(a) Express $3x^2 - 12x + 14$ in the form $3(x + a)^2 + b$, where $a$ and $b$ are constants to be found. (b) The function $f(x) = 3x^2 - 12x + 14$ is defined for $x \ge k$, where $k$ is a constant. Find the least value of $k$ for which the function $f^{-1}$ exists. (c) For the rest of this question, you should assume that $k$ has the value found in part (b). Find an expression for $f^{-1}(x)$. (d) Hence or otherwise solve the equation $ff(x) = 29$.

(a) 3(x - 2)^2 + 2 · (b) k = 2 · (c) f^-1(x) = sqrt((x-2)/3) + 2 · (d) x = 3

The equation of a curve is $y = kx^{\frac{1}{2}} - 4x^2 + 2$, where $k$ is a constant. (a) Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ in terms of $k$. (b) It is given that $k = 2$. Find the coordinates of the stationary point and determine its nature. (c) Points $A$ and $B$ on the curve have $x$-coordinates 0.25 and 1 respectively. For a different value of $k$, the tangents to the curve at the points $A$ and $B$ meet at a point with $x$-coordinate 0.6. Find this value of…

(a) dy/dx = (1/2)kx^(-1/2) - 8x; d^2y/dx^2 = -(1/4)kx^(-3/2) - 8 · (b) (1/4, 11/4), maximum · (c) k = 3

9709/13 Oct/Nov 2024 — Mathematics Solved Past Paper

Q13MMedium-EasySeries

An arithmetic progression has fourth term 15 and eighth term 25.

Find the 30th term of the progression.

DifficultyMedium-Easy

Worked solution

Approach

Let the first term be $a$ and the common difference be $d$ . Use the nth term formula $u_n = a + (n-1)d$ to write equations for the fourth and eighth terms, then solve the simultaneous equations to find $a$ and $d$ . Finally substitute $n = 30$ .

Working

The fourth term is $u_4 = a + 3d = 15$ and the eighth term is $u_8 = a + 7d = 25$ . Subtract the first equation from the second:

a + 7d - (a + 3d) = 25 - 15

4d = 10

d = \frac{5}{2}

Substitute $d = \frac{5}{2}$ into $a + 3d = 15$ :

a + 3\left(\frac{5}{2}\right) = 15

a + \frac{15}{2} = 15

a = \frac{15}{2}

Now find the 30th term:

u_{30} = a + 29d = \frac{15}{2} + 29\left(\frac{5}{2}\right)

= \frac{15 + 145}{2} = \frac{160}{2} = 80

Answer

The 30th term is $80$ .

Final answer

80

Detailed explanation

Walkthrough

This is an arithmetic progression, so each term is obtained by adding a fixed common difference $d$ to the previous term. The nth term formula is $u_n = a + (n-1)d$ , where $a$ is the first term.

We are told the fourth term is 15. Since the fourth term corresponds to $n = 4$ , we have $a + 3d = 15$ . Similarly, the eighth term corresponds to $n = 8$ , so $a + 7d = 25$ .

These two equations have the same $a$ and $d$ , so we solve them simultaneously. Subtracting the first equation from the second removes $a$ and gives $4d = 10$ , so $d = \frac{5}{2}$ . Substituting back gives $a = \frac{15}{2}$ .

Finally, the 30th term uses $n = 30$ , so $u_{30} = a + 29d = \frac{15}{2} + 29 \cdot \frac{5}{2} = 80$ .

Key Takeaways

The nth term of an arithmetic progression is $u_n = a + (n-1)d$ . Two known terms can be converted into two equations in $a$ and $d$ , and once these are found, any term can be calculated directly.

Common Mistakes

Using $a + 30d$ instead of $a + 29d$ for the 30th term. The first term is $n = 1$ , so the 30th term uses $n - 1 = 29$ .
Solving the equations incorrectly, for example subtracting the wrong way round or making a sign error.
Treating $4d = 10$ as $d = 10$ instead of dividing by 4 to get $d = \frac{5}{2}$ .
Giving an unsupported final answer. The mark scheme awards method marks for forming the equations and for a valid method to find $a$ , $d$ , or $u_{30}$ directly.

Things to Be Careful About

The fourth term is $a + 3d$ , not $a + 4d$ ; the eighth term is $a + 7d$ , not $a + 8d$ . Keep the fractions exact: $d = \frac{5}{2}$ and $a = \frac{15}{2}$ . Check with the eighth term: $u_8 = \frac{15}{2} + 7\left(\frac{5}{2}\right) = \frac{15 + 35}{2} = 25$ , which confirms the values are correct.

Techniques used

write the nth term of an arithmetic progressionform simultaneous equations from given termssolve the equations to find the first term and common differenceevaluate a specified term

Similar questions

Q22MMedium-EasyTrigonometry

Find the exact solution of the equation

\cos \frac{1}{6}\pi + \tan 2x + \frac{\sqrt{3}}{2} = 0

for $-\frac{1}{4}\pi < x < \frac{1}{4}\pi$ .

DifficultyMedium-Easy

Worked solution

Approach

Recognise the exact value of $\cos\frac{\pi}{6}$ , substitute it into the equation, isolate $\tan 2x$ , and then solve for $x$ within the interval.

Working

\cos\frac{\pi}{6} + \tan 2x + \frac{\sqrt{3}}{2} = 0

Since $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ :

\frac{\sqrt{3}}{2} + \tan 2x + \frac{\sqrt{3}}{2} = 0

\tan 2x = -\sqrt{3}

The interval for $x$ is $-\frac{\pi}{4} < x < \frac{\pi}{4}$ , so the interval for $2x$ is $-\frac{\pi}{2} < 2x < \frac{\pi}{2}$ . In this interval the tangent function is one-to-one, and $\tan\left(-\frac{\pi}{3}\right) = -\sqrt{3}$ . Therefore

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

x = -\frac{\pi}{6}

This lies in the given range $-\frac{\pi}{4} < x < \frac{\pi}{4}$ .

Answer

x = -\frac{\pi}{6}

Final answer

x = -pi/6

Detailed explanation

Walkthrough

Start by identifying the exact value $\cos\frac{\pi}{6}$ . Since $\frac{\pi}{6}$ radians is $30^\circ$ , we have $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ . Substitute this into the given equation. The term $\frac{\sqrt{3}}{2}$ appears twice, so they combine to give $\sqrt{3}$ , leaving $\tan 2x + \sqrt{3} = 0$ . Rearranging gives $\tan 2x = -\sqrt{3}$ .

Next, solve $\tan 2x = -\sqrt{3}$ . The interval for $x$ is $-\frac{\pi}{4} < x < \frac{\pi}{4}$ , so the interval for $2x$ is $-\frac{\pi}{2} < 2x < \frac{\pi}{2}$ . On this interval the tangent function is one-to-one, so there is exactly one solution. Since $\tan\frac{\pi}{3} = \sqrt{3}$ , the value $-\sqrt{3}$ occurs at $2x = -\frac{\pi}{3}$ . Dividing by 2 gives $x = -\frac{\pi}{6}$ , which lies inside the given range.

Key Takeaways

This question tests the ability to recall exact trigonometric values and to solve a trigonometric equation within a restricted interval. It also highlights the importance of adjusting the interval when the variable is multiplied by a constant, here $2x$ , before finding the solution.

Common Mistakes

Confusing $\cos\frac{\pi}{6}$ with $\cos\frac{\pi}{3}$ : $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ , not $\frac{1}{2}$ .
Forgetting to include the negative sign when rearranging, giving $\tan 2x = \sqrt{3}$ instead of $\tan 2x = -\sqrt{3}$ .
Selecting $2x = \frac{\pi}{3}$ rather than $2x = -\frac{\pi}{3}$ , which would give an answer outside the intended interval.
The mark scheme notes that simply writing $\tan 2x = 0$ earns no method mark.

Things to Be Careful About

Always convert the interval for $x$ into an interval for $2x$ before solving.
The tangent function is one-to-one on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ , so within this interval there is only one solution.
The final answer must be exact, so keep the working in radians and use exact values rather than decimals.
Check that the final value of $x$ satisfies the original inequality.

Techniques used

evaluate the exact trigonometric valuerearrange the equation to isolate the tangent termsolve the trigonometric equation within the specified interval

Similar questions

Q3MediumSeries

(a)

Find the coefficients of $x^3$ and $x^4$ in the expansion of $(3 - ax)^5$ , where $a$ is a constant. Give your answers in terms of $a$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the general term of the binomial expansion of $(3 - ax)^5$ , then choose the value of $r$ that gives the required power of $x$ .

