(a) By sketching a suitable pair of graphs, show that the equation $\cot 2x = \sec x$ has exactly one root in the interval $0 < x < \frac{1}{2}\pi$. (b) Show that if a sequence of real values given by the iterative formula $$ x_{n+1} = \frac{1}{2}\tan^{-1}(\cos x_n) $$ converges, then it converges to the root in part (a).

(a) Exactly one root in 0 < x < π/2 · (b) The sequence converges to the root of cot 2x = sec x in 0 < x < π/2

(a) The complex number $u$ is given by $$ u = \frac{(\cos \frac{1}{7}\pi + i\sin \frac{1}{7}\pi)^4}{\cos \frac{1}{7}\pi - i\sin \frac{1}{7}\pi} $$ Find the exact value of $\arg u$. (b) The complex numbers $u$ and $u^ $ are plotted on an Argand diagram. Describe the single geometrical transformation that maps $u$ onto $u^ $ and state the exact value of $\arg u^ $.

(a) arg u = 5π/7 · (b) Reflection in the real axis; arg u = -5π/7

(a) Show that the equation $\tan^3 x + 2\tan 2x - \tan x = 0$ may be expressed as $$ \tan^4 x - 2\tan^2 x - 3 = 0 $$ for $\tan x \neq 0$. (b) Hence solve the equation $\tan^3 2\theta + 2\tan 4\theta - \tan 2\theta = 0$ for $0 < \theta < \pi$. Give your answers in exact form.

(a) tan^4 x - 2 tan^2 x - 3 = 0 for tan x ≠ 0 · (b) θ = π/6, π/3, 2π/3, 5π/6

The parametric equations of a curve are $$ x = \tan^2 2t, \quad y = \cos 2t, $$ for $0 < t < \frac{1}{4}\pi$. (a) Show that $\frac{dy}{dx} = -\frac{1}{2}\cos^3 2t$. (b) Hence find the equation of the normal to the curve at the point where $t = \frac{1}{8}\pi$. Give your answer in the form $y = mx + c$.

(a) -1/2 cos^3 2t · (b) y = 4√2x - 7√2/2

Let $f(x) = \frac{2e^{2x}}{e^{2x} - 3e^x + 2}$. (a) Find $f'(x)$ and hence find the exact coordinates of the stationary point of the curve with equation $y = f(x)$. (b) Use the substitution $u = e^x$ and partial fractions to find the exact value of $\int_{\ln 3}^{\ln 5} f(x) \, dx$. Give your answer in the form $\ln a$, where $a$ is a rational number in its simplest form.

(a) x = ln(4/3), y = -16 · (b) ln(81/4)

9709/32 Oct/Nov 2024 — Mathematics Solved Past Paper

Q14MMedium-EasyAlgebra

Expand $(9 - 3x)^{\frac{1}{2}}$ in ascending powers of $x$ , up to and including the term in $x^2$ , simplifying the coefficients.

DifficultyMedium-Easy

Worked solution

Approach

Factor out $9$ so that the bracket becomes $1 + u$ , then apply the binomial expansion for the rational power $n = \frac{1}{2}$ , keeping terms up to $u^2$ . Finally multiply by the factor $3$ and simplify the coefficients.

Working

Write the expression as:

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

Use the binomial expansion:

(1+u)^{\frac{1}{2}} = 1 + \frac{1}{2}u + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}u^2 + \cdots

Substitute $u = -\frac{x}{3}$ :

3\left[1 + \frac{1}{2}\left(-\frac{x}{3}\right) + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}\left(-\frac{x}{3}\right)^2\right]

Simplify the $x$ term:

3\left(1 - \frac{x}{6}\right) = 3 - \frac{x}{2}

Simplify the $x^2$ term:

3 \cdot \frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2} \cdot \frac{x^2}{9} = 3 \cdot \left(-\frac{1}{8}\right)\cdot \frac{x^2}{9} = -\frac{x^2}{24}

Therefore:

(9 - 3x)^{\frac{1}{2}} = 3 - \frac{1}{2}x - \frac{1}{24}x^2 + \cdots

Answer

3 - \frac{1}{2}x - \frac{1}{24}x^2

Final answer

3 - x/2 - x^2/24

Detailed explanation

Walkthrough

Start by noticing that the base $9 - 3x$ is not of the form $1 + u$ , so factor out $9$ :

9 - 3x = 9\left(1 - \frac{x}{3}\right)

Since the whole expression is raised to the power $\frac{1}{2}$ , the factor $9$ comes out as $9^{\frac{1}{2}} = 3$ . This is the key first step: it turns the problem into the standard binomial form $3(1+u)^{\frac{1}{2}}$ with $u = -\frac{x}{3}$ .

Then apply the binomial expansion for a rational index:

(1+u)^{\frac{1}{2}} = 1 + \frac{1}{2}u + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}u^2 + \cdots

This formula is valid for $|u|<1$ , i.e. $|x|<3$ , but only the first three terms are needed. Substitute $u = -\frac{x}{3}$ and multiply every term by the factor $3$ . The $x$ term becomes $3 \cdot \frac{1}{2}\cdot\left(-\frac{x}{3}\right) = -\frac{x}{2}$ , and the $x^2$ term becomes $3 \cdot \left(-\frac{1}{8}\right)\cdot \frac{x^2}{9} = -\frac{x^2}{24}$ .

Finally collect the constant term $3$ with the two simplified terms to write:

3 - \frac{1}{2}x - \frac{1}{24}x^2

This is the expansion in ascending powers of $x$ , up to and including the term in $x^2$ .

Key Takeaways

To expand a binomial whose first term is not $1$ , factor out the constant so it becomes $(1+u)^n$ .
For rational $n$ , use the general binomial coefficients such as $\frac{n(n-1)}{2!}$ , and simplify them numerically.
The expansion is valid only when $|u|<1$ ; here that means $|x|<3$ .

Common Mistakes

Forgetting to multiply by the factor $3$ after factoring out $9$ . The expansion of $(1 - \frac{x}{3})^{\frac{1}{2}}$ alone is $1 - \frac{x}{6} - \frac{x^2}{72}$ , not the required answer.
Sign errors when substituting $u = -\frac{x}{3}$ , especially in the $x^2$ term where the square makes the sign positive but the binomial coefficient is negative.
Leaving coefficients as symbolic binomial coefficients instead of simplifying them; the question asks for simplified coefficients.
Omitting the unsimplified working needed for the method mark; the mark scheme requires a correct unsimplified $x$ or $x^2$ term to be seen.

Things to Be Careful About

The constant term is $9^{\frac{1}{2}} = 3$ , not $1$ .
The $x^2$ term: $\left(-\frac{x}{3}\right)^2 = \frac{x^2}{9}$ , and the binomial coefficient is $\frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2} = -\frac{1}{8}$ , so the term is $-\frac{x^2}{24}$ .
The expansion is only valid for $|x|<3$ ; although the question does not ask for the range, it is good practice to note it.
If you choose to expand $(1 - \frac{x}{3})^{\frac{1}{2}}$ first, remember to multiply the final expansion by $3$ before giving the answer.

Techniques used

factor out a constant to obtain the form (1+u)^napply the binomial expansion for a rational indexsimplify fractional coefficients

Similar questions

Q2Medium-EasyTrigonometryNumerical Solution of Equations

(a)

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

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Sketch $y = \cot 2x$ and $y = \sec x$ on the same axes for $0 < x < \frac{\pi}{2}$ . Analyse the behaviour of each curve — asymptotes, intercepts, and monotonicity — to show they intersect exactly once.

Working

Graph of $y = \cot 2x$ :

\cot 2x = \frac{\cos 2x}{\sin 2x}

Vertical asymptotes occur where $\sin 2x = 0$ , i.e., $2x = 0, \pi, 2\pi, \ldots$ , giving $x = 0, \frac{\pi}{2}, \pi, \ldots$

In the interval $0 < x < \frac{\pi}{2}$ , there are vertical asymptotes at $x = 0$ and $x = \frac{\pi}{2}$ .

The $x$ -intercept occurs where $\cos 2x = 0$ , i.e., $2x = \frac{\pi}{2}$ , so $x = \frac{\pi}{4}$ .

As $x \to 0^+$ , $\cot 2x \to +\infty$ .
As $x \to \frac{\pi}{2}^-$ , $\cot 2x \to -\infty$ .

Since $\cot 2x$ is continuous and strictly decreasing on $\left(0, \frac{\pi}{2}\right)$ , it falls from $+\infty$ to $-\infty$ , crossing zero at $x = \frac{\pi}{4}$ .

Graph of $y = \sec x$ :

\sec x = \frac{1}{\cos x}

In the interval $0 < x < \frac{\pi}{2}$ , $\cos x > 0$ , so $\sec x > 0$ .

At $x = 0$ , $\sec 0 = 1$ .
As $x \to \frac{\pi}{2}^-$ , $\sec x \to +\infty$ .

Since $\cos x$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$ , $\sec x$ is strictly increasing on this interval, rising from $1$ to $+\infty$ .

Number of intersections:

As $x \to 0^+$ :

\cot 2x \to +\infty \quad \text{and} \quad \sec x \to 1

So $\cot 2x > \sec x$ near $x = 0$ .

