the variables $h$ and $t$ in the differential equation from part (a). Integrate both sides, using the standard integral form for $\frac{1}{a^2 - x^2}$ . Apply the initial condition $h = 0$ when $t = 0$ to find the constant of integration, then substitute $h = 20$ to find the time $t$ .

Working

Separate variables:

\int \frac{1}{500 - h^2} \, dh = \int \frac{1}{250} \, dt

Integrate the right-hand side:

\int \frac{1}{250} \, dt = \frac{t}{250}

For the left-hand side, use the standard integral $\int \frac{1}{a^2 - x^2} \, dx = \frac{1}{2a} \ln \left| \frac{a + x}{a - x} \right|$ with $a = \sqrt{500} = 10\sqrt{5}$ :

\int \frac{1}{500 - h^2} \, dh = \frac{1}{2\sqrt{500}} \ln \left| \frac{\sqrt{500} + h}{\sqrt{500} - h} \right| = \frac{1}{20\sqrt{5}} \ln \left| \frac{10\sqrt{5} + h}{10\sqrt{5} - h} \right|

Equating the integrals and adding the constant of integration $C$ :

\frac{1}{20\sqrt{5}} \ln \left| \frac{10\sqrt{5} + h}{10\sqrt{5} - h} \right| = \frac{t}{250} + C

Use the initial condition $t = 0, h = 0$ :

\frac{1}{20\sqrt{5}} \ln \left| \frac{10\sqrt{5}}{10\sqrt{5}} \right| = 0 + C \implies C = 0

The particular solution is:

\frac{t}{250} = \frac{1}{20\sqrt{5}} \ln \left( \frac{10\sqrt{5} + h}{10\sqrt{5} - h} \right)

Find the time when $h = 20$ :

\frac{t}{250} = \frac{1}{20\sqrt{5}} \ln \left( \frac{10\sqrt{5} + 20}{10\sqrt{5} - 20} \right)

t = \frac{250}{20\sqrt{5}} \ln \left( \frac{10\sqrt{5} + 20}{10\sqrt{5} - 20} \right) = \frac{25}{2\sqrt{5}} \ln \left( \frac{\sqrt{5} + 2}{\sqrt{5} - 2} \right)

Rationalise the argument of the logarithm:

\frac{\sqrt{5} + 2}{\sqrt{5} - 2} \times \frac{\sqrt{5} + 2}{\sqrt{5} + 2} = \frac{(\sqrt{5} + 2)^2}{5 - 4} = 9 + 4\sqrt{5}

t = \frac{5\sqrt{5}}{2} \ln(9 + 4\sqrt{5}) \approx 16.14

Answer

t = 16.1 \text{ s}

Final answer

16.1 s

Detailed explanation

Walkthrough

We begin by separating the variables in the differential equation $\frac{dh}{dt} = \frac{500 - h^2}{250}$ to get $\frac{1}{500 - h^2} \, dh = \frac{1}{250} \, dt$ . Integrating the right side gives $\frac{t}{250} + C$ . For the left side, we recognize the standard integral form $\int \frac{1}{a^2 - x^2} \, dx = \frac{1}{2a} \ln \left| \frac{a + x}{a - x} \right|$ where $a = \sqrt{500} = 10\sqrt{5}$ .

Applying this formula gives $\frac{1}{20\sqrt{5}} \ln \left| \frac{10\sqrt{5} + h}{10\sqrt{5} - h} \right|$ . Equating the two integrated sides and using the initial condition $h = 0$ when $t = 0$ shows that the constant of integration $C = 0$ because $\ln(1) = 0$ .

To find the time when $h = 20$ , we substitute $h = 20$ into the equation and solve for $t$ . This involves simplifying the logarithmic argument, which can be done by rationalising the denominator: $\frac{10\sqrt{5} + 20}{10\sqrt{5} - 20} = \frac{\sqrt{5} + 2}{\sqrt{5} - 2} = 9 + 4\sqrt{5}$ . Multiplying by $\frac{250}{20\sqrt{5}}$ gives the final numerical value $t \approx 16.1$ seconds.

Key Takeaways

Separation of variables is a powerful technique for solving first-order differential equations.
Recognising standard integral forms, particularly $\int \frac{1}{a^2 - x^2} \, dx$ , is essential for solving these equations.
Applying initial conditions correctly determines the particular solution from the general solution.