Working

The general term in $(3 - ax)^5$ is

\binom{5}{r}3^{5-r}(-ax)^r

For the coefficient of $x^3$ , take $r = 3$ :

\binom{5}{3}3^{2}(-a)^3 = 10 \cdot 3^2 \cdot (-a^3) = -90a^3

For the coefficient of $x^4$ , take $r = 4$ :

\binom{5}{4}3^{1}(-a)^4 = 5 \cdot 3 \cdot a^4 = 15a^4

Answer

x^3: -90a^3, \qquad x^4: 15a^4

Final answer

coefficient of x^3 is -90a^3 and coefficient of x^4 is 15a^4

Detailed explanation

Walkthrough

The expansion of $(3 - ax)^5$ has terms of the form

\binom{5}{r}3^{5-r}(-ax)^r.

Here $r$ is the power of $x$ in the term. For the coefficient of $x^3$ , set $r = 3$ . The term is

\binom{5}{3}3^{2}(-ax)^3 = 10 \cdot 9 \cdot (-a^3)x^3 = -90a^3x^3.

The negative sign appears because $(-a)^3 = -a^3$ . For the coefficient of $x^4$ , set $r = 4$ :

\binom{5}{4}3^{1}(-ax)^4 = 5 \cdot 3 \cdot a^4x^4 = 15a^4x^4.

This time $(-a)^4 = a^4$ , so the coefficient is positive.

Key Takeaways

This question uses the binomial theorem in the form $\binom{n}{k}a^{n-k}b^k$ . The key is to identify which value of $r$ gives the required power of $x$ and to handle the sign of $(-a)^r$ correctly.

Common Mistakes

Forgetting the binomial coefficient or the factor $3^{5-r}$ .
Sign errors: $(-a)^3$ is $-a^3$ , while $(-a)^4$ is $a^4$ .
Using the wrong value of $r$ for the required power of $x$ .

Things to Be Careful About

The coefficient is in terms of $a$ ; do not substitute a value for $a$ in part (a).
The mark scheme allows the coefficient to be found inside the full expansion, but the direct calculation should clearly show the coefficient.

Techniques used

apply the binomial theorem to expand a binomial poweridentify the coefficient of the x^3 termidentify the coefficient of the x^4 term

(b)

Given that the coefficient of $x^4$ in the expansion of $(ax + 7)(3 - ax)^5$ is 240, find the positive value of $a$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the coefficients found in part (a). When multiplying by $(ax + 7)$ , the $x^4$ coefficient comes from $ax$ times the $x^3$ coefficient plus $7$ times the $x^4$ coefficient. Set this equal to $240$ and solve for the positive value of $a$ .

Working

From part (a):

[x^3](3 - ax)^5 = -90a^3, \qquad [x^4](3 - ax)^5 = 15a^4.

Now multiply by $(ax + 7)$ :

\begin{aligned} [x^4](ax + 7)(3 - ax)^5 &= a(-90a^3) + 7(15a^4) \\ &= -90a^4 + 105a^4 \\ &= 15a^4. \end{aligned}

Given that this coefficient is $240$ :

15a^4 = 240

a^4 = 16

Since $a$ is positive:

a = 2

Answer

a = 2

Final answer

a = 2

Detailed explanation

Walkthrough

Write the product as

(ax + 7)(3 - ax)^5 = ax(3 - ax)^5 + 7(3 - ax)^5.

To obtain an $x^4$ term, the first part uses $ax$ times the $x^3$ term from $(3 - ax)^5$ , because $x \cdot x^3 = x^4$ . The second part uses $7$ times the $x^4$ term. From part (a):

[x^3](3 - ax)^5 = -90a^3, \qquad [x^4](3 - ax)^5 = 15a^4.

Therefore

[x^4](ax + 7)(3 - ax)^5 = a(-90a^3) + 7(15a^4) = -90a^4 + 105a^4 = 15a^4.

Given that this coefficient is $240$ :

15a^4 = 240 \quad\Rightarrow\quad a^4 = 16.

Since the question asks for the positive value, $a = 2$ .

Key Takeaways

When multiplying a binomial expansion by a linear factor, the coefficient of a given power is formed by summing contributions from pairs of terms whose powers add to that power. This question also shows how to solve a simple quartic equation by taking fourth roots and selecting the positive value.

Common Mistakes

Forgetting to multiply the $x^3$ coefficient by the factor $a$ from $ax$ .
Selecting the wrong pairs of terms, such as using the $x^2$ coefficient instead of the $x^3$ coefficient.
Including $a = -2$ as a solution; the question asks for the positive value, and the mark scheme does not accept $-2$ .

Things to Be Careful About

The coefficient of $x^3$ is negative, so the contribution from $ax$ is $-90a^4$ .
The equation $a^4 = 16$ has two real solutions, $a = \pm 2$ , but only $a = 2$ is positive.
Keep the coefficient in terms of $a$ until the final equation is formed.

Techniques used

multiply a linear factor by a binomial expansionselect the terms that contribute to x^4solve a simple quartic equation for the positive value of a

Q44MMediumQuadraticsTrigonometry

Solve the equation $4\sin^4\theta + 12\sin^2\theta - 7 = 0$ for $0^\circ \le \theta \le 360^\circ$ .

DifficultyMedium

Worked solution

Approach

Let $x = \sin^2 \theta$ . Then $\sin^4 \theta = x^2$ , so the equation becomes a quadratic in $x$ . Factorise the quadratic, discard the impossible root because $\sin^2 \theta$ cannot be negative, then solve $\sin \theta = \pm \frac{1}{\sqrt{2}}$ and list all angles in the given interval.

Working

Let

x = \sin^2 \theta

The equation becomes

4x^2 + 12x - 7 = 0

Factorise:

(2x + 7)(2x - 1) = 0

So

x = -\frac{7}{2} \quad \text{or} \quad x = \frac{1}{2}

Since $x = \sin^2 \theta \ge 0$ , discard $x = -\frac{7}{2}$ . Therefore

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

Taking square roots:

\sin \theta = \pm \frac{1}{\sqrt{2}}

For $\sin \theta = \frac{1}{\sqrt{2}}$ , the angles in $0^\circ \le \theta \le 360^\circ$ are

\theta = 45^\circ, 135^\circ

For $\sin \theta = -\frac{1}{\sqrt{2}}$ , the angles are

\theta = 225^\circ, 315^\circ

Answer

\theta = 45^\circ, 135^\circ, 225^\circ, 315^\circ

Final answer

θ = 45°, 135°, 225°, 315°

Detailed explanation

Walkthrough

The equation contains $\sin^4 \theta$ and $\sin^2 \theta$ . This is a hidden quadratic in $\sin^2 \theta$ . Let $x = \sin^2 \theta$ ; then $\sin^4 \theta = x^2$ . Substituting gives $4x^2 + 12x - 7 = 0$ .

Factorise the quadratic: $(2x + 7)(2x - 1) = 0$ . This gives $x = -\frac{7}{2}$ or $x = \frac{1}{2}$ . Since $\sin^2 \theta$ is always non-negative, $x = -\frac{7}{2}$ is impossible. So $\sin^2 \theta = \frac{1}{2}$ .

Now take square roots: $\sin \theta = \pm \frac{1}{\sqrt{2}}$ . The positive value occurs in the first and second quadrants, giving $45^\circ$ and $135^\circ$ . The negative value occurs in the third and fourth quadrants, giving $225^\circ$ and $315^\circ$ . These are all the solutions in $0^\circ \le \theta \le 360^\circ$ .

Key Takeaways

Recognise a hidden quadratic: an equation with $\sin^4 \theta$ and $\sin^2 \theta$ is quadratic in $\sin^2 \theta$ .
Use the fact that $\sin^2 \theta$ is always nonnegative to discard impossible roots.
When solving $\sin \theta = k$ , consider both the positive and negative cases and use the symmetry of the sine graph to find all angles in the interval.

Common Mistakes

Keeping $x = -\frac{7}{2}$ as a possible solution, even though $\sin^2 \theta$ cannot be negative.
Forgetting the $\pm$ sign when taking the square root of $\sin^2 \theta = \frac{1}{2}$ .
Only giving two angles instead of four. For example, writing only $45^\circ$ and $135^\circ$ , or only $45^\circ$ and $225^\circ$ .
Giving answers in radians when the question asks for degrees. The mark scheme allows radians only if all four correct angles are given in radians.

Things to Be Careful About

The interval is inclusive: $0^\circ \le \theta \le 360^\circ$ , so solutions at $0^\circ$ or $360^\circ$ would be allowed, but none occur here.
The exact value $\frac{1}{\sqrt{2}}$ corresponds to $\sin 45^\circ$ . Use exact values rather than decimals where possible.
The mark scheme awards the factorisation mark for the method, then a mark for obtaining $\sin^2 \theta = \frac{1}{2}$ and $\sin \theta = \pm \frac{1}{\sqrt{2}}$ , then marks for the angles: any two correct angles gain one mark, and all four correct with no extra angles gain the final mark.