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

\cot\left(2 \cdot \frac{\pi}{4}\right) = \cot\frac{\pi}{2} = 0 \quad \text{and} \quad \sec\frac{\pi}{4} = \sqrt{2} \approx 1.414

So $\cot 2x < \sec x$ at $x = \frac{\pi}{4}$ .

Since $\cot 2x$ is continuous and strictly decreasing, and $\sec x$ is continuous and strictly increasing on $\left(0, \frac{\pi}{4}\right)$ , and they swap relative order, there is exactly one point of intersection in $\left(0, \frac{\pi}{4}\right)$ .

For $x \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ :

\cot 2x < 0 \quad \text{and} \quad \sec x > 0

So there is no intersection in this sub-interval.

Therefore, $\cot 2x = \sec x$ has exactly one root in $0 < x < \frac{\pi}{2}$ .

Answer

The equation $\cot 2x = \sec x$ has exactly one root in $0 < x < \frac{\pi}{2}$ , as shown by the single intersection of the two curves.

Final answer

Exactly one root in 0 < x < π/2

Detailed explanation

Walkthrough

We are asked to show that $\cot 2x = \sec x$ has exactly one root in the interval $0 < x < \frac{\pi}{2}$ by sketching suitable graphs.

Step 1: Sketch $y = \cot 2x$ .

The function $\cot 2x = \frac{\cos 2x}{\sin 2x}$ has vertical asymptotes where the denominator is zero: $\sin 2x = 0$ gives $2x = 0, \pi, 2\pi, \ldots$ , so $x = 0, \frac{\pi}{2}, \pi, \ldots$ . In our interval $(0, \frac{\pi}{2})$ , the asymptotes are at both endpoints. The $x$ -intercept is where $\cos 2x = 0$ , giving $x = \frac{\pi}{4}$ . As $x$ approaches $0$ from the right, $\cot 2x$ goes to $+\infty$ ; as $x$ approaches $\frac{\pi}{2}$ from the left, $\cot 2x$ goes to $-\infty$ . The function is continuous and strictly decreasing on $(0, \frac{\pi}{2})$ .

Step 2: Sketch $y = \sec x$ .

The function $\sec x = \frac{1}{\cos x}$ is positive on $(0, \frac{\pi}{2})$ since $\cos x > 0$ there. At $x = 0$ , $\sec 0 = 1$ . As $x \to \frac{\pi}{2}^-$ , $\cos x \to 0^+$ so $\sec x \to +\infty$ . Since $\cos x$ is strictly decreasing on this interval, $\sec x$ is strictly increasing from $1$ to $+\infty$ .

Step 3: Compare the two curves.

Near $x = 0^+$ , $\cot 2x \to +\infty$ while $\sec x \to 1$ , so $\cot 2x > \sec x$ . At $x = \frac{\pi}{4}$ , $\cot 2x = 0$ while $\sec x = \sqrt{2} > 0$ , so $\cot 2x < \sec x$ . By the Intermediate Value Theorem (since both functions are continuous on $(0, \frac{\pi}{4})$ ), there is at least one intersection. Since $\cot 2x$ is strictly decreasing and $\sec x$ is strictly increasing, their difference $\cot 2x - \sec x$ is strictly decreasing, so there is exactly one intersection in $(0, \frac{\pi}{4})$ .

For $x \in (\frac{\pi}{4}, \frac{\pi}{2})$ , $\cot 2x < 0$ while $\sec x > 0$ , so no intersection is possible.

Therefore, there is exactly one root in $(0, \frac{\pi}{2})$ .

Key Takeaways

When asked to show an equation has a given number of roots by sketching, identify two functions whose intersection points are the roots.
Analyse asymptotes, intercepts, and monotonicity of each graph to determine the number of intersections.
Use the fact that a strictly decreasing function and a strictly increasing function can intersect at most once to prove uniqueness.

Common Mistakes

Forgetting the vertical asymptotes of $\cot 2x$ at $x = 0$ and $x = \frac{\pi}{2}$ , leading to an incorrect sketch.
Not justifying why there is exactly one intersection — merely sketching is not enough; you must explain why the curves cross only once.
Marking the intersection at $x = \frac{\pi}{2}$ — this is outside the open interval $(0, \frac{\pi}{2})$ .

Things to Be Careful About

The interval is open: $0 < x < \frac{\pi}{2}$ . The asymptotes at the endpoints are not roots.
At $x = \frac{\pi}{4}$ , $\cot 2x = 0$ but $\sec x = \sqrt{2}$ , so this is not a root.
For $x > \frac{\pi}{4}$ in the interval, $\cot 2x$ is negative while $\sec x$ is positive, so no further intersections can occur.
The mark scheme also accepts the alternative approach of sketching $y = \tan 2x$ and $y = \cos x$ , which gives $\tan 2x = \cos x$ as the equivalent equation.

Techniques used

sketch graph of cot 2xsketch graph of sec xanalyse asymptotes and monotonicityjustify single intersection

(b)

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

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

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

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

If the sequence $x_{n+1} = \frac{1}{2}\tan^{-1}(\cos x_n)$ converges, let the limit be $L$ . Substitute $x_n = L$ and $x_{n+1} = L$ into the iterative formula, then rearrange to show that $L$ satisfies the equation $\cot 2L = \sec L$ .

Working

Suppose the sequence converges to a limit $L$ . Then as $n \to \infty$ :

x_{n+1} \to L \quad \text{and} \quad x_n \to L

Substituting into the iterative formula:

L = \frac{1}{2}\tan^{-1}(\cos L)

Multiply both sides by 2:

2L = \tan^{-1}(\cos L)

Take the tangent of both sides:

\tan 2L = \cos L

Rearrange by dividing both sides by $\cos L$ (which is non-zero since $L \in (0, \frac{\pi}{2})$ ):

\frac{\tan 2L}{\cos L} = 1

Alternatively, write $\tan 2L = \frac{\sin 2L}{\cos 2L}$ :

\frac{\sin 2L}{\cos 2L} = \cos L

Rearranging:

\frac{\cos 2L}{\sin 2L} = \frac{1}{\cos L}

\cot 2L = \sec L

This is exactly the equation from part (a). Therefore, if the sequence converges, it converges to the root of $\cot 2x = \sec x$ in $0 < x < \frac{\pi}{2}$ .

Answer

If the sequence converges, it converges to the root in part (a).

Final answer

The sequence converges to the root of cot 2x = sec x in 0 < x < π/2

Detailed explanation

Walkthrough

We are given the iterative formula $x_{n+1} = \frac{1}{2}\tan^{-1}(\cos x_n)$ and asked to show that if it converges, it converges to the root found in part (a).

Step 1: Assume convergence.

If the sequence converges, let the limit be $L$ . This means $\lim_{n \to \infty} x_n = L$ and $\lim_{n \to \infty} x_{n+1} = L$ .

Step 2: Substitute into the iterative formula.

Taking the limit on both sides of $x_{n+1} = \frac{1}{2}\tan^{-1}(\cos x_n)$ :

L = \frac{1}{2}\tan^{-1}(\cos L)

Step 3: Rearrange to recover the original equation.

Multiply by 2: $2L = \tan^{-1}(\cos L)$ .

Take the tangent of both sides: $\tan 2L = \cos L$ .

Now we need to show this is equivalent to $\cot 2L = \sec L$ . Write $\tan 2L = \frac{\sin 2L}{\cos 2L}$ :

\frac{\sin 2L}{\cos 2L} = \cos L

Rearrange by cross-multiplying or inverting:

\frac{\cos 2L}{\sin 2L} = \frac{1}{\cos L}

\cot 2L = \sec L

This is exactly the equation from part (a), so $L$ must be the root of that equation in $(0, \frac{\pi}{2})$ .

Key Takeaways

If a sequence defined by $x_{n+1} = F(x_n)$ converges to $L$ , then $L = F(L)$ — this is the fixed-point equation.
Rearranging the fixed-point equation can recover the original equation whose root is being approximated.
The iterative formula is designed so that its fixed point is the solution to the target equation.

Common Mistakes

Forgetting to state the assumption that the sequence converges before taking the limit.
Not showing the rearrangement steps clearly — the mark scheme requires seeing $\tan 2x = \cos x$ or the equivalent $\cot 2x = \sec L$ .
Dividing by zero or assuming $\cos L \neq 0$ without justification (though $L \in (0, \frac{\pi}{2})$ guarantees this).

Things to Be Careful About

The question says "if a sequence ... converges" — you must state this assumption before taking the limit.
The rearrangement from $\tan 2L = \cos L$ to $\cot 2L = \sec L$ requires $\cos 2L \neq 0$ and $\sin 2L \neq 0$ , which holds since $L \in (0, \frac{\pi}{2})$ and $L \neq \frac{\pi}{4}$ (as we showed $\cot 2L = 0 \neq \sec L$ at $x = \frac{\pi}{4}$ ).
The mark scheme accepts stopping at $\tan 2x = \cos x$ if the alternative approach from part (a) was used.

Techniques used

take limit of iterative formularearrange to original equation

Q35MMedium-HardComplex Numbers

The square roots of $6 - 8i$ can be expressed in the Cartesian form $x + iy$ , where $x$ and $y$ are real and exact.

By first forming a quartic equation in $x$ or $y$ , find the square roots of $6 - 8i$ in exact Cartesian form.