Common Mistakes

Forgetting the absolute value signs inside the logarithm (though they can be omitted if the argument is known to be positive).
Incorrectly applying the standard integral formula, especially the coefficient $\frac{1}{2a}$ .
Algebraic errors when rationalising the fraction inside the logarithm.

Things to Be Careful About

Ensure you use the correct form of the standard integral; do not confuse it with $\int \frac{1}{a^2 + x^2} \, dx$ which gives an arctan result.
When calculating the final numerical answer, keep at least 3 significant figures during intermediate steps to avoid rounding errors.

Techniques used

separate variablesintegrate using standard logarithmic formapply initial conditionssolve for time

Q11MediumVectors

Fig. 3

Relative to the origin $O$ , the position vectors of the points $A$ , $B$ and $C$ are

\vec{OA} = 4\mathbf{i} - 2\mathbf{j}, \quad \vec{OB} = 2\mathbf{i} + 8\mathbf{j} + 4\mathbf{k}, \quad \text{and} \quad \vec{OC} = -2\mathbf{i} + 6\mathbf{k}.

The midpoint of $AB$ is $M$ , as shown in the diagram.

(a)

Find the vectors $\vec{MB}$ and $\vec{MC}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Find the position vector of the midpoint $M$ using the average of $\vec{OA}$ and $\vec{OB}$ . Then subtract $\vec{OM}$ from $\vec{OB}$ and $\vec{OC}$ to obtain $\vec{MB}$ and $\vec{MC}$ .

Working

The position vector of $M$ , the midpoint of $AB$ , is:

\vec{OM} = \frac{1}{2}(\vec{OA} + \vec{OB})

Substitute the given vectors:

\vec{OM} = \frac{1}{2}\left((4\mathbf{i} - 2\mathbf{j}) + (2\mathbf{i} + 8\mathbf{j} + 4\mathbf{k})\right)

\vec{OM} = \frac{1}{2}(6\mathbf{i} + 6\mathbf{j} + 4\mathbf{k}) = 3\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}

Now find $\vec{MB}$ :

\vec{MB} = \vec{OB} - \vec{OM} = (2\mathbf{i} + 8\mathbf{j} + 4\mathbf{k}) - (3\mathbf{i} + 3\mathbf{j} + 2\mathbf{k})

\vec{MB} = -\mathbf{i} + 5\mathbf{j} + 2\mathbf{k}

Now find $\vec{MC}$ :

\vec{MC} = \vec{OC} - \vec{OM} = (-2\mathbf{i} + 6\mathbf{k}) - (3\mathbf{i} + 3\mathbf{j} + 2\mathbf{k})

\vec{MC} = -5\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}

Answer

\vec{MB} = -\mathbf{i} + 5\mathbf{j} + 2\mathbf{k}

\vec{MC} = -5\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}

Final answer

MB = -i + 5j + 2k, MC = -5i - 3j + 4k

Detailed explanation

Walkthrough

First, we locate the midpoint $M$ of the segment $AB$ . The position vector of a midpoint is the average of the position vectors of the endpoints, so we compute $\vec{OM} = \frac{1}{2}(\vec{OA} + \vec{OB})$ . Adding the components of $\vec{OA}$ and $\vec{OB}$ gives $6\mathbf{i} + 6\mathbf{j} + 4\mathbf{k}$ , and halving this yields $\vec{OM} = 3\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}$ .

Next, to find the displacement vector from $M$ to $B$ , we subtract the position vector of the start point from the end point: $\vec{MB} = \vec{OB} - \vec{OM}$ . Subtracting component by component gives $-\mathbf{i} + 5\mathbf{j} + 2\mathbf{k}$ . Similarly, $\vec{MC} = \vec{OC} - \vec{OM}$ gives $-5\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}$ .

Key Takeaways

The position vector of a midpoint is the arithmetic mean of the position vectors of the endpoints.
A displacement vector from $P$ to $Q$ is found by $\vec{OQ} - \vec{OP}$ .

Common Mistakes

Forgetting to divide by 2 when finding the midpoint.
Subtracting in the wrong order (e.g., computing $\vec{OM} - \vec{OB}$ instead of $\vec{OB} - \vec{OM}$ ).
Dropping a component (like $\mathbf{k}$ ) when vectors have different numbers of non-zero components.