Techniques used

substitute x = sin^2 thetafactorise the resulting quadraticdiscard the impossible roottake square roots to solve for sin thetafind all angles in the given interval

Similar questions

Q5MediumFunctions

Fig. 5.1

In the diagram, the graph with equation $y = f(x)$ is shown with solid lines and the graph with equation $y = g(x)$ is shown with broken lines.

(a)

Describe fully a sequence of three transformations which transforms the graph of $y = f(x)$ to the graph of $y = g(x)$ .

6M Similar

DifficultyMedium

Worked solution

Approach

Identify key coordinates on both graphs to determine the sequence of transformations. Track how the vertices of $y = f(x)$ map to the vertices of $y = g(x)$ through reflection, stretch, and translation.

Working

Key points on $y = f(x)$ :
$A(0, 3)$ , $B(1, 1)$ , $C(2, 1)$ , $D(4, 2)$

Key points on $y = g(x)$ :
$A'(-5, 4)$ , $B'(-3, 2)$ , $C'(-2, 2)$ , $D'(-1, 6)$

Step 1: Reflection in the y-axis
Reflecting $f(x)$ in the y-axis maps $(x, y) \to (-x, y)$ :
$A(0, 3) \to (0, 3)$
$B(1, 1) \to (-1, 1)$
$C(2, 1) \to (-2, 1)$
$D(4, 2) \to (-4, 2)$

Step 2: Stretch parallel to the y-axis
Applying a stretch with scale factor 2 parallel to the y-axis maps $(x, y) \to (x, 2y)$ :
$(0, 3) \to (0, 6)$
$(-1, 1) \to (-1, 2)$
$(-2, 1) \to (-2, 2)$
$(-4, 2) \to (-4, 4)$

Step 3: Translation
Comparing the intermediate points to $g(x)$ :
Intermediate: $(0, 6)$ , $(-1, 2)$ , $(-2, 2)$ , $(-4, 4)$
Target $g(x)$ : $(-1, 6)$ , $(-2, 2)$ , $(-3, 2)$ , $(-5, 4)$

The y-coordinates match exactly. The x-coordinates are each decreased by 1. This corresponds to a translation by the vector $\begin{pmatrix} -1 \\ 0 \end{pmatrix}$ .

The sequence of three transformations is:

Reflection in the y-axis
Stretch with scale factor 2 parallel to the y-axis
Translation by $\begin{pmatrix} -1 \\ 0 \end{pmatrix}$

Answer

Reflection in the y-axis, stretch with scale factor 2 parallel to the y-axis, translation by $\begin{pmatrix} -1 \\ 0 \end{pmatrix}$ .

Final answer

Reflection in the y-axis, stretch scale factor 2 parallel to the y-axis, translation by (-1, 0)

Detailed explanation

Walkthrough

First, we extract the key coordinates (vertices) from both graphs. For $f(x)$ , the vertices are $(0, 3)$ , $(1, 1)$ , $(2, 1)$ , and $(4, 2)$ . For $g(x)$ , the vertices are $(-5, 4)$ , $(-3, 2)$ , $(-2, 2)$ , and $(-1, 6)$ .

By comparing the y-coordinates, we see that every y-value on $g(x)$ is exactly double the corresponding y-value on $f(x)$ (e.g., $4 = 2 \times 2$ , $2 = 2 \times 1$ , $6 = 2 \times 3$ ). This immediately tells us there is a stretch with scale factor 2 parallel to the y-axis.

Next, we look at the x-coordinates. The original x-values are $0, 1, 2, 4$ , and the target x-values are $-5, -3, -2, -1$ . The target values are negative, suggesting a reflection in the y-axis. Reflecting the original points gives x-values $0, -1, -2, -4$ .

Finally, comparing the reflected x-values $(0, -1, -2, -4)$ with the target x-values $(-1, -2, -3, -5)$ , we see that each target x-value is exactly 1 less than the reflected x-value. This means we need a translation by the vector $\begin{pmatrix} -1 \\ 0 \end{pmatrix}$ .

The order of transformations matters. We applied reflection first, then stretch, then translation. This is a valid sequence that correctly maps $f(x)$ to $g(x)$ .

Key Takeaways

Always identify key points (vertices, intercepts) on both graphs to determine transformations.
Compare coordinates systematically: y-coordinates often reveal vertical stretches/compressions, while x-coordinates reveal horizontal reflections/stretches/translations.
The order of transformations is critical; applying them in a different order may yield a different result or require different translation vectors.

Common Mistakes

Giving the translation as 'left 1 unit' instead of the vector $\begin{pmatrix} -1 \\ 0 \end{pmatrix}$ or 'shift left by 1'. Mark schemes often require the vector or precise terminology.
Incorrectly ordering the transformations. For example, translating first and then reflecting will result in a different final position unless the translation vector is adjusted.
Forgetting to specify the axis of reflection (e.g., just saying 'reflection' instead of 'reflection in the y-axis').

Things to Be Careful About

Ensure the stretch is described as 'parallel to the y-axis' with the correct scale factor. A stretch parallel to the x-axis would affect the x-coordinates, not the y-coordinates.
The translation vector must be written as $\begin{pmatrix} -1 \\ 0 \end{pmatrix}$ . Writing just $-1$ or 'left' may not be accepted depending on the specific mark scheme guidance.
Verify the sequence by checking all key points. If one point doesn't map correctly, the sequence or vector is wrong.

Techniques used

identify key points on graphsdetermine reflection transformationdetermine stretch transformationdetermine translation vector

(b)

Find an expression for $g(x)$ in the form $af(bx + c)$ , where $a$ , $b$ and $c$ are integers.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the sequence of transformations found in part (a) to build the algebraic expression for $g(x)$ . Apply the transformations in order to $f(x)$ , remembering how each transformation affects the function notation.

Working

Starting with $y = f(x)$ :

1. Reflection in the y-axis
Replace $x$ with $-x$ :
$y = f(-x)$

2. Stretch with scale factor 2 parallel to the y-axis
Multiply the entire function by 2:
$y = 2f(-x)$

3. Translation by $\begin{pmatrix} -1 \\ 0 \end{pmatrix}$
Replace $x$ with $(x - (-1))$ , which is $(x + 1)$ :
$y = 2f(-(x + 1))$
$y = 2f(-x - 1)$

Thus, $g(x) = 2f(-x - 1)$ .

Comparing this to the form $af(bx + c)$ :
$a = 2$ , $b = -1$ , $c = -1$ .

Answer

g(x) = 2f(-x - 1)

where $a = 2$ , $b = -1$ , $c = -1$ .

Final answer

g(x) = 2f(-x - 1)

Detailed explanation

Walkthrough

To find the expression for $g(x)$ , we apply the transformations from part (a) sequentially to the function $f(x)$ .

Step 1: Reflection in the y-axis
A reflection in the y-axis negates the input variable $x$ . So, $f(x)$ becomes $f(-x)$ .

Step 2: Stretch with scale factor 2 parallel to the y-axis
A vertical stretch by a factor of 2 multiplies the output of the function by 2. So, $f(-x)$ becomes $2f(-x)$ .

Step 3: Translation by $\begin{pmatrix} -1 \\ 0 \end{pmatrix}$
A translation by a vector $\begin{pmatrix} h \\ k \end{pmatrix}$ replaces $x$ with $(x - h)$ and $y$ with $(y - k)$ . Here, $h = -1$ and $k = 0$ . So we replace $x$ with $(x - (-1)) = (x + 1)$ .

Substituting $(x + 1)$ into our current expression $2f(-x)$ :
$y = 2f(-(x + 1))$
Expanding the inner term:
$y = 2f(-x - 1)$

This matches the required form $af(bx + c)$ with $a = 2$ , $b = -1$ , and $c = -1$ .

Key Takeaways

Reflection in the y-axis: $f(x) \to f(-x)$
Vertical stretch by scale factor $k$ : $f(x) \to kf(x)$
Translation by $\begin{pmatrix} h \\ 0 \end{pmatrix}$ : $f(x) \to f(x - h)$
Always apply transformations in the correct order, and remember that horizontal transformations (reflections, stretches, translations) affect the $x$ inside the function argument, while vertical transformations affect the entire function output.

Common Mistakes

Forgetting to apply the translation to the $x$ inside the function argument. For example, writing $2f(-x) - 1$ instead of $2f(-x - 1)$ . The translation $\begin{pmatrix} -1 \\ 0 \end{pmatrix}$ is a horizontal shift, so it must be applied to $x$ .
Incorrectly handling the negative sign in the reflection. Reflecting $f(-x)$ and then translating by $\begin{pmatrix} -1 \\ 0 \end{pmatrix}$ requires replacing $x$ with $(x + 1)$ , giving $f(-(x + 1)) = f(-x - 1)$ . A common error is writing $f(-x + 1)$ .
Confusing the order of operations. If you translate first and then reflect, the algebraic expression will be different.