DifficultyMedium-Hard

Worked solution

Approach

Let a square root be $x + iy$ . Square it and equate real and imaginary parts. This gives two equations. Eliminate $y$ (or $x$ ) to form a quartic in $x$ (or $y$ ). Solve as a quadratic in $x^2$ (or $y^2$ ), reject non-real values, then find the corresponding $y$ (or $x$ ) values and pair them correctly.

Working

Let $z = x + iy$ . Then

z^2 = (x+iy)^2 = x^2 - y^2 + 2xy\,i.

Equating real and imaginary parts with $6 - 8i$ :

x^2 - y^2 = 6, \qquad 2xy = -8.

Thus $xy = -4$ . Using $y = -\frac{4}{x}$ :

x^2 - \frac{16}{x^2} = 6

x^4 - 16 = 6x^2

x^4 - 6x^2 - 16 = 0.

Let $u = x^2$ :

u^2 - 6u - 16 = 0

(u-8)(u+2) = 0

so $u = 8$ or $u = -2$ . Since $x$ is real, $u = x^2 = 8$ , hence $x = \pm 2\sqrt{2}$ .

For $x = 2\sqrt{2}$ :

y = -\frac{4}{x} = -\frac{4}{2\sqrt{2}} = -\sqrt{2}.

For $x = -2\sqrt{2}$ :

y = -\frac{4}{x} = -\frac{4}{-2\sqrt{2}} = \sqrt{2}.

Answer

The square roots are

2\sqrt{2} - \sqrt{2}\,i \quad \text{and} \quad -2\sqrt{2} + \sqrt{2}\,i,

i.e.

\pm(2\sqrt{2} - \sqrt{2}\,i).

Final answer

±(2√2 - √2i)

Detailed explanation

Walkthrough

We are looking for numbers of the form $x + iy$ whose square equals $6 - 8i$ . The first step is to square $x + iy$ . Because $i^2 = -1$ , the square is $(x^2 - y^2) + 2xy\,i$ . We then compare real parts: $x^2 - y^2$ must equal $6$ , and imaginary parts: $2xy$ must equal $-8$ . This gives two equations in two unknowns.

Next, we eliminate one variable. From $2xy = -8$ , $xy = -4$ , so $y = -4/x$ . Substituting into $x^2 - y^2 = 6$ gives $x^2 - 16/x^2 = 6$ . Multiplying by $x^2$ gives $x^4 - 6x^2 - 16 = 0$ , a quartic in $x$ . Although it is quartic, it is quadratic in the variable $u = x^2$ . Factorising gives $(u - 8)(u + 2) = 0$ , so $u = 8$ or $u = -2$ . Since $u = x^2$ must be non-negative for real $x$ , we reject $u = -2$ . Thus $x^2 = 8$ , so $x = \pm 2\sqrt{2}$ .

Finally, use $y = -4/x$ to find the corresponding $y$ values. If $x = 2\sqrt{2}$ , then $y = -\sqrt{2}$ . If $x = -2\sqrt{2}$ , then $y = \sqrt{2}$ . These pairings give the two square roots. The answer can be written $\pm(2\sqrt{2} - \sqrt{2}\,i)$ .

Key Takeaways

This question combines complex number arithmetic with solving a polynomial equation. The key idea is that equating real and imaginary parts turns one complex equation into two real equations. Eliminating one variable produces a quartic, but because only even powers of $x$ appear, it can be solved as a quadratic in $x^2$ . It is also important to reject solutions that would make $x$ or $y$ non-real, and to pair the signs correctly using $xy = -4$ .

Common Mistakes

Forgetting that $i^2 = -1$ , so $(x + iy)^2 = x^2 - y^2 + 2xy\,i$ , not $x^2 + y^2 + 2xy\,i$ .
Using the wrong sign for the imaginary part: $2xy = -8$ , not $+8$ .
Making a sign error when dividing by 2: $xy = -4$ . A wrong relation here often leads to an incorrect quartic.
Rejecting the negative value of $u = x^2$ is necessary, but forgetting to reject it and trying to use $x^2 = -2$ would give non-real $x$ .
Pairing $x$ and $y$ incorrectly. The pairs must satisfy $xy = -4$ , so $x = 2\sqrt{2}$ goes with $y = -\sqrt{2}$ , not with $y = \sqrt{2}$ .

Things to Be Careful About

The mark scheme allows a minor slip in the elimination step, but not seriously incorrect algebra. In particular, a sign error in $xy = -4$ would be serious and would not earn the method mark.
Both square roots are needed; give both $\pm(2\sqrt{2} - \sqrt{2}\,i)$ or state the two separate values.
If you solve using $y$ instead of $x$ , the quartic becomes $y^4 + 6y^2 - 16 = 0$ . The same care is needed: $y^2 = 2$ , so $y = \pm\sqrt{2}$ , and then use $x = -4/y$ to pair.
Exact surd forms are required; do not give decimal approximations.

Techniques used

square a complex number and equate real and imaginary partseliminate one variable to form a quarticsolve a quadratic in x^2pair signs using xy = -4

Similar questions

Q43MMedium-EasyLogarithmic and Exponential Functions

Solve the equation $5^x = 5^{x+2} - 10$ . Give your answer correct to 3 decimal places.

DifficultyMedium-Easy

Worked solution

Approach

Rewrite $5^{x+2}$ using the index law $a^{m+n}=a^m a^n$ , then isolate $5^x$ . Once $5^x$ is equal to a constant, take natural logarithms of both sides and solve for $x$ .

Working

Using $5^{x+2}=5^x \cdot 5^2 = 25 \cdot 5^x$ , the equation becomes

5^x = 25 \cdot 5^x - 10

Rearrange to collect the terms in $5^x$ :

10 = 24 \cdot 5^x

So

5^x = \frac{10}{24} = \frac{5}{12}

Take natural logarithms:

x \ln 5 = \ln\left(\frac{5}{12}\right)

Therefore

x = \frac{\ln\left(\frac{5}{12}\right)}{\ln 5}

Evaluating gives

x \approx -0.544

Answer

x = -0.544 \quad (3\text{ dp})

Final answer

-0.544

Detailed explanation

Walkthrough

Start by making the two powers of 5 comparable. Since $5^{x+2}=5^x \times 5^2=25 \times 5^x$ , the equation $5^x=5^{x+2}-10$ becomes $5^x=25 \times 5^x - 10$ . This is now linear in the single unknown quantity $5^x$ .

Collect the $5^x$ terms on one side. Subtracting $25 \times 5^x$ from both sides gives $-24 \times 5^x = -10$ , or equivalently $24 \times 5^x = 10$ . Dividing by 24 gives $5^x = \frac{10}{24} = \frac{5}{12}$ .

Now the unknown is in the exponent. Taking natural logarithms of both sides is the standard way to bring it down: $\ln(5^x)=x\ln 5$ , so $x\ln 5 = \ln\left(\frac{5}{12}\right)$ . Dividing by $\ln 5$ gives $x = \frac{\ln\left(\frac{5}{12}\right)}{\ln 5}$ . Evaluating this on a calculator gives $x \approx -0.543959$ , which rounds to $-0.544$ to 3 decimal places.

Key Takeaways

Use the index law $a^{m+n}=a^m a^n$ to rewrite terms with the same base in terms of one unknown power.
An equation of the form $a^x=b$ can be solved by taking logarithms of both sides.
The logarithm brings the exponent down: $\ln(a^x)=x\ln a$ .
Keep the exact fraction until the final step, then round only at the end.

Common Mistakes

Writing $5^{x+2}$ as $5^x+2$ or $5^x \times 2$ instead of $25 \times 5^x$ .
Forgetting to collect like terms in $5^x$ before taking logarithms.
Taking logarithms before isolating the exponential term, which makes the algebra harder.
Giving an unsupported answer: the mark scheme awards 0/3 if no working is shown, even if the final value is correct.

Things to Be Careful About

The final answer must be given to 3 decimal places, so use $-0.544$ , not $-0.54$ or an unrounded value.
Since $\ln\left(\frac{5}{12}\right)$ is negative, the value of $x$ is negative; check that this makes sense.
The fraction can be written as $\frac{5}{12}$ or $\frac{10}{24}$ ; both are equivalent and acceptable.
The logarithm of a positive number is required; $5^x$ is always positive, so there is no domain issue here.

Techniques used

apply index laws to rewrite powers of 5isolate the exponential termtake logarithms of both sidessolve the resulting linear equation

Similar questions

Q5Medium-EasyComplex Numbers

(a)

The complex number $u$ is given by

u = \frac{(\cos \frac{1}{7}\pi + i\sin \frac{1}{7}\pi)^4}{\cos \frac{1}{7}\pi - i\sin \frac{1}{7}\pi}

Find the exact value of $\arg u$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Write the numerator in polar form using De Moivre's theorem, identify the argument of the denominator as the conjugate, then divide by subtracting arguments.

Working

By De Moivre's theorem,

\left(\cos \frac{\pi}{7} + i\sin \frac{\pi}{7}\right)^4 = \cos \frac{4\pi}{7} + i\sin \frac{4\pi}{7}

so the argument of the numerator is $\frac{4\pi}{7}$ .

The denominator is the conjugate of $\cos \frac{\pi}{7} + i\sin \frac{\pi}{7}$ , so

\cos \frac{\pi}{7} - i\sin \frac{\pi}{7} = \cos\left(-\frac{\pi}{7}\right) + i\sin\left(-\frac{\pi}{7}\right)

and its argument is $-\frac{\pi}{7}$ .