Things to Be Careful About

Ensure all three components ( $\mathbf{i}, \mathbf{j}, \mathbf{k}$ ) are accounted for, even if one is zero in the original vectors.
The mark scheme accepts column vector notation as an equivalent answer.

Techniques used

find midpoint position vectorsubtract position vectors to find displacement vectors

(b)

Calculate the exact value of the cosine of angle $CMB$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the scalar product formula $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$ to find $\cos(\angle CMB)$ . Calculate the scalar product and the magnitudes of $\vec{MB}$ and $\vec{MC}$ , then divide and simplify the surds.

Working

Compute the scalar product $\vec{MB} \cdot \vec{MC}$ :

\vec{MB} \cdot \vec{MC} = (-1)(-5) + (5)(-3) + (2)(4)

\vec{MB} \cdot \vec{MC} = 5 - 15 + 8 = -2

Compute the magnitudes:

|\vec{MB}| = \sqrt{(-1)^2 + 5^2 + 2^2} = \sqrt{1 + 25 + 4} = \sqrt{30}

|\vec{MC}| = \sqrt{(-5)^2 + (-3)^2 + 4^2} = \sqrt{25 + 9 + 16} = \sqrt{50}

Apply the cosine formula:

\cos(\angle CMB) = \frac{\vec{MB} \cdot \vec{MC}}{|\vec{MB}| |\vec{MC}|}

\cos(\angle CMB) = \frac{-2}{\sqrt{30} \cdot \sqrt{50}}

\cos(\angle CMB) = \frac{-2}{\sqrt{1500}}

Simplify $\sqrt{1500}$ :

\sqrt{1500} = \sqrt{100 \times 15} = 10\sqrt{15}

So:

\cos(\angle CMB) = \frac{-2}{10\sqrt{15}} = \frac{-1}{5\sqrt{15}}

Rationalise the denominator:

\cos(\angle CMB) = \frac{-1}{5\sqrt{15}} \times \frac{\sqrt{15}}{\sqrt{15}} = -\frac{\sqrt{15}}{75}

Answer

\cos(\angle CMB) = -\frac{\sqrt{15}}{75}

Final answer

-sqrt(15)/75

Detailed explanation

Walkthrough

To find the cosine of the angle between two vectors, we use the identity $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$ . First, we calculate the scalar (dot) product of $\vec{MB}$ and $\vec{MC}$ by multiplying corresponding components and summing: $(-1)(-5) + (5)(-3) + (2)(4) = -2$ .

Next, we find the magnitudes of each vector by taking the square root of the sum of the squares of their components: $|\vec{MB}| = \sqrt{30}$ and $|\vec{MC}| = \sqrt{50}$ .

Dividing the scalar product by the product of the magnitudes gives $\frac{-2}{\sqrt{1500}}$ . Simplifying $\sqrt{1500}$ to $10\sqrt{15}$ and reducing the fraction yields $-\frac{1}{5\sqrt{15}}$ . Finally, rationalising the denominator by multiplying top and bottom by $\sqrt{15}$ gives the exact answer $-\frac{\sqrt{15}}{75}$ .

Key Takeaways

The scalar product provides a direct way to find the angle between two vectors.
Simplifying surds step-by-step prevents arithmetic errors in the final answer.

Common Mistakes

Sign errors in the scalar product (e.g., $5 - 15 + 8$ computed as $18$ instead of $-2$ ).
Forgetting to rationalise the denominator when an exact value is required.
Not stating $\cos\theta = \dots$ explicitly before giving the final value.

Things to Be Careful About

Ensure the angle is correctly identified as $\angle CMB$ (the angle at $M$ between $MC$ and $MB$ ).
The mark scheme allows equivalent exact forms such as $-\frac{1}{5\sqrt{15}}$ , but the rationalised form is preferred.

Techniques used

compute scalar product of two vectorsfind magnitudes of vectorsapply cosine formula for angle between vectors

(c)

Hence or otherwise find the exact area of triangle $ABC$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Since $M$ is the midpoint of $AB$ , the median $CM$ divides $\triangle ABC$ into two triangles of equal area. Thus, the area of $\triangle ABC$ is twice the area of $\triangle MBC$ . Use the sine formula for the area of a triangle: $\text{Area} = \frac{1}{2}ab\sin C$ . Find $\sin(\angle CMB)$ using $\sin^2\theta + \cos^2\theta = 1$ .

Working

Since $M$ is the midpoint of $AB$ , $\text{Area}(\triangle ABC) = 2 \times \text{Area}(\triangle MBC)$ .