Things to Be Careful About

The form requested is $af(bx + c)$ . Ensure your final expression is fully expanded inside the function argument so it matches this form exactly. $2f(-(x + 1))$ is correct but should be simplified to $2f(-x - 1)$ to clearly show $b = -1$ and $c = -1$ .
Check the signs carefully. $b$ is the coefficient of $x$ , and $c$ is the constant term. In $-x - 1$ , $b = -1$ and $c = -1$ .

Techniques used

apply transformation sequence to function notationsubstitute into af(bx+c) form

Q65MMediumSeries

The first term of a convergent geometric progression is 10. The sum of the first 4 terms of the progression is $p$ and the sum of the first 8 terms of the progression is $q$ . It is given that $\frac{q}{p} = \frac{17}{16}$ .

Find the two possible values of the sum to infinity.

DifficultyMedium

Worked solution

Approach

Write the sums of the first 4 and first 8 terms using the geometric progression sum formula. Form the ratio $\frac{q}{p}$ , simplify it using the factorisation $1-r^8=(1-r^4)(1+r^4)$ , solve for $r$ , then use the sum to infinity formula on the convergent values of $r$ .

Working

Let the common ratio be $r$ . Since the progression is convergent, $|r| < 1$ .

The sum of the first $n$ terms is

S_n = \frac{10(1-r^n)}{1-r}

Therefore

\frac{q}{p} = \frac{S_8}{S_4} = \frac{\frac{10(1-r^8)}{1-r}}{\frac{10(1-r^4)}{1-r}} = \frac{1-r^8}{1-r^4}

Given $\frac{q}{p} = \frac{17}{16}$ :

\frac{1-r^8}{1-r^4} = \frac{17}{16}

Since $1-r^8 = (1-r^4)(1+r^4)$ , this becomes

1+r^4 = \frac{17}{16}

So

r^4 = \frac{1}{16}

Hence $r = \pm \frac{1}{2}$ . (The factorised form $(16r^4-1)(r^4-1)=0$ also gives $r^4=1$ , i.e. $r=\pm 1$ , but these are not convergent and are rejected.)

For a convergent geometric progression,

S_\infty = \frac{10}{1-r}

If $r = \frac{1}{2}$ :

S_\infty = \frac{10}{1-\frac{1}{2}} = 20

If $r = -\frac{1}{2}$ :

S_\infty = \frac{10}{1+\frac{1}{2}} = \frac{10}{\frac{3}{2}} = \frac{20}{3}

Answer

S_\infty = 20 \quad \text{or} \quad S_\infty = \frac{20}{3}

Final answer

20 and 20/3

Detailed explanation

Walkthrough

We know the first term is $10$ and the common ratio is some value $r$ . The sum of the first $n$ terms of a geometric progression is

S_n = \frac{10(1-r^n)}{1-r}

Here $p = S_4$ and $q = S_8$ . The condition $\frac{q}{p} = \frac{17}{16}$ means we should form the ratio of these two sums.

When we divide $S_8$ by $S_4$ , the common factor $\frac{10}{1-r}$ cancels, leaving

\frac{1-r^8}{1-r^4}

This is the key simplification: we do not need to know $r$ yet, but we can simplify the expression using the difference of squares factorisation

1-r^8 = (1-r^4)(1+r^4)

so the ratio becomes $1+r^4$ . Setting this equal to $\frac{17}{16}$ gives $r^4 = \frac{1}{16}$ , so $r = \pm \frac{1}{2}$ . The factorisation also produces $r^4 = 1$ , but $r = \pm 1$ would make the progression non-convergent and the sum to infinity formula invalid, so those values are rejected.

Finally, the sum to infinity of a convergent geometric progression is

S_\infty = \frac{a}{1-r}

with $a = 10$ . Substituting $r = \frac{1}{2}$ gives $20$ , and substituting $r = -\frac{1}{2}$ gives $\frac{20}{3}$ .

Key Takeaways

This question tests the formula for the sum of the first $n$ terms of a geometric progression, the sum to infinity formula, and the convergence condition $|r| < 1$ . It also tests algebraic simplification: recognising that $1-r^8$ can be factorised as $(1-r^4)(1+r^4)$ makes the ratio much easier to solve.

Common Mistakes

Forming the ratio as $\frac{p}{q}$ instead of $\frac{q}{p}$ .
Forgetting that the factor $\frac{10}{1-r}$ cancels, and trying to solve a more complicated equation.
Writing $\frac{1-r^8}{1-r^4}$ as $r^4$ directly without using the factorisation.
Only taking the positive root $r = \frac{1}{2}$ and missing $r = -\frac{1}{2}$ .
Including $r = \pm 1$ in the final answer; these are not convergent and do not give valid sums to infinity.
Giving only one value of $S_\infty$ when two are required.

Things to Be Careful About

The condition is $\frac{q}{p} = \frac{17}{16}$ , so $q$ is the sum of the first 8 terms and $p$ is the sum of the first 4 terms.
The equation $r^4 = \frac{1}{16}$ has two real solutions, $r = \pm \frac{1}{2}$ . Do not confuse this with $r^2 = \frac{1}{16}$ , which would give $r = \pm \frac{1}{4}$ .
The sum to infinity formula is only valid when $|r| < 1$ . Both $\frac{1}{2}$ and $-\frac{1}{2}$ satisfy this, but $\pm 1$ do not.
The mark scheme allows $6.67$ or better for $\frac{20}{3}$ , but the exact value is preferred.

Techniques used

form the ratio of geometric progression partial sumssimplify using the difference of powers factorisationsolve the resulting quartic equationapply the sum to infinity formula with a convergence check

Similar questions

Q7MediumCircular MeasureTrigonometry

Fig. 7.1

The diagram shows a metal plate $ABCDEF$ consisting of five parts. The parts $BCD$ and $DEF$ are semicircles. The part $BAFO$ is a sector of a circle with centre $O$ and radius 20 cm, and $D$ lies on this circle. The parts $OBD$ and $ODF$ are triangles. Angles $BOD$ and $DOF$ are both $\theta$ radians.

(a)

Given that $\theta = 1.2$ , find the area of the metal plate. Give your answer correct to 3 significant figures.

5M Similar

DifficultyMedium

Worked solution

Approach

The metal plate ABCDEF is composed of four distinct regions:

The major sector BAFO (the large circular sector with angle $2\pi - 2\theta$ )
Two triangles $\triangle OBD$ and $\triangle ODF$
Two semicircles on diameters $BD$ and $DF$

We calculate each area separately and sum them.

Working

Step 1: Area of sector BAFO

The angle $\angle BOD = \theta$ and $\angle DOF = \theta$ , so $\angle BOF = 2\theta$ . The sector BAFO is the major sector, so its angle is $2\pi - 2\theta$ .

With $\theta = 1.2$ and radius $r = 20$ :

\text{Area of sector} = \frac{1}{2} r^2 (2\pi - 2\theta) = \frac{1}{2} \times 20^2 \times (2\pi - 2.4)

= 200(2\pi - 2.4) \approx 776.63 \text{ cm}^2

Step 2: Length $BD$ (and $DF$ )

In $\triangle OBD$ , $OB = OD = 20$ and $\angle BOD = \theta = 1.2$ . Using the cosine rule:

BD^2 = 20^2 + 20^2 - 2 \times 20 \times 20 \times \cos 1.2

Alternatively, using the half-angle identity:

BD = 2 \times 20 \sin\frac{\theta}{2} = 40 \sin 0.6 \approx 22.585 \text{ cm}

Since $\angle DOF = \theta$ as well, $DF = BD = 40\sin 0.6$ .

Step 3: Area of the two semicircles

The radius of each semicircle is $\frac{BD}{2} = 20\sin 0.6$ .

\text{Area of two semicircles} = 2 \times \frac{1}{2}\pi (20\sin 0.6)^2 = \pi \times 400\sin^2 0.6

= 400\pi \sin^2 0.6 \approx 400.64 \text{ cm}^2

Step 4: Area of the two triangles

Using the formula $\frac{1}{2}ab\sin C$ for $\triangle OBD$ and $\triangle ODF$ :

\text{Area of } \triangle OBD = \frac{1}{2} \times 20 \times 20 \times \sin 1.2 = 200\sin 1.2

\text{Total area of triangles} = 2 \times 200\sin 1.2 = 400\sin 1.2 \approx 372.81 \text{ cm}^2

Step 5: Total area

\text{Total area} = 776.63 + 400.64 + 372.81 = 1550.08 \text{ cm}^2

Rounding to 3 significant figures:

\text{Total area} \approx 1550 \text{ cm}^2

Answer

1550 \text{ cm}^2

Final answer

1550

Detailed explanation

Walkthrough

Step 1: The plate is a composite shape. The large sector BAFO is the major sector of a circle of radius 20 cm, with central angle $2\pi - 2\theta$ (since $\angle BOD + \angle DOF = 2\theta$ and the sector goes the long way around through A). With $\theta = 1.2$ , this angle is $2\pi - 2.4$ . We use the sector area formula $\frac{1}{2}r^2\alpha$ .