For division in polar form, arguments subtract:

\arg u = \frac{4\pi}{7} - \left(-\frac{\pi}{7}\right) = \frac{5\pi}{7}

Answer

\arg u = \frac{5\pi}{7}

Final answer

arg u = 5π/7

Detailed explanation

Walkthrough

We are dividing two complex numbers given in modulus-argument form. The numerator is a power of $\cos \frac{\pi}{7} + i\sin \frac{\pi}{7}$ . De Moivre's theorem tells us that raising this to the 4th power multiplies the argument by 4, so the numerator has argument $\frac{4\pi}{7}$ .

The denominator is $\cos \frac{\pi}{7} - i\sin \frac{\pi}{7}$ , which is the conjugate. Its argument is the negative of $\frac{\pi}{7}$ , i.e. $-\frac{\pi}{7}$ .

When dividing two complex numbers in polar form, we subtract the argument of the denominator from the argument of the numerator. Hence $\arg u = \frac{4\pi}{7} - \left(-\frac{\pi}{7}\right) = \frac{5\pi}{7}$ .

Key Takeaways

De Moivre's theorem: $(\cos \theta + i\sin \theta)^n = \cos n\theta + i\sin n\theta$ .
The conjugate $\cos \theta - i\sin \theta$ has argument $-\theta$ .
Division in polar form: arguments subtract.

Common Mistakes

Forgetting that the denominator has argument $-\frac{\pi}{7}$ , not $+\frac{\pi}{7}$ .
Subtracting in the wrong order, giving $-\frac{5\pi}{7}$ instead of $\frac{5\pi}{7}$ .
Giving the answer in degrees; the question requires an exact value in radians.

Things to Be Careful About

The principal argument is conventionally in $(-\pi, \pi]$ , and $\frac{5\pi}{7}$ is in this range.
Keep the fractions in terms of $\pi$ ; do not convert to decimals.
The mark scheme requires the argument of a complex number to be stated or implied to earn the method mark.

Techniques used

apply De Moivre's theorem to a poweridentify the argument of a conjugatesubtract arguments when dividing in polar form

(b)

The complex numbers $u$ and $u^*$ are plotted on an Argand diagram.

Describe the single geometrical transformation that maps $u$ onto $u^*$ and state the exact value of $\arg u^*$ .

2M Similar

DifficultyEasy

Worked solution

Approach

Use the fact that $u^*$ is the complex conjugate of $u$ . On an Argand diagram, conjugation reflects a point in the real axis. Therefore the argument of $u^*$ is the negative of the argument of $u$ .

Working

If $u = r(\cos \theta + i\sin \theta)$ , then

u^* = r(\cos \theta - i\sin \theta) = r\left(\cos(-\theta) + i\sin(-\theta)\right)

so $\arg u^* = -\theta$ .

From part (a), $\arg u = \frac{5\pi}{7}$ . Therefore

\arg u^* = -\frac{5\pi}{7}

Answer

Reflection in the real axis; $\arg u^* = -\frac{5\pi}{7}$ .

Final answer

Reflection in the real axis; arg u* = -5π/7

Detailed explanation

Walkthrough

A complex number $u = x + iy$ has conjugate $u^* = x - iy$ . On the Argand diagram, the point $(x, y)$ is reflected in the real axis to $(x, -y)$ . This is exactly the same as reflecting the line from the origin to $u$ across the real axis, so the angle with the positive real axis changes from $\theta$ to $-\theta$ . Using $\arg u = \frac{5\pi}{7}$ from part (a), we get $\arg u^* = -\frac{5\pi}{7}$ .

Key Takeaways

Conjugation corresponds to reflection in the real axis.
The argument of the conjugate is the negative of the argument of the original number.

Common Mistakes

Saying reflection in the imaginary axis instead of the real axis.
Writing $\arg u^* = \arg u$ instead of $-\arg u$ .
Giving only $\arg u^* = -\arg u$ without substituting the exact value from part (a); the question asks for the exact value.

Things to Be Careful About

The mark scheme requires the word 'reflection'; 'mirror' or 'flip' is not enough.
The answer must be exact; do not give a decimal approximation.
Both $\frac{5\pi}{7}$ and $-\frac{5\pi}{7}$ are principal arguments. If using a non-principal form, it should be consistent, but the standard exact value is $-\frac{5\pi}{7}$ .

Techniques used

recognise conjugation as reflection in the real axisnegate the argument for the conjugate

Q64MMedium-EasyLogarithmic and Exponential Functions

Fig. 1

The variables $x$ and $y$ satisfy the equation $ay = b^x$ , where $a$ and $b$ are constants. The graph of $\ln y$ against $x$ is a straight line passing through the points $(0.50, 2.24)$ and $(3.40, 8.27)$ , as shown in the diagram.

Find the values of $a$ and $b$ . Give each value correct to 1 significant figure.

DifficultyMedium-Easy

Worked solution

Approach

Take the natural logarithm of both sides of the equation $ay = b^x$ to transform it into a linear form $\ln y = (\ln b)x - \ln a$ . This matches the equation of a straight line $Y = mX + c$ , where the gradient is $\ln b$ and the y-intercept is $-\ln a$ . Use the two given points on the graph to find the gradient and y-intercept, then solve for $a$ and $b$ .

Working

Given $ay = b^x$ , take $\ln$ of both sides:

\ln(ay) = \ln(b^x)

Apply logarithm laws:

\ln a + \ln y = x \ln b

Rearranging for $\ln y$ :

\ln y = (\ln b)x - \ln a

This is a straight line with gradient $m = \ln b$ and y-intercept $c = -\ln a$ .

The line passes through $(0.50, 2.24)$ and $(3.40, 8.27)$ . The gradient is:

\ln b = \frac{8.27 - 2.24}{3.40 - 0.50} = \frac{6.03}{2.90} \approx 2.07931

So:

b = e^{2.07931} \approx 7.999

Rounding to 1 significant figure, $b = 8$ .

To find $a$ , substitute the gradient and one point, e.g., $(0.50, 2.24)$ , into the linear equation:

2.24 = (2.07931)(0.50) - \ln a

2.24 = 1.03965 - \ln a

\ln a = 1.03965 - 2.24 = -1.20035

So:

a = e^{-1.20035} \approx 0.301

Rounding to 1 significant figure, $a = 0.3$ .

Answer

a = 0.3, \quad b = 8

Final answer

a = 0.3, b = 8

Detailed explanation

Walkthrough

First, we need to convert the exponential equation $ay = b^x$ into a linear form so we can use the straight-line graph. Taking the natural logarithm of both sides gives $\ln(ay) = \ln(b^x)$ . Using logarithm laws, this becomes $\ln a + \ln y = x \ln b$ . Rearranging for $\ln y$ gives $\ln y = (\ln b)x - \ln a$ , which is in the form $Y = mX + c$ with $Y = \ln y$ , $X = x$ , gradient $m = \ln b$ , and y-intercept $c = -\ln a$ .

Next, we calculate the gradient $m$ using the two coordinates provided on the graph: $(0.50, 2.24)$ and $(3.40, 8.27)$ . The gradient is $\frac{8.27 - 2.24}{3.40 - 0.50} = \frac{6.03}{2.90} \approx 2.07931$ . Since $m = \ln b$ , we find $b = e^{2.07931} \approx 7.999$ , which rounds to $8$ to 1 significant figure.

Finally, we use the y-intercept relationship $c = -\ln a$ . Substituting the gradient and one of the points into the linear equation gives $2.24 = (2.07931)(0.50) - \ln a$ , leading to $\ln a = -1.20035$ . Exponentiating gives $a = e^{-1.20035} \approx 0.301$ , which rounds to $0.3$ to 1 significant figure.

Key Takeaways

Taking logarithms of both sides of an exponential equation is a standard technique to linearise it for graph-based analysis.
Carefully identifying the gradient and y-intercept in terms of the original constants is crucial; here, the y-intercept is $-\ln a$ , not $\ln a$ .
Always round final answers to the specified number of significant figures only after all intermediate calculations are complete.

Common Mistakes

Forgetting the $\ln a$ term when linearising, incorrectly writing $\ln y = x \ln b$ instead of $\ln y = (\ln b)x - \ln a$ .
Misidentifying the y-intercept as $\ln a$ instead of $-\ln a$ , which leads to an incorrect value for $a$ .
Rounding intermediate values (like the gradient) too early, which can cause small errors in the final answer.

Things to Be Careful About

The question asks for values correct to 1 significant figure, not decimal places. $7.999$ rounds to $8$ (1 s.f.), and $0.301$ rounds to $0.3$ (1 s.f.).
Ensure that the coordinates are used correctly in the gradient formula $\frac{y_2 - y_1}{x_2 - x_1}$ , keeping in mind that the vertical axis is $\ln y$ , not $y$ .

Techniques used

linearise equation using logarithmscalculate gradient from two pointsfind y-intercept and solve for constants

Similar questions

Q7MediumTrigonometry

(a)

Show that the equation $\tan^3 x + 2\tan 2x - \tan x = 0$ may be expressed as

\tan^4 x - 2\tan^2 x - 3 = 0

for $\tan x \neq 0$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Let $t = \tan x$ . Replace $\tan 2x$ using the double-angle formula, clear the fraction, then factor out $t$ and divide by $t$ (valid because $t \neq 0$ ).