First, find $\sin(\angle CMB)$ using the identity $\sin^2\theta + \cos^2\theta = 1$ :

\sin^2(\angle CMB) = 1 - \cos^2(\angle CMB)

\sin^2(\angle CMB) = 1 - \left(-\frac{\sqrt{15}}{75}\right)^2 = 1 - \frac{15}{5625}

Simplify $\frac{15}{5625}$ :

\frac{15}{5625} = \frac{1}{375}

So:

\sin^2(\angle CMB) = 1 - \frac{1}{375} = \frac{374}{375}

Since $\angle CMB$ is an angle in a triangle, $\sin(\angle CMB) > 0$ :

\sin(\angle CMB) = \sqrt{\frac{374}{375}}

Now calculate the area of $\triangle MBC$ :

\text{Area}(\triangle MBC) = \frac{1}{2} |\vec{MB}| |\vec{MC}| \sin(\angle CMB)

\text{Area}(\triangle MBC) = \frac{1}{2} \sqrt{30} \cdot \sqrt{50} \cdot \sqrt{\frac{374}{375}}

Combine the square roots:

\text{Area}(\triangle MBC) = \frac{1}{2} \sqrt{30 \times 50 \times \frac{374}{375}}

\text{Area}(\triangle MBC) = \frac{1}{2} \sqrt{1500 \times \frac{374}{375}}

Simplify $\frac{1500}{375} = 4$ :

\text{Area}(\triangle MBC) = \frac{1}{2} \sqrt{4 \times 374} = \frac{1}{2} \sqrt{1496} = \frac{1}{2} \times 2\sqrt{374} = \sqrt{374}

Finally, the area of $\triangle ABC$ is:

\text{Area}(\triangle ABC) = 2 \times \sqrt{374} = 2\sqrt{374}

Answer

\text{Area}(\triangle ABC) = 2\sqrt{374}

Final answer

2sqrt(374)

Detailed explanation

Walkthrough

The key geometric insight is that a median of a triangle divides it into two smaller triangles of equal area. Since $CM$ is the median to side $AB$ , the area of $\triangle ABC$ is exactly twice the area of $\triangle MBC$ .

To find the area of $\triangle MBC$ , we use the formula $\frac{1}{2}ab\sin C$ , where $a$ and $b$ are the lengths of two sides and $C$ is the included angle. We already know $|\vec{MB}| = \sqrt{30}$ , $|\vec{MC}| = \sqrt{50}$ , and we found $\cos(\angle CMB) = -\frac{1}{5\sqrt{15}}$ in part (b).

Using $\sin^2\theta = 1 - \cos^2\theta$ , we compute $\sin^2(\angle CMB) = 1 - \frac{1}{375} = \frac{374}{375}$ , so $\sin(\angle CMB) = \sqrt{\frac{374}{375}}$ .

Substituting into the area formula: $\frac{1}{2} \sqrt{30} \sqrt{50} \sqrt{\frac{374}{375}} = \frac{1}{2} \sqrt{\frac{1500 \times 374}{375}} = \frac{1}{2} \sqrt{4 \times 374} = \sqrt{374}$ .

Doubling this gives the area of $\triangle ABC$ as $2\sqrt{374}$ .

Key Takeaways

A median divides a triangle into two equal-area sub-triangles.
The sine area formula $\frac{1}{2}ab\sin C$ is highly efficient when two sides and the included angle are known.
Combining square roots before simplifying often leads to cleaner arithmetic.

Common Mistakes

Forgetting to double the area of $\triangle MBC$ to get the area of $\triangle ABC$ .
Sign errors when computing $\sin^2\theta = 1 - \cos^2\theta$ (e.g., forgetting the square on the cosine).
Incorrectly simplifying $\frac{1500}{375}$ ; noting that $375 \times 4 = 1500$ is a helpful check.

Things to Be Careful About

Always take the positive root for $\sin\theta$ when $\theta$ is an angle in a triangle, since angles in a triangle are between $0^\circ$ and $180^\circ$ .
The mark scheme explicitly warns against changing the sign of the cosine from part (b); if done, the final answer must also be adjusted accordingly, but this is not recommended.

Techniques used

use Pythagorean identity to find sine from cosineapply sine formula for triangle areause midpoint property to relate triangle areas

Mathematics 9709/32 — October/November 2025

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