Step 2: To find the semicircle areas, we need the diameter $BD$ . In triangle OBD, we know two sides ( $OB = OD = 20$ ) and the included angle ( $\theta = 1.2$ ). The chord length formula $BD = 2r\sin(\theta/2)$ gives $BD = 40\sin 0.6$ . This is equivalent to applying the cosine rule and simplifying using $1 - \cos\theta = 2\sin^2(\theta/2)$ .

Step 3: Each semicircle has radius $r_{\text{small}} = BD/2 = 20\sin 0.6$ . Two semicircles together form a full circle, so their combined area is $\pi r_{\text{small}}^2 = 400\pi\sin^2 0.6$ .

Step 4: The triangles OBD and ODF each have two sides of length 20 and included angle $\theta$ . Using $\frac{1}{2}ab\sin C$ , each has area $200\sin 1.2$ , giving a total of $400\sin 1.2$ .

Step 5: Sum all three contributions: sector + triangles + semicircles $= 776.63 + 372.81 + 400.64 = 1550.08$ , which rounds to $1550$ cm² to 3 significant figures.

Key Takeaways

Composite areas can be decomposed into simpler geometric shapes (sectors, triangles, circles).
The chord length formula $2r\sin(\theta/2)$ is a useful shortcut when two radii and the central angle are known.
The area formula $\frac{1}{2}ab\sin C$ is essential for triangles where two sides and the included angle are given.
Always verify which sector (major or minor) is being referenced by checking the diagram.

Common Mistakes

Using the minor sector angle $2\theta$ instead of the major sector angle $2\pi - 2\theta$ . The sector BAFO goes through A, which is the long way around.
Forgetting that two semicircles on equal diameters combine to a full circle, or miscalculating the radius as $BD$ instead of $BD/2$ .
Using $\cos$ instead of $\sin$ in the triangle area formula.
Not converting the final answer to the required number of significant figures.

Things to Be Careful About

The angle $\theta$ is in radians throughout — ensure your calculator is in radian mode.
The sector BAFO is the major sector (angle $> \pi$ ), not the minor sector between OB and OF.
When computing $BD = 40\sin(\theta/2)$ , note that $\theta/2 = 0.6$ , not $1.2$ .
The mark scheme accepts $1550.09$ rounded to 3 sf as $1550$ , but be careful with intermediate rounding — carry at least 4–5 significant figures through the calculation.

Techniques used

identify components of composite areaapply sector area formulause sine rule for triangle areaapply cosine rule for chord lengthcombine and evaluate numerically

(b)

Given instead that the area of each semicircle is $50\pi\text{ cm}^2$ , find the exact perimeter of the metal plate.

5M Similar

DifficultyMedium

Worked solution

Approach

Given the area of each semicircle is $50\pi$ cm², we first find the radius of the semicircles, then use this to determine $\theta$ . Finally, we calculate the perimeter of the metal plate from the arc lengths.

Working

Step 1: Find the radius of the semicircles

Each semicircle has area $50\pi$ :

\frac{1}{2}\pi r^2 = 50\pi

r^2 = 100 \implies r = 10 \text{ cm}

Step 2: Find $\theta$

The radius of each semicircle is half the chord length $BD$ :

r = \frac{BD}{2} = 20\sin\frac{\theta}{2}

Setting this equal to 10:

20\sin\frac{\theta}{2} = 10

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

Since $\theta$ is a positive angle less than $\pi$ :

\frac{\theta}{2} = \frac{\pi}{6} \implies \theta = \frac{\pi}{3}

Step 3: Arc length of sector BAFO

The central angle of the major sector BAFO is $2\pi - 2\theta = 2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}$ .

\text{Arc length} = r\alpha = 20 \times \frac{4\pi}{3} = \frac{80\pi}{3} \text{ cm}

Step 4: Arc lengths of the semicircles

Each semicircular arc has length $\pi r = 10\pi$ cm. There are two such arcs (BCD and DEF):

\text{Total semicircular arc length} = 2 \times 10\pi = 20\pi \text{ cm}

Step 5: Total perimeter

The perimeter of the metal plate consists of the major arc BAFO and the two semicircular arcs:

\text{Perimeter} = \frac{80\pi}{3} + 20\pi = \frac{80\pi}{3} + \frac{60\pi}{3} = \frac{140\pi}{3} \text{ cm}

Answer

\frac{140\pi}{3} \text{ cm}

Final answer

140π/3

Detailed explanation

Walkthrough

Step 1: From the semicircle area formula $\frac{1}{2}\pi r^2 = 50\pi$ , we solve directly for $r = 10$ cm. This is the radius of each small semicircle.

Step 2: The chord $BD$ is the diameter of the semicircle, so $BD = 2r = 20$ cm. In triangle OBD with $OB = OD = 20$ and $\angle BOD = \theta$ , the chord length is $BD = 2 \times 20 \times \sin(\theta/2) = 40\sin(\theta/2)$ . Setting $40\sin(\theta/2) = 20$ gives $\sin(\theta/2) = 1/2$ , so $\theta/2 = \pi/6$ and $\theta = \pi/3$ .

Step 3: The perimeter of the plate is formed by three curved arcs only — the straight lines $OB$ , $OD$ , $OF$ , $BD$ , and $DF$ are all internal to the plate. The boundary consists of:

The major arc from B through A to F: length $= 20(2\pi - 2\theta) = 20(2\pi - 2\pi/3) = 20(4\pi/3) = 80\pi/3$
The semicircular arc from B through C to D: length $= \pi r = 10\pi$
The semicircular arc from D through E to F: length $= \pi r = 10\pi$

Step 4: Adding these: $80\pi/3 + 10\pi + 10\pi = 80\pi/3 + 20\pi = 80\pi/3 + 60\pi/3 = 140\pi/3$ cm.

Key Takeaways

The perimeter of a composite shape may consist entirely of curved arcs if straight edges are internal.
Working backwards from area to find an angle, then using that angle for perimeter calculations, is a common two-stage approach.
Exact values like $\sin(\pi/6) = 1/2$ should be used to give exact answers in terms of $\pi$ .
Always identify which edges form the boundary versus which are internal to the shape.

Common Mistakes

Including straight internal edges (like $BD$ , $DF$ , $OB$ , $OF$ ) in the perimeter calculation. These are not on the boundary of the plate.
Using the wrong sector angle — the arc BAFO is the major arc with angle $2\pi - 2\theta$ , not $2\theta$ .
Forgetting that the perimeter of a semicircular arc (not the full semicircle) is $\pi r$ , not $\pi r + 2r$ .
Not expressing the final answer as a single exact term involving $\pi$ .

Things to Be Careful About

The question asks for the perimeter of the metal plate, which is the outer boundary only. All internal lines (radii and chords) are not part of the perimeter.
The answer must be a single exact term — combine the fractions over a common denominator.
$\theta = \pi/3$ is in radians, so all arc length calculations use radians directly: $s = r\alpha$ .
The mark scheme requires the answer as $\frac{140\pi}{3}$ or equivalently $46\frac{2}{3}\pi$ ; decimal approximations are not accepted.

Techniques used

solve for radius from semicircle areause inverse sine to find thetacalculate arc length of major sectorsum arc lengths for total perimeter

Q8MediumQuadraticsFunctions

(a)

Express $3x^2 - 12x + 14$ in the form $3(x + a)^2 + b$ , where $a$ and $b$ are constants to be found.

2M Similar

(b)

The function $f(x) = 3x^2 - 12x + 14$ is defined for $x \ge k$ , where $k$ is a constant.

Find the least value of $k$ for which the function $f^{-1}$ exists.

1M Similar

(c)

For the rest of this question, you should assume that $k$ has the value found in part (b).

Find an expression for $f^{-1}(x)$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Set $y = f(x)$ using the completed square form, rearrange to make $x$ the subject, and use the restricted domain $x \geq 2$ to choose the positive square root. Finally replace $y$ by $x$ to write the inverse function.

Working

Let

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

Subtract $2$ and divide by $3$ :

(x-2)^2 = \frac{y-2}{3}

Since the domain is $x \geq 2$ , take the positive square root:

x - 2 = \sqrt{\frac{y-2}{3}}

So

x = \sqrt{\frac{y-2}{3}} + 2

Interchanging $x$ and $y$ gives the inverse function:

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

Answer

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

Final answer

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

Detailed explanation

Walkthrough

To find the inverse, write $y=f(x)$ using the completed square form. Rearrange to isolate $x$ . Subtract $2$ from both sides and divide by $3$ to get $(x-2)^2=(y-2)/3$ . Since the domain was restricted to $x\ge2$ , $x-2$ is non-negative, so take the positive square root: $x-2=\sqrt{(y-2)/3}$ . Add $2$ to get $x$ in terms of $y$ . Finally, replace $y$ with $x$ to write the inverse function in the usual notation.