Working

Let $t = \tan x$ . The double-angle formula gives

\tan 2x = \frac{2t}{1 - t^2}

Substitute into the equation:

t^3 + 2\left(\frac{2t}{1 - t^2}\right) - t = 0

Multiply through by $1 - t^2$ :

t^3(1 - t^2) + 4t - t(1 - t^2) = 0

Expand and simplify:

t^3 - t^5 + 4t - t + t^3 = 0

- t^5 + 2t^3 + 3t = 0

Factor:

- t(t^4 - 2t^2 - 3) = 0

Since $t = \tan x \neq 0$ , divide by $-t$ :

t^4 - 2t^2 - 3 = 0

Hence

\tan^4 x - 2\tan^2 x - 3 = 0

Answer

The equation reduces to $\tan^4 x - 2\tan^2 x - 3 = 0$ for $\tan x \neq 0$ .

Final answer

tan^4 x - 2 tan^2 x - 3 = 0 for tan x ≠ 0

Detailed explanation

Walkthrough

Start with the equation $\tan^3 x + 2\tan 2x - \tan x = 0$ . The only term not already in terms of $\tan x$ is $\tan 2x$ . Use the double-angle formula $\tan 2x = \frac{2\tan x}{1 - \tan^2 x}$ to rewrite everything in terms of $t = \tan x$ . Then multiply the whole equation by $1 - t^2$ to remove the fraction. Expand and collect like terms to obtain a polynomial. Factor out $-t$ . Because the question specifies $\tan x \neq 0$ , dividing by $-t$ is valid and leaves the required quartic. Each of these steps earns a mark: the correct double-angle substitution, a correct fraction-free equation, and the final reduction.

Key Takeaways

This question tests the double-angle formula for tangent and algebraic manipulation of rational expressions. It also shows why a condition like $\tan x \neq 0$ is needed before dividing by a factor.

Common Mistakes

Forgetting to multiply the first and last terms by $1 - t^2$ when clearing the fraction.
Making sign errors when expanding $-t(1 - t^2)$ ; the middle term becomes $+t^3$ , not $-t^3$ .
Dividing by $\tan x$ without stating that $\tan x \neq 0$ .
In a 'show that' question, skipping lines or writing an incorrect intermediate equation.

Things to Be Careful About

The mark scheme allows the intermediate equation to be missing ' $=0$ ', but every line that is written must be correct. It also condones not mentioning $\tan x \neq 0$ , but mentioning it is good practice. The final answer must be exactly $\tan^4 x - 2\tan^2 x - 3 = 0$ .

Techniques used

use the double-angle formula for tanclear denominators by multiplying throughfactor and divide by a nonzero factor

(b)

Hence solve the equation $\tan^3 2\theta + 2\tan 4\theta - \tan 2\theta = 0$ for $0 < \theta < \pi$ . Give your answers in exact form.

3M Similar

DifficultyMedium

Worked solution

Approach

Apply the result of part (a) with $x = 2\theta$ . This gives a quartic in $\tan 2\theta$ . Let $u = \tan^2 2\theta$ , solve the resulting quadratic, then find all values of $2\theta$ in $(0, 2\pi)$ and halve them.

Working

Using part (a) with $x = 2\theta$ (for $\tan 2\theta \neq 0$ ):

\tan^4 2\theta - 2\tan^2 2\theta - 3 = 0

Let $u = \tan^2 2\theta$ . Then

u^2 - 2u - 3 = 0

(u - 3)(u + 1) = 0

So $u = 3$ or $u = -1$ . Since $u = \tan^2 2\theta \geq 0$ , only $u = 3$ is possible:

\tan^2 2\theta = 3

\tan 2\theta = \pm \sqrt{3}

Since $0 < \theta < \pi$ , we have $0 < 2\theta < 2\pi$ .

For $\tan 2\theta = \sqrt{3}$ :

2\theta = \frac{\pi}{3}, \frac{4\pi}{3}

For $\tan 2\theta = -\sqrt{3}$ :

2\theta = \frac{2\pi}{3}, \frac{5\pi}{3}

Halving gives

\theta = \frac{\pi}{6}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{6}

Note: the original equation also has $\tan 2\theta = 0$ , giving $\theta = \frac{\pi}{2}$ , but this is not required by the mark scheme.

Answer

\theta = \frac{\pi}{6}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{6}

Final answer

θ = π/6, π/3, 2π/3, 5π/6

Detailed explanation

Walkthrough

Since part (a) showed the reduction for $\tan x \neq 0$ , replace $x$ by $2\theta$ . This gives $\tan^4 2\theta - 2\tan^2 2\theta - 3 = 0$ . To solve this, set $u = \tan^2 2\theta$ ; the quartic becomes a quadratic $u^2 - 2u - 3 = 0$ . Factorise to get $u = 3$ or $u = -1$ . Since a square cannot be negative, reject $u = -1$ . Thus $\tan^2 2\theta = 3$ , so $\tan 2\theta = \pm \sqrt{3}$ . Now use the range: $0 < \theta < \pi$ means $0 < 2\theta < 2\pi$ . Find the four angles in this interval with tangent $\sqrt{3}$ or $-\sqrt{3}$ , then halve each to get $\theta$ . The mark scheme requires exact values and does not require the extra solution $\theta = \pi/2$ that comes from $\tan 2\theta = 0$ in the original equation.

Key Takeaways

This part combines a substitution from a previous result, solving a quadratic in $\tan^2$ , and solving a trigonometric equation over a doubled interval. It is important to consider both positive and negative square roots and to halve the angles at the end.

Common Mistakes

Forgetting to halve the values of $2\theta$ ; this loses the method mark.
Only taking the positive square root, missing $\tan 2\theta = -\sqrt{3}$ .
Giving answers outside $0 < \theta < \pi$ .
Giving decimal answers instead of exact multiples of $\pi$ .
Not using the result from part (a) and instead trying to solve the original equation from scratch.

Things to Be Careful About

The mark scheme accepts equivalent fractions such as $\frac{2\pi}{6}$ for $\frac{\pi}{3}$ . It ignores answers outside the interval. If a candidate makes a copying slip but still has a complete method to obtain a value of $\theta$ , the method mark can still be awarded. The solution $\theta = \frac{\pi}{2}$ from $\tan 2\theta = 0$ is not required.

Techniques used

apply the previous result by substitutionsolve a quadratic in tan^2find all angles in the doubled intervalhalve the angles to obtain theta

Q8MediumDifferentiation

The parametric equations of a curve are

x = \tan^2 2t, \quad y = \cos 2t,

for $0 < t < \frac{1}{4}\pi$ .

(a)

Show that $\frac{dy}{dx} = -\frac{1}{2}\cos^3 2t$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Differentiate $x$ and $y$ with respect to the parameter $t$ , then use the parametric formula

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

and simplify using $\tan 2t = \frac{\sin 2t}{\cos 2t}$ and $\sec 2t = \frac{1}{\cos 2t}$ .

Working

Differentiate $x = \tan^2 2t$ using the chain rule:

\frac{dx}{dt} = 2\tan 2t \cdot \sec^2 2t \cdot 2 = 4\tan 2t \sec^2 2t

Differentiate $y = \cos 2t$ :

\frac{dy}{dt} = -2\sin 2t

Therefore,

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2\sin 2t}{4\tan 2t \sec^2 2t}

Using $\tan 2t = \frac{\sin 2t}{\cos 2t}$ and $\sec 2t = \frac{1}{\cos 2t}$ :

\frac{dy}{dx} = \frac{-2\sin 2t}{4 \cdot \frac{\sin 2t}{\cos 2t} \cdot \frac{1}{\cos^2 2t}} = \frac{-2\sin 2t}{\frac{4\sin 2t}{\cos^3 2t}} = -\frac{1}{2}\cos^3 2t

Answer

\frac{dy}{dx} = -\frac{1}{2}\cos^3 2t

Final answer

-1/2 cos^3 2t

Detailed explanation

Walkthrough

We have $x$ and $y$ given as functions of the parameter $t$ . To find $\frac{dy}{dx}$ , differentiate both $x$ and $y$ with respect to $t$ , then use

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

First, $x = \tan^2 2t$ . Think of it as $(\tan 2t)^2$ . The chain rule gives $2\tan 2t$ times the derivative of $\tan 2t$ , which is $2\sec^2 2t$ , so $\frac{dx}{dt} = 4\tan 2t \sec^2 2t$ .

Next, $y = \cos 2t$ differentiates to $\frac{dy}{dt} = -2\sin 2t$ .

Divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ . The expression contains $\tan 2t$ and $\sec^2 2t$ ; rewrite $\tan 2t = \frac{\sin 2t}{\cos 2t}$ and $\sec 2t = \frac{1}{\cos 2t}$ . The denominator becomes $\frac{4\sin 2t}{\cos^3 2t}$ , so the $\sin 2t$ factors cancel and the result is $-\frac{1}{2}\cos^3 2t$ .

Key Takeaways

This question tests parametric differentiation: when $x$ and $y$ are both functions of a parameter, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ . It also tests the chain rule and the ability to simplify trigonometric fractions.

Common Mistakes

Forgetting the factor $2$ from differentiating $2t$ when differentiating $\tan^2 2t$ .
Using $\frac{dx}{dt} \div \frac{dy}{dt}$ instead of $\frac{dy}{dt} \div \frac{dx}{dt}$ .
Stopping before fully simplifying to the required $\cos^3 2t$ form.

Things to Be Careful About

The domain $0 < t < \frac{1}{4}\pi$ ensures $\sin 2t \neq 0$ , so cancelling $\sin 2t$ is valid. Keep the exact trigonometric form rather than converting to decimals.

Techniques used

apply the chain rule to differentiate a squared tangent functionapply the chain rule to differentiate a cosine functioncombine parametric derivatives using dy/dx = (dy/dt)/(dx/dt)simplify using tangent and secant identities

(b)

Hence find the equation of the normal to the curve at the point where $t = \frac{1}{8}\pi$ . Give your answer in the form $y = mx + c$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Substitute $t = \frac{1}{8}\pi$ into the parametric equations to find the point. Use the derivative from part (a) to find the gradient of the tangent, then take its negative reciprocal to get the gradient of the normal. Form the equation of the normal using point-gradient form and rearrange into $y = mx + c$ .

Working

At $t = \frac{1}{8}\pi$ :

x = \tan^2\left(2 \cdot \frac{\pi}{8}\right) = \tan^2\left(\frac{\pi}{4}\right) = 1^2 = 1

y = \cos\left(2 \cdot \frac{\pi}{8}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}

So the point is $\left(1, \frac{\sqrt{2}}{2}\right)$ .

From part (a),

\frac{dy}{dx} = -\frac{1}{2}\cos^3 2t

At $t = \frac{\pi}{8}$ :

\frac{dy}{dx} = -\frac{1}{2}\cos^3\left(\frac{\pi}{4}\right) = -\frac{1}{2}\left(\frac{\sqrt{2}}{2}\right)^3 = -\frac{\sqrt{2}}{8}

The gradient of the normal is the negative reciprocal:

m = -\frac{1}{-\frac{\sqrt{2}}{8}} = \frac{8}{\sqrt{2}} = 4\sqrt{2}

Equation of the normal through $\left(1, \frac{\sqrt{2}}{2}\right)$ :

y - \frac{\sqrt{2}}{2} = 4\sqrt{2}(x - 1)

Rearrange:

y = 4\sqrt{2}x - 4\sqrt{2} + \frac{\sqrt{2}}{2} = 4\sqrt{2}x - \frac{7\sqrt{2}}{2}

Answer

y = 4\sqrt{2}x - \frac{7\sqrt{2}}{2}

Final answer

y = 4√2x - 7√2/2

Detailed explanation

Walkthrough

First find the coordinates of the point. Since $t = \frac{1}{8}\pi$ , we have $2t = \frac{\pi}{4}$ . Thus $x = \tan^2\left(\frac{\pi}{4}\right) = 1$ and $y = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$ .

Next, use the derivative from part (a): $\frac{dy}{dx} = -\frac{1}{2}\cos^3 2t$ . At $t = \frac{\pi}{8}$ , $\cos 2t = \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$ , so the tangent gradient is

-\frac{1}{2}\left(\frac{\sqrt{2}}{2}\right)^3 = -\frac{\sqrt{2}}{8}.

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal:

m = -\frac{1}{-\frac{\sqrt{2}}{8}} = \frac{8}{\sqrt{2}} = 4\sqrt{2}.

Now use the point-gradient equation $y - y_1 = m(x - x_1)$ with $(x_1, y_1) = \left(1, \frac{\sqrt{2}}{2}\right)$ and $m = 4\sqrt{2}$ :

y - \frac{\sqrt{2}}{2} = 4\sqrt{2}(x - 1).

Expand and collect the constant terms:

y = 4\sqrt{2}x - 4\sqrt{2} + \frac{\sqrt{2}}{2} = 4\sqrt{2}x - \frac{7\sqrt{2}}{2}.

Key Takeaways

This part combines parametric coordinates with the geometric meaning of the derivative: the tangent gradient is $\frac{dy}{dx}$ at the point, and the normal gradient is its negative reciprocal. It also reviews writing a line in the form $y = mx + c$ .

Common Mistakes

Using the tangent gradient instead of the normal gradient.
Taking the reciprocal instead of the negative reciprocal.
Making a sign error when combining $-4\sqrt{2}$ and $\frac{\sqrt{2}}{2}$ .
Leaving the answer in a form other than $y = mx + c$ .

Things to Be Careful About

The mark scheme accepts equivalent decimals such as $y = 5.66x - 4.95$ , but the exact surd form is preferred. Be careful that $\frac{8}{\sqrt{2}}$ simplifies to $4\sqrt{2}$ , not $4/\sqrt{2}$ .

Techniques used

substitute the parameter value into parametric equationsevaluate the derivative at the parameter valuetake the negative reciprocal to find the normal gradientform the line equation using a point and gradientrearrange into y = mx + c

Q9MediumVectors

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

\vec{OA} = \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix}, \quad \vec{OB} = \begin{pmatrix} 0 \\ 4 \\ 1 \end{pmatrix} \quad \text{and} \quad \vec{OC} = \begin{pmatrix} -3 \\ -2 \\ 2 \end{pmatrix}.

(a)

The point $D$ is such that $ABCD$ is a trapezium with $\vec{DC} = 3\vec{AB}$ .

Find the position vector of $D$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the given relation $\vec{DC} = 3\vec{AB}$ together with the link $\vec{DC} = \vec{OC} - \vec{OD}$ to express $\vec{OD}$ directly in terms of the given position vectors.

Working

First compute $\vec{AB}$ :

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

Therefore

\vec{DC} = 3\vec{AB} = \begin{pmatrix} -6 \\ 9 \\ 12 \end{pmatrix}.

Since $\vec{DC} = \vec{OC} - \vec{OD}$ , rearrange to get $\vec{OD} = \vec{OC} - \vec{DC}$ :

\vec{OD} = \begin{pmatrix} -3 \\ -2 \\ 2 \end{pmatrix} - \begin{pmatrix} -6 \\ 9 \\ 12 \end{pmatrix} = \begin{pmatrix} 3 \\ -11 \\ -10 \end{pmatrix}

Answer

\vec{OD} = \begin{pmatrix} 3 \\ -11 \\ -10 \end{pmatrix}

Final answer

(3, -11, -10)

Detailed explanation

Walkthrough

We need to find the position vector of D given that ABCD is a trapezium in which the side DC is parallel to AB and exactly three times as long. The condition $\vec{DC} = 3\vec{AB}$ ties D and C to A and B.

Step 1: Compute $\vec{AB}$ as the difference of position vectors, $\vec{OB} - \vec{OA}$ . This gives the displacement from A to B.

Step 2: Form $\vec{DC} = 3\vec{AB}$ . The factor 3 stretches the vector, so $\vec{DC}$ is three times the length of $\vec{AB}$ and points in the same direction.

Step 3: The defining relation $\vec{DC} = \vec{OC} - \vec{OD}$ comes from the fact that the vector from D to C equals the position vector of the head C minus the position vector of the tail D. Rearranging gives $\vec{OD} = \vec{OC} - \vec{DC}$ , and we substitute the known values.

Key Takeaways

$\vec{AB} = \vec{OB} - \vec{OA}$ is the displacement from A to B.
The relation $\vec{DC} = \vec{OC} - \vec{OD}$ can always be rearranged to isolate any one position vector.
The relation $\vec{DC} = k\vec{AB}$ gives both the direction and the magnitude of the side $\vec{DC}$ in terms of $\vec{AB}$ .

Common Mistakes

Writing $\vec{AB}$ as $\vec{OA} - \vec{OB}$ instead of $\vec{OB} - \vec{OA}$ (this would actually be $\vec{BA}$ ).
Using $\vec{OD} - \vec{OC}$ instead of $\vec{OC} - \vec{OD}$ when rearranging, which gives the wrong sign for $\vec{DC}$ .
Sign errors when subtracting two column vectors component by component.

Things to Be Careful About

The order of letters in a vector matters: $\vec{AB} \neq \vec{BA}$ — they are negatives of each other.
A scalar multiple preserves direction; $\vec{DC} = 3\vec{AB}$ means DC is parallel to AB and points the same way, with D on the opposite side of A from B (since the trapezium is ABCD rather than ABDC).

Techniques used

compute vector AB from position vectorsform vector DC as a scalar multiple of ABrearrange OD = OC - DC to find the position vector of D

(b)

The diagonals of the trapezium intersect at the point $P$ .

Find the position vector of $P$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Write parametric vector equations for both diagonals AC and BD, equate their components to find the parameters, then substitute back to obtain the position vector of P.