Key Takeaways

Finding an inverse involves swapping the roles of input and output. The restricted domain determines which square root to take. The completed-square form makes the rearrangement straightforward.

Common Mistakes

Taking the negative square root, which would give the inverse for the wrong branch. Forgetting to swap $x$ and $y$ at the end. Omitting the $+2$ when writing the final expression.

Things to Be Careful About

The positive root is required because $x\ge2$ . The final expression can also be written as $\frac{\sqrt{3x-6}}{3}+2$ , which is equivalent to the answer above.

Techniques used

set y equal to the completed-square formrearrange to isolate x - 2take the positive square root using the restricted domainswap variables to express the inverse

(d)

Hence or otherwise solve the equation $ff(x) = 29$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Since $f$ has an inverse on the restricted domain, apply $f^{-1}$ to both sides of $ff(x)=29$ . This gives $f(x)=f^{-1}(29)$ . Evaluate $f^{-1}(29)$ , then solve the resulting quadratic equation, discarding any solution outside the domain $x\ge2$ .

Working

Using the inverse from part (c):

f^{-1}(29) = \sqrt{\frac{29-2}{3}} + 2 = \sqrt{9} + 2 = 3 + 2 = 5

Therefore

f(x) = 5

Using the completed square form:

3(x-2)^2 + 2 = 5

Subtract $2$ and divide by $3$ :

(x-2)^2 = 1

So

x - 2 = \pm 1

giving $x = 3$ or $x = 1$ . Since the domain of $f$ is $x \ge 2$ , discard $x = 1$ .

Answer

x = 3

Final answer

x = 3

Detailed explanation

Walkthrough

The equation $ff(x)=29$ means apply $f$ twice to $x$ and get $29$ . Because $f$ is one-one on the restricted domain, we can apply $f^{-1}$ to both sides: $f(x)=f^{-1}(29)$ . Evaluate $f^{-1}(29)$ using the formula from part (c): $\sqrt{(29-2)/3}+2 = \sqrt{9}+2 = 5$ . So we need $f(x)=5$ . Using $f(x)=3(x-2)^2+2$ , solve $3(x-2)^2+2=5$ , giving $(x-2)^2=1$ , so $x=3$ or $x=1$ . Since $f$ is only defined for $x\ge2$ , $x=1$ is not in the domain and must be discarded. The solution is $x=3$ .

Key Takeaways

Applying an inverse function to both sides of an equation is a powerful way to solve composite equations. Always check solutions against the domain of the original function.

Common Mistakes

Forgetting to discard $x=1$ when using the direct $f(x)$ method. Alternatively, expanding $ff(x)$ and solving a quartic without noticing the domain restriction can produce extra roots. The mark scheme allows either method but requires $x=1$ to be discarded if it appears.

Things to Be Careful About

The domain restriction is essential: $x=1$ is algebraically valid but outside the domain. When using the inverse method, the inverse formula is only valid for $x\ge2$ . Be careful with the order of operations when evaluating $f^{-1}(29)$ .

Techniques used

apply the inverse function to both sidesevaluate the inverse at a numbersolve the resulting quadratic equationdiscard a solution outside the domain

Q9MediumQuadraticsIntegration

Fig. 9.1

The diagram shows the curves with equations $y = x^3 - 3x + 3$ and $y = 2x^3 - 4x^2 + 3$ .

(a)

Find the $x$ -coordinates of the points of intersection of the curves.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Set the two curve equations equal to each other and solve the resulting equation for $x$ .

Working

x^3 - 3x + 3 = 2x^3 - 4x^2 + 3

Rearranging all terms to one side:

x^3 - 4x^2 + 3x = 0

Factor out the common factor $x$ :

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

Factor the quadratic expression:

x(x - 1)(x - 3) = 0

Setting each factor to zero gives:

x = 0, \quad x = 1, \quad x = 3

Answer

x = 0, \; x = 1, \; x = 3

Final answer

x = 0, x = 1, x = 3

Detailed explanation

Walkthrough

To find the points of intersection of two curves, we set their equations equal to each other. This gives $x^3 - 3x + 3 = 2x^3 - 4x^2 + 3$ . Moving all terms to one side simplifies to $x^3 - 4x^2 + 3x = 0$ . We notice every term contains a factor of $x$ , so we factor it out to get $x(x^2 - 4x + 3) = 0$ . The quadratic $x^2 - 4x + 3$ factors into $(x-1)(x-3)$ since $1 \times 3 = 3$ and $1 + 3 = 4$ . Setting each factor to zero yields the three intersection x-coordinates: $x = 0$ , $x = 1$ , and $x = 3$ .

Key Takeaways

Intersection points of curves are found by equating their equations.
Cubic equations with a common factor can be reduced to a linear times a quadratic.
Factorising quadratics by finding two numbers that multiply to the constant term and add to the coefficient of $x$ .

Common Mistakes

Forgetting to move all terms to one side before factoring.
Not factoring out the common $x$ term, leading to an incomplete solution (missing $x = 0$ ).
Incorrect factorisation of the quadratic $x^2 - 4x + 3$ .

Things to Be Careful About

The mark scheme notes that $x = 0$ may be seen in the working and is a valid answer even if not explicitly stated as a final answer.
Ensure all three roots are found; awarding only $x = 1$ and $x = 3$ without the method marks is not acceptable.

Techniques used

set curve equations equal to find intersectionsfactor out common term from cubicsolve resulting quadratic equation

(b)

Find the area of the shaded region.

4M Similar

DifficultyMedium

Worked solution

Approach

From the diagram, the shaded region lies between $x = 1$ and $x = 3$ . The curve $y = x^3 - 3x + 3$ is above $y = 2x^3 - 4x^2 + 3$ in this interval. The area is found by integrating the difference (upper curve minus lower curve) from $x = 1$ to $x = 3$ .

Working

\text{Area} = \int_1^3 \left[(x^3 - 3x + 3) - (2x^3 - 4x^2 + 3)\right] dx

Simplify the integrand:

= \int_1^3 (-x^3 + 4x^2 - 3x) \, dx

Integrate term by term:

= \left[-\frac{x^4}{4} + \frac{4x^3}{3} - \frac{3x^2}{2}\right]_1^3

Substitute the upper limit $x = 3$ :

-\frac{3^4}{4} + \frac{4 \cdot 3^3}{3} - \frac{3 \cdot 3^2}{2} = -\frac{81}{4} + \frac{108}{3} - \frac{27}{2}

= -\frac{81}{4} + 36 - \frac{27}{2} = \frac{-81 + 144 - 54}{4} = \frac{9}{4}

Substitute the lower limit $x = 1$ :

-\frac{1^4}{4} + \frac{4 \cdot 1^3}{3} - \frac{3 \cdot 1^2}{2} = -\frac{1}{4} + \frac{4}{3} - \frac{3}{2}

= \frac{-3 + 16 - 18}{12} = -\frac{5}{12}

Compute the area:

\text{Area} = \frac{9}{4} - \left(-\frac{5}{12}\right) = \frac{9}{4} + \frac{5}{12} = \frac{27}{12} + \frac{5}{12} = \frac{32}{12} = \frac{8}{3}

Answer

\frac{8}{3}

Final answer

8/3

Detailed explanation

Walkthrough

The shaded region is bounded by the two curves between their intersection points at $x = 1$ and $x = 3$ . From the diagram, $y = x^3 - 3x + 3$ is the upper curve and $y = 2x^3 - 4x^2 + 3$ is the lower curve in this interval. The area between two curves $f(x)$ and $g(x)$ where $f(x) \geq g(x)$ on $[a, b]$ is $\int_a^b [f(x) - g(x)] \, dx$ .

Subtracting the lower curve from the upper curve: $(x^3 - 3x + 3) - (2x^3 - 4x^2 + 3) = -x^3 + 4x^2 - 3x$ .

Integrating term by term: $\int (-x^3 + 4x^2 - 3x) \, dx = -\frac{x^4}{4} + \frac{4x^3}{3} - \frac{3x^2}{2} + C$ .

Evaluating at $x = 3$ : $-\frac{81}{4} + 36 - \frac{27}{2} = \frac{-81 + 144 - 54}{4} = \frac{9}{4}$ .

Evaluating at $x = 1$ : $-\frac{1}{4} + \frac{4}{3} - \frac{3}{2} = \frac{-3 + 16 - 18}{12} = -\frac{5}{12}$ .

Subtracting: $\frac{9}{4} - \left(-\frac{5}{12}\right) = \frac{27}{12} + \frac{5}{12} = \frac{32}{12} = \frac{8}{3}$ .