Working

The direction of diagonal AC is

\vec{AC} = \vec{OC} - \vec{OA} = \begin{pmatrix} -3 \\ -2 \\ 2 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix} = \begin{pmatrix} -5 \\ -3 \\ 5 \end{pmatrix}

Using D from part (a), the direction of diagonal BD is

\vec{BD} = \vec{OD} - \vec{OB} = \begin{pmatrix} 3 \\ -11 \\ -10 \end{pmatrix} - \begin{pmatrix} 0 \\ 4 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ -15 \\ -11 \end{pmatrix}

The parametric equations of the two diagonals are therefore

\begin{aligned} \text{AC:} \quad \vec{r} &= \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix} + \lambda \begin{pmatrix} -5 \\ -3 \\ 5 \end{pmatrix} \\ \text{BD:} \quad \vec{r} &= \begin{pmatrix} 0 \\ 4 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ -15 \\ -11 \end{pmatrix} \end{aligned}

At the intersection P the two expressions are equal. Equating the $x$ - and $y$ -components:

\begin{aligned} 2 - 5\lambda &= 3\mu \quad &(1)\\ 1 - 3\lambda &= 4 - 15\mu \quad &(2) \end{aligned}

From (1), $\mu = \dfrac{2 - 5\lambda}{3}$ . Substitute into (2):

1 - 3\lambda = 4 - 15 \cdot \frac{2 - 5\lambda}{3} = 4 - 5(2 - 5\lambda) = -6 + 25\lambda

So $1 - 3\lambda = -6 + 25\lambda$ , which gives $28\lambda = 7$ , hence $\lambda = \dfrac{1}{4}$ .

Check using the $z$ -components: $-3 + 5 \cdot \dfrac{1}{4} = -\dfrac{7}{4}$ and $1 - 11 \cdot \dfrac{1}{4} = -\dfrac{7}{4}$ . ✓

Substitute $\lambda = \dfrac{1}{4}$ back into the equation of line AC:

\vec{OP} = \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix} + \frac{1}{4}\begin{pmatrix} -5 \\ -3 \\ 5 \end{pmatrix} = \begin{pmatrix} 2 - \frac{5}{4} \\ 1 - \frac{3}{4} \\ -3 + \frac{5}{4} \end{pmatrix} = \begin{pmatrix} \frac{3}{4} \\ \frac{1}{4} \\ -\frac{7}{4} \end{pmatrix}

Answer

\vec{OP} = \begin{pmatrix} 3/4 \\ 1/4 \\ -7/4 \end{pmatrix}

Final answer

(3/4, 1/4, -7/4)

Detailed explanation

Walkthrough

The two diagonals of trapezium ABCD are AC and BD, and they meet at the interior point P. To locate P, we use the standard technique of writing each diagonal as a parametric line and equating the two expressions.

Step 1: Compute the direction $\vec{AC} = \vec{OC} - \vec{OA}$ . This is the displacement from A to C.

Step 2: Compute the direction $\vec{BD} = \vec{OD} - \vec{OB}$ using the position vector of D from part (a).

Step 3: Write each diagonal in the form $\vec{r} = \vec{a} + t\vec{d}$ , using A as the basepoint of line AC and B as the basepoint of line BD. We use different parameter names ( $\lambda$ and $\mu$ ) so we can keep track of which line a parameter belongs to.

Step 4: At the intersection, the two expressions are equal, so the $x$ -, $y$ -, and $z$ -components must all match. This yields three equations in two unknowns, of which we use any two to solve. We pick $x$ and $y$ because they are simplest.

Step 5: Eliminate one parameter, solve for the other, and substitute back to get the position vector of P. The third component should agree automatically — this is a useful consistency check that confirms the lines really do meet.

An alternative observation: in a trapezium with $AB \parallel DC$ and $DC = 3AB$ , the diagonals divide each other in the ratio $1:3$ , so $\vec{OP} = \vec{OA} + \tfrac{1}{4}\vec{AC}$ — the same answer obtained more directly.

Key Takeaways

A line through two points $P$ and $Q$ can be written as $\vec{r} = \vec{OP} + t\vec{PQ}$ for some real parameter $t$ .
To find where two lines meet in 3D, write both in parametric form, equate components, and solve. The third component acts as a check that the lines are not skew.
In a trapezium with parallel sides in ratio $1:3$ , the diagonals cut each other in the same ratio (similar-triangles argument).

Common Mistakes

Using the wrong direction vector (e.g. $\vec{CA}$ instead of $\vec{AC}$ ): this still works but the parameter value at P has the opposite sign.
Failing to verify the third component after solving — if the lines are skew rather than intersecting, the third equation will be inconsistent.
Sign slips in $\vec{OD} - \vec{OB}$ or in substituting $\mu$ back into line BD.

Things to Be Careful About

The mark scheme uses the equivalent form $\vec{r} = (2, 1, -3) + \lambda(5, 3, -5)$ for line AC, which has $\vec{CA}$ as the direction. This gives $\lambda = -\tfrac{1}{4}$ instead of our $\lambda = \tfrac{1}{4}$ but the same P.
Always substitute the found parameter back and confirm all three components agree before reporting the answer.

Techniques used

form parametric vector equation of a line through two pointsequate components of two parametric linessolve simultaneous equations for the parameterssubstitute the parameter back into a line equation

(c)

Using a scalar product, calculate angle $ABC$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Form vectors $\vec{BA}$ and $\vec{BC}$ from the common vertex B, then apply the scalar product formula $\cos\theta = \dfrac{\vec{BA} \cdot \vec{BC}}{|\vec{BA}|\,|\vec{BC}|}$ to obtain $\angle ABC$ .

Working

Find the two direction vectors from B:

\begin{aligned} \vec{BA} &= \vec{OA} - \vec{OB} = \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix} - \begin{pmatrix} 0 \\ 4 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ -4 \end{pmatrix} \\ \vec{BC} &= \vec{OC} - \vec{OB} = \begin{pmatrix} -3 \\ -2 \\ 2 \end{pmatrix} - \begin{pmatrix} 0 \\ 4 \\ 1 \end{pmatrix} = \begin{pmatrix} -3 \\ -6 \\ 1 \end{pmatrix} \end{aligned}

Compute the scalar product:

\vec{BA} \cdot \vec{BC} = (2)(-3) + (-3)(-6) + (-4)(1) = -6 + 18 - 4 = 8

Compute the magnitudes:

\begin{aligned} |\vec{BA}| &= \sqrt{2^2 + (-3)^2 + (-4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \\ |\vec{BC}| &= \sqrt{(-3)^2 + (-6)^2 + 1^2} = \sqrt{9 + 36 + 1} = \sqrt{46} \end{aligned}

Apply the formula:

\cos(\angle ABC) = \frac{\vec{BA} \cdot \vec{BC}}{|\vec{BA}|\,|\vec{BC}|} = \frac{8}{\sqrt{29}\,\sqrt{46}} = \frac{8}{\sqrt{1334}}

Therefore

\angle ABC = \cos^{-1}\!\left(\frac{8}{\sqrt{1334}}\right) = 77.3°\ \text{(3 s.f.)}

Equivalently, $\angle ABC \approx 1.35$ radians.

Answer

\angle ABC = 77.3°\ \text{(or 1.35 rad)}

Final answer

77.3° (or 1.35 rad)

Detailed explanation

Walkthrough

The angle ABC is the angle at vertex B between the two sides BA and BC. The scalar product formula gives this directly from the two direction vectors:

\cos(\angle ABC) = \frac{\vec{BA} \cdot \vec{BC}}{|\vec{BA}|\,|\vec{BC}|}

Step 1: Compute $\vec{BA}$ and $\vec{BC}$ . These are the vectors from B to A and from B to C respectively, each obtained by subtracting the position vector of B from that of the other vertex.

Step 2: Compute the scalar product by multiplying corresponding components and summing. The negative times negative $(-3)(-6) = +18$ is the dominant positive term, and the final result is $8$ .

Step 3: Compute each magnitude using $\sqrt{x^2 + y^2 + z^2}$ . Squaring removes sign issues, and the three terms are then summed and square-rooted.

Step 4: Divide the scalar product by the product of the magnitudes, then take the inverse cosine. A small positive cosine gives a large angle; here $8/\sqrt{1334} \approx 0.219$ corresponds to about $77.3°$ .

Key Takeaways

The scalar product gives angles between vectors via $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$ .
In 3D, magnitudes use $\sqrt{x^2 + y^2 + z^2}$ and scalar products use $a_1 b_1 + a_2 b_2 + a_3 b_3$ .
The angle between two vectors is always in $[0°, 180°]$ , so $\cos^{-1}$ returns a unique value.

Common Mistakes

Using $\vec{AB}$ and $\vec{CB}$ (instead of $\vec{BA}$ and $\vec{BC}$ ): these point into B rather than out of B and would give the supplementary angle. Both vectors must emanate from the same vertex.
Forgetting to take the inverse cosine, leaving the answer as $\tfrac{8}{\sqrt{1334}}$ instead of converting to degrees.
Sign errors in the scalar product (e.g. treating $-3 \cdot -4$ as $-12$ instead of $+12$ ).

Things to Be Careful About

The mark scheme accepts either $77.3°$ or $1.35$ rad — both come from the same numerical value.
The exact value is $77.347\dots°$ ; round at the end, not part-way through the calculation.
Using $\vec{BA}$ and $\vec{BC}$ vs. $\vec{AB}$ and $\vec{CB}$ gives the same final angle because the scalar product and the product of magnitudes are both unchanged when both vectors are negated.

Techniques used

compute direction vectors from the common vertex Bevaluate the scalar product componentwisecompute magnitudes of 3D vectorsapply inverse cosine to obtain the angle

Q10Medium-HardDifferential EquationsAlgebra

A balloon in the shape of a sphere has volume $V$ and radius $r$ . Air is pumped into the balloon at a constant rate of $40\pi$ starting when time $t = 0$ and $r = 0$ . At the same time, air begins to flow out of the balloon at a rate of $0.8\pi r$ . The balloon remains a sphere at all times.