Key Takeaways

The area between two curves is $\int_a^b [\text{upper} - \text{lower}] \, dx$ .
Always verify which curve is on top in the interval of integration.
Careful fraction arithmetic is essential when evaluating at multiple limits.
The constant of integration cancels out in definite integrals.

Common Mistakes

Using the wrong limits (e.g., including $x = 0$ which is outside the shaded region).
Subtracting the curves in the wrong order, giving a negative area.
Arithmetic errors when evaluating fractions at the limits.
Forgetting to substitute both limits and subtract.

Things to Be Careful About

The mark scheme explicitly states not to allow $x = 0$ as a limit; only $x = 1$ and $x = 3$ are correct for the shaded region.
If two separate integrals are used, substitution must be applied to both.
One sign error is allowed in each expression if brackets are shown, but the final answer must be correct.
Accept AWRT 2.67 as equivalent to $\frac{8}{3}$ .

Techniques used

identify upper and lower curves from diagramset up definite integral of difference of functionsevaluate antiderivative at upper and lower limitssimplify fraction arithmetic

Q10MediumCoordinate GeometryQuadratics

Points $A$ and $B$ have coordinates $(4, 3)$ and $(8, -5)$ respectively. A circle with radius 10 passes through the points $A$ and $B$ .

(a)

Show that the centre of the circle lies on the line $y = \frac{1}{2}x - 4$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

The centre of a circle is equidistant from $A$ and $B$ , so it lies on the perpendicular bisector of $AB$ . Find the midpoint and gradient of $AB$ , then use the perpendicular gradient to write the equation of the perpendicular bisector.

Working

Gradient of $AB$ :

m_{AB} = \frac{-3 - 3}{8 - 4} = \frac{-8}{4} = -2

Midpoint of $AB$ :

M = \left( \frac{3 + 4}{2}, \frac{-5 + 3}{2} \right) = (6, -1)

The perpendicular bisector has gradient the negative reciprocal of $m_{AB}$ :

m_{\perp} = -\frac{1}{-2} = \frac{1}{2}

Using point-slope form through $(6, -1)$ :

y + 1 = \frac{1}{2}(x - 6)

Simplify:

y = \frac{1}{2}x - 3 - 1 = \frac{1}{2}x - 4

Answer

The centre of the circle lies on the line $y = \frac{1}{2}x - 4$ .

Final answer

y = 1/2 x - 4

Detailed explanation

Walkthrough

Since $A$ and $B$ are both on the circle, the centre is the same distance from both points. The set of points equidistant from $A$ and $B$ is the perpendicular bisector of the segment $AB$ . We first find the gradient of $AB$ using the coordinates. Then we find the midpoint of $AB$ , because the perpendicular bisector passes through the midpoint. The perpendicular bisector is perpendicular to $AB$ , so its gradient is the negative reciprocal of the gradient of $AB$ . Using the point-slope form of a line with this gradient and the midpoint gives the required line. Simplifying the equation shows that the centre lies on $y = \frac{1}{2}x - 4$ .

Key Takeaways

The perpendicular bisector of any chord of a circle passes through the centre of the circle. To find it, find the gradient and midpoint of the chord, then use the negative reciprocal gradient through the midpoint.

Common Mistakes

Forgetting to take the negative reciprocal of the gradient of $AB$ .
Using the midpoint incorrectly or not finding it.
Substituting the midpoint into the wrong line form.
For part (a), the answer is given, so the working must be shown; an unsupported answer is not accepted.

Things to Be Careful About

The coordinates of $A$ and $B$ must be substituted in the correct order: $A = (4,3)$ and $B = (8,-5)$ . When simplifying $y + 1 = \frac{1}{2}(x - 6)$ , remember to subtract $3$ from $1$ correctly to obtain $\frac{1}{2}x - 4$ .

Techniques used

calculate the gradient of a line segmentfind the midpoint of a segmentuse the negative reciprocal gradient for a perpendicular linewrite the equation of a line through a given point

(b)

Find the two possible equations of the circle.

5M Similar

DifficultyMedium

Worked solution

Approach

From part (a), the centre lies on $y = \frac{1}{2}x - 4$ . Write the centre as $C\left(a, \frac{1}{2}a - 4\right)$ . Since the radius is 10, the square of the distance from $C$ to $A$ is 100. Substitute the coordinates of $A$ into the circle equation, solve the resulting quadratic for $a$ , and then write the two possible circle equations.

Working

Let the centre be $C\left(a, \frac{1}{2}a - 4\right)$ . Using point $A(4,3)$ :

(4 - a)^2 + \left(3 - \left(\frac{1}{2}a - 4\right)\right)^2 = 100

Simplify the second bracket:

3 - \left(\frac{1}{2}a - 4\right) = 7 - \frac{1}{2}a

So:

(4 - a)^2 + \left(7 - \frac{1}{2}a\right)^2 = 100

Expand:

(4 - a)^2 = a^2 - 8a + 16

\left(7 - \frac{1}{2}a\right)^2 = 49 - 7a + \frac{1}{4}a^2

Therefore:

a^2 - 8a + 16 + 49 - 7a + \frac{1}{4}a^2 = 100

\frac{5}{4}a^2 - 15a + 65 = 100

\frac{5}{4}a^2 - 15a - 35 = 0

Multiply through by 4:

5a^2 - 60a - 140 = 0

Divide by 5:

a^2 - 12a - 28 = 0

Factorise:

(a - 14)(a + 2) = 0

So $a = 14$ or $a = -2$ .

Corresponding $y$ -coordinates:

a = 14: \quad y = \frac{1}{2}(14) - 4 = 3

a = -2: \quad y = \frac{1}{2}(-2) - 4 = -5

Thus the centres are $(14,3)$ and $(-2,-5)$ . With radius 10, the two circle equations are:

(x - 14)^2 + (y - 3)^2 = 100

and

(x + 2)^2 + (y + 5)^2 = 100

Answer

The two possible equations are $(x - 14)^2 + (y - 3)^2 = 100$ and $(x + 2)^2 + (y + 5)^2 = 100$ .

Final answer

(x - 14)^2 + (y - 3)^2 = 100 and (x + 2)^2 + (y + 5)^2 = 100

Detailed explanation

Walkthrough

From part (a), the centre lies on the line $y = \frac{1}{2}x - 4$ , so any centre can be written as $\left(a, \frac{1}{2}a - 4\right)$ . The circle has radius 10, so the square of the distance from the centre to any point on the circle is 100. Using point $A(4,3)$ , we get an equation in $a$ . Expand both squared brackets, combine like terms, and simplify to a quadratic. Solve the quadratic to find two values of $a$ . For each value, compute the corresponding $y$ -coordinate from the line. Then write the equation of a circle with each centre and radius 10.

Key Takeaways

The equation of a circle with centre $(h,k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$ . When the centre is constrained to lie on a line, substitute the line equation into the circle equation to get a single-variable equation. A quadratic produces two possible circles because two circles of the same radius can pass through the same two points.

Common Mistakes

Forgetting to use the radius squared: the right-hand side must be $100$ , not $10$ .
Sign errors when expanding $(4 - a)^2$ or $\left(7 - \frac{1}{2}a\right)^2$ .
Losing one of the two solutions when solving the quadratic.
Not using the line equation to find the $y$ -coordinate of each centre.
Writing the final equation with the wrong signs for the centre coordinates.

Things to Be Careful About

Check that both centres satisfy the line from part (a): $(14,3)$ and $(-2,-5)$ both give $y = \frac{1}{2}x - 4$ . When writing the final circle equations, match the signs to the centre: for centre $(-2,-5)$ , the equation is $(x + 2)^2 + (y + 5)^2 = 100$ , not $(x - 2)^2 + (y - 5)^2 = 100$ .

Techniques used

parametrize the centre using the equation of a linesubstitute a point into the circle equationexpand and simplify a quadratic equationsolve a quadratic by factorisationwrite circle equations from centres and radius

Q11Medium-HardDifferentiation

The equation of a curve is $y = kx^{\frac{1}{2}} - 4x^2 + 2$ , where $k$ is a constant.

(a)

Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ in terms of $k$ .

2M Similar

DifficultyEasy

Worked solution

Approach

Differentiate $y = kx^{1/2} - 4x^2 + 2$ term by term using the power rule, then differentiate the result to find the second derivative.

Working

Given

y = kx^{1/2} - 4x^2 + 2.

Differentiate with respect to $x$ :

\frac{dy}{dx} = \frac{1}{2}kx^{-1/2} - 8x.

Differentiate again:

\frac{d^2y}{dx^2} = -\frac{1}{4}kx^{-3/2} - 8.

Answer

\frac{dy}{dx} = \frac{1}{2}kx^{-1/2} - 8x, \qquad \frac{d^2y}{dx^2} = -\frac{1}{4}kx^{-3/2} - 8.

Final answer

dy/dx = (1/2)kx^(-1/2) - 8x; d^2y/dx^2 = -(1/4)kx^(-3/2) - 8

Detailed explanation

Walkthrough

The curve is a sum of three terms. Differentiate term by term: the derivative of $kx^{1/2}$ is $\frac{1}{2}kx^{-1/2}$ because the constant $k$ is carried along; the derivative of $-4x^2$ is $-8x$ ; and the derivative of the constant $2$ is $0$ . Then differentiate $\frac{dy}{dx}$ to get the second derivative: the derivative of $\frac{1}{2}kx^{-1/2}$ is $\frac{1}{2}k\left(-\frac{1}{2}\right)x^{-3/2} = -\frac{1}{4}kx^{-3/2}$ , and the derivative of $-8x$ is $-8$ .

Key Takeaways

This question tests the power rule for fractional powers, the fact that constants multiply through derivatives, and the idea that the second derivative is just the derivative of the first derivative.

Common Mistakes

A common error is forgetting that the derivative of the constant $2$ is $0$ . Another is subtracting the exponent incorrectly: $x^{1/2}$ becomes $x^{-1/2}$ , not $x^{3/2}$ . Sign errors are also common when differentiating $x^{-1/2}$ to $x^{-3/2}$ .

Things to Be Careful About

Keep $k$ as an unknown constant throughout. Use fractional powers consistently. The mark scheme awards one mark for each correct derivative, so both $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ must be stated clearly.

Techniques used

differentiate powers of x term by termapply constant factor ruledifferentiate the first derivative to obtain the second derivative

(b)

It is given that $k = 2$ .

Find the coordinates of the stationary point and determine its nature.

4M Similar

DifficultyMedium

Worked solution

Approach

Set $k=2$ , so $\frac{dy}{dx} = x^{-1/2} - 8x$ . At a stationary point the first derivative is zero. Solve this equation for $x$ , find the corresponding $y$ , then use the sign of the second derivative to determine the nature.

Working

With $k=2$ , the curve is

y = 2x^{1/2} - 4x^2 + 2.

From part (a),

\frac{dy}{dx} = x^{-1/2} - 8x.

Set $\frac{dy}{dx}=0$ :

x^{-1/2} - 8x = 0.

Multiply through by $x^{1/2}$ :

1 - 8x^{3/2} = 0.

So

x^{3/2} = \frac{1}{8}.

Raise both sides to the power $\frac{2}{3}$ :

x = \left(\frac{1}{8}\right)^{2/3} = \frac{1}{4}.

Now find $y$ :

y = 2\left(\frac{1}{4}\right)^{1/2} - 4\left(\frac{1}{4}\right)^2 + 2 = 2\left(\frac{1}{2}\right) - 4\left(\frac{1}{16}\right) + 2 = 1 - \frac{1}{4} + 2 = \frac{11}{4}.

For the nature, with $k=2$ ,

\frac{d^2y}{dx^2} = -\frac{1}{2}x^{-3/2} - 8.

At $x=\frac{1}{4}$ :

\frac{d^2y}{dx^2} = -\frac{1}{2}\left(\frac{1}{4}\right)^{-3/2} - 8.

Since $\left(\frac{1}{4}\right)^{-3/2} = 8$ ,

\frac{d^2y}{dx^2} = -\frac{1}{2}(8) - 8 = -4 - 8 = -12 < 0.

Therefore the stationary point is a maximum.

Answer

The stationary point is $\left(\frac{1}{4}, \frac{11}{4}\right)$ and it is a maximum.

Final answer

(1/4, 11/4), maximum

Detailed explanation

Walkthrough

Substitute $k=2$ into the first derivative found in part (a). At a stationary point, $\frac{dy}{dx}=0$ , so set $x^{-1/2} - 8x = 0$ . Multiplying by $x^{1/2}$ gives $1 - 8x^{3/2}=0$ , which avoids introducing an unwanted $x=0$ solution. Then $x^{3/2}=\frac{1}{8}$ , so raising both sides to the power $\frac{2}{3}$ gives $x=\frac{1}{4}$ . Substitute this into the curve equation to get $y=\frac{11}{4}$ . Finally, substitute $x=\frac{1}{4}$ into $\frac{d^2y}{dx^2} = -\frac{1}{2}x^{-3/2}-8$ ; the result is $-12<0$ , so the point is a maximum.

Key Takeaways

This question combines solving a fractional-power equation with the second-derivative test for stationary points. It also reinforces that a stationary point requires both coordinates, and that the sign of the second derivative determines whether it is a maximum or minimum.

Common Mistakes

The mark scheme warns that squaring $x^{-1/2} - 8x = 0$ directly to $x^{-1} - 64x^2 = 0$ is not a valid method and scores M0. Including $x=0$ as a solution is also penalised because $x^{-1/2}$ is undefined at $x=0$ . Another common mistake is stopping after finding $x$ and not calculating $y$ or determining the nature.

Things to Be Careful About

Use the multiplication by $x^{1/2}$ method rather than squaring. Remember that $x=0$ is not in the domain of the original derivative. The mark scheme allows follow-through on the second derivative using your own $x$ -value, but only if $x=0$ is not the only solution.

Techniques used

set the first derivative equal to zerosolve an equation with fractional powersevaluate the y-coordinate of the stationary pointuse the second derivative to determine the nature

(c)

Points $A$ and $B$ on the curve have $x$ -coordinates 0.25 and 1 respectively. For a different value of $k$ , the tangents to the curve at the points $A$ and $B$ meet at a point with $x$ -coordinate 0.6.

Find this value of $k$ .

6M Similar

DifficultyMedium-Hard

Worked solution

Approach

For each tangent, find the y-coordinate and gradient at the given x-coordinate, write the tangent equation, then substitute $x=0.6$ into both equations and solve for $k$ .

Working

The curve is

y = kx^{1/2} - 4x^2 + 2, \qquad \frac{dy}{dx} = \frac{1}{2}kx^{-1/2} - 8x.

At $x=1$ :

y = k - 2, \qquad \frac{dy}{dx} = \frac{1}{2}k - 8.

Tangent at $A$ :

y - (k-2) = \left(\frac{1}{2}k - 8\right)(x - 1).

Equivalently,

y = \left(\frac{1}{2}k - 8\right)x + \frac{1}{2}k + 6.

At $x=\frac{1}{4}$ :

y = \frac{1}{2}k + \frac{7}{4}, \qquad \frac{dy}{dx} = k - 2.

Tangent at $B$ :

y - \left(\frac{1}{2}k + \frac{7}{4}\right) = (k-2)\left(x - \frac{1}{4}\right).

Equivalently,

y = (k-2)x + \frac{1}{4}k + \frac{9}{4}.

The tangents meet at $x = 0.6$ . Substitute $x = 0.6$ into both tangent equations and equate:

\left(\frac{1}{2}k - 8\right)(0.6) + \frac{1}{2}k + 6 = (k-2)(0.6) + \frac{1}{4}k + \frac{9}{4}.

Simplify:

0.3k - 4.8 + 0.5k + 6 = 0.6k - 1.2 + 0.25k + 2.25.

0.8k + 1.2 = 0.85k + 1.05.

0.15 = 0.05k.

k = 3.

Answer

k = 3.

Final answer

k = 3

Detailed explanation

Walkthrough

For each of the two points, compute the $y$ -coordinate from the curve and the gradient from $\frac{dy}{dx}$ . At $x=1$ , the point is $(1, k-2)$ and the gradient is $\frac{1}{2}k - 8$ . Use point-slope form to write the tangent. At $x=\frac{1}{4}$ , the point is $\left(\frac{1}{4}, \frac{1}{2}k + \frac{7}{4}\right)$ and the gradient is $k-2$ ; write its tangent similarly. Because the two tangents meet at $x=0.6$ , substitute $x=0.6$ into both tangent equations and set the right-hand sides equal. This gives a linear equation in $k$ ; solving it gives $k=3$ .

Key Takeaways

This problem tests the application of differentiation to tangents: finding the gradient at a point, writing the tangent equation, and using the intersection of two lines to set up an equation for an unknown parameter.

Common Mistakes

The mark scheme warns that if the constants in both tangent equations are taken to be the same, the method mark is lost. It is also a common error to forget to substitute $x=0.6$ after equating, or to make arithmetic slips with decimals and fractions. Unsupported answers are not accepted, so all tangent equations and the substitution must be shown.

Things to Be Careful About

Check that each tangent uses the correct point and gradient. Simplify the tangent equations carefully before substituting. Use consistent arithmetic with $0.6 = \frac{3}{5}$ if preferred. The final value $k=3$ must be reached by a clear method; the mark scheme awards method marks for each tangent equation and for equating them at $x=0.6$ .

Techniques used

evaluate the y-coordinate and gradient at a given x-coordinatewrite the equation of a tangent using point-slope formsubstitute a common x-coordinate into both tangent equationssolve a linear equation for the unknown parameter

Mathematics 9709/13 — October/November 2024

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