(a)

Show that $r$ and $t$ satisfy the differential equation

\frac{dr}{dt} = \frac{50 - r}{5r^2}.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Let $V$ be the volume of the spherical balloon. The net rate at which volume changes is inflow minus outflow. Relate $\frac{dV}{dt}$ to $\frac{dr}{dt}$ using the sphere volume formula and the chain rule.

Working

The net rate of change of volume is

\frac{dV}{dt} = 40\pi - 0.8\pi r.

For a sphere,

V = \frac{4}{3}\pi r^3,

so

\frac{dV}{dr} = 4\pi r^2.

By the chain rule,

\frac{dV}{dt} = \frac{dV}{dr}\frac{dr}{dt} = 4\pi r^2\frac{dr}{dt}.

Equating the two expressions for $\frac{dV}{dt}$ :

4\pi r^2\frac{dr}{dt} = 40\pi - 0.8\pi r.

Divide by $4\pi r^2$ :

\frac{dr}{dt} = \frac{40 - 0.8r}{4r^2}.

Since $40 - 0.8r = 0.8(50-r)$ and $0.8 = \frac{4}{5}$ ,

\frac{dr}{dt} = \frac{0.8(50-r)}{4r^2} = \frac{50-r}{5r^2}.

Answer

\frac{dr}{dt} = \frac{50-r}{5r^2}

Final answer

dr/dt = (50-r)/(5r^2)

Detailed explanation

Walkthrough

We start from the physical statement: volume is being added at $40\pi$ and removed at $0.8\pi r$ , so the net rate of change of volume is $\frac{dV}{dt}=40\pi-0.8\pi r$ . Next, because the balloon is always a sphere, $V=\frac{4}{3}\pi r^3$ , so differentiating with respect to $r$ gives $\frac{dV}{dr}=4\pi r^2$ . The chain rule links the two rates: $\frac{dV}{dt}=\frac{dV}{dr}\frac{dr}{dt}$ . Substituting and dividing by $4\pi r^2$ gives the required differential equation.

Key Takeaways

This part tests the ability to translate a rate-of-change statement into a differential equation, and to use the chain rule when the intermediate variable is the radius. The key idea is that $\frac{dV}{dt}$ can be expressed both from the given rates and from the geometry of the sphere.

Common Mistakes

A common mistake is to ignore the outflow and write $\frac{dV}{dt}=40\pi$ only. Another is to confuse $\frac{dV}{dr}$ with $\frac{dV}{dt}$ . The mark scheme requires a complete correct statement for each of the three steps.

Things to Be Careful About

The factor $\pi$ cancels, so it is easy to lose it incorrectly. Also, the outflow term is $0.8\pi r$ , not $0.8\pi$ , so the radius must remain in the term. Since $r=0$ initially, the expression is not valid at the exact starting instant, but the differential equation is derived for $r>0$ .

Techniques used

write the net rate of change of volumedifferentiate the sphere volume formulaapply the chain rule to relate rates

(b)

Find the quotient and remainder when $5r^2$ is divided by $50 - r$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Divide $5r^2$ by $50-r$ . Because the divisor has leading term $-r$ , the quotient will have negative coefficients. Express the result as $5r^2=(50-r)Q(r)+R$ .

Working

Long division:

5r^2 \div (50-r)

The first term of the quotient is

\frac{5r^2}{-r} = -5r.

Multiplying:

(50-r)(-5r) = -250r + 5r^2.

Subtracting gives $250r$ . The next term of the quotient is

\frac{250r}{-r} = -250.

Multiplying:

(50-r)(-250) = -12500 + 250r.

Subtracting leaves remainder $12500$ . Hence

5r^2 = (50-r)(-5r-250) + 12500.

Answer

Quotient: $-5r-250$ ; remainder: $12500$ .

Final answer

quotient = -5r - 250, remainder = 12500

Detailed explanation

Walkthrough

We divide $5r^2$ by $50-r$ . Since the divisor is $50-r$ , its leading term is $-r$ . The first quotient term is found by dividing $5r^2$ by $-r$ , giving $-5r$ . Multiplying back and subtracting removes the $5r^2$ term and leaves $250r$ . Then $250r$ divided by $-r$ gives $-250$ . Multiplying and subtracting leaves the constant remainder $12500$ . Therefore $5r^2=(50-r)(-5r-250)+12500$ .

Key Takeaways

This part practises polynomial division by a linear factor written with a negative leading coefficient. It is important to keep the signs correct when the divisor is $50-r$ rather than $r-50$ .

Common Mistakes

A common mistake is to divide by $r-50$ and obtain quotient $5r+250$ ; this gives the same remainder but the quotient differs by a sign. The mark scheme allows this only if the quotient and remainder are clearly identified consistently.

Things to Be Careful About

The divisor is $50-r$ , not $r-50$ . The quotient is $-5r-250$ , while the remainder is positive $12500$ . If using the remainder theorem, it can only give the remainder, not the quotient.

Techniques used

divide a quadratic polynomial by a linear divisorhandle a divisor with negative leading coefficientstate the quotient and remainder

(c)

Solve the differential equation in part (a), obtaining an expression for $t$ in terms of $r$ .

6M Similar

DifficultyMedium-Hard

Worked solution

Approach

Invert the differential equation to write $\frac{dt}{dr}$ , separate the variables, use the division from part (b) to rewrite the integrand, integrate, and then use $t=0,\ r=0$ to find the constant.

Working

From part (a),

\frac{dr}{dt} = \frac{50-r}{5r^2}.

Therefore

\frac{dt}{dr} = \frac{5r^2}{50-r}.

Using the division in part (b),

\frac{5r^2}{50-r} = -5r - 250 + \frac{12500}{50-r}.

So

dt = \left(-5r - 250 + \frac{12500}{50-r}\right) dr.

Integrate:

t = \int \left(-5r - 250 + \frac{12500}{50-r}\right) dr.

t = -\frac{5}{2}r^2 - 250r - 12500\ln(50-r) + C.

Use $t=0$ when $r=0$ :

0 = 0 - 0 - 12500\ln(50) + C,

so

C = 12500\ln 50.

Hence

t = -\frac{5}{2}r^2 - 250r - 12500\ln(50-r) + 12500\ln 50.

Answer

t = -\frac{5}{2}r^2 - 250r - 12500\ln(50-r) + 12500\ln 50

Final answer

t = -5/2 r^2 - 250r - 12500 ln(50-r) + 12500 ln 50

Detailed explanation

Walkthrough

We need $t$ in terms of $r$ , so we invert the given differential equation to get $\frac{dt}{dr}=\frac{5r^2}{50-r}$ . The integrand is not a simple standard form, but part (b) rewrites it as $-5r-250+\frac{12500}{50-r}$ . Integrating term by term: $-5r$ gives $-\frac{5}{2}r^2$ , $-250$ gives $-250r$ , and $\frac{12500}{50-r}$ gives $-12500\ln(50-r)$ because the derivative of $\ln(50-r)$ is $-\frac{1}{50-r}$ . Finally, the condition $t=0$ when $r=0$ determines the constant $C=12500\ln 50$ .

Key Takeaways

This part combines separable differential equations with algebraic preparation of the integrand. The division in part (b) is not an isolated exercise; it is exactly what makes the integral possible. The integral of $\frac{k}{50-r}$ is $-k\ln(50-r)$ , not $+k\ln(50-r)$ .

Common Mistakes

A common mistake is to forget the minus sign when integrating $\frac{12500}{50-r}$ . Another is to omit the constant of integration or to apply the initial condition incorrectly. The mark scheme allows missing $dr$ or $dt$ or integral signs, but not both.

Things to Be Careful About

The expression is valid for $0 \le r < 50$ because the logarithm requires $50-r>0$ . When using the initial condition, substitute $r=0$ into $\ln(50-r)$ to get $\ln 50$ , not $\ln 0$ . Also, the constant $12500\ln 50$ may be left as it is or evaluated numerically.

Techniques used

separate variablesrewrite the integrand using polynomial divisionintegrate a rational expressionapply the initial condition to determine the constant

(d)

Find the value of $t$ when the radius of the balloon is 12.

1M Similar

Q11Medium-HardDifferentiationIntegration

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

(a)

Find $f'(x)$ and hence find the exact coordinates of the stationary point of the curve with equation $y = f(x)$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Differentiate $f(x)$ using the quotient rule, set $f'(x)=0$ to locate the stationary point, solve the resulting exponential equation, and then substitute back to find the $y$ -coordinate.

Working

Let $N(x)=2e^{2x}$ and $D(x)=e^{2x}-3e^x+2$ . Then

\begin{aligned} N'(x) &= 4e^{2x}, \\ D'(x) &= 2e^{2x} - 3e^x. \end{aligned}

Using the quotient rule,

\begin{aligned} f'(x) &= \frac{N'D - ND'}{D^2} \\ &= \frac{4e^{2x}\left(e^{2x}-3e^x+2\right) - 2e^{2x}\left(2e^{2x}-3e^x\right)}{\left(e^{2x}-3e^x+2\right)^2}. \end{aligned}

Simplify the numerator:

\begin{aligned} 4e^{2x}\left(e^{2x}-3e^x+2\right) - 2e^{2x}\left(2e^{2x}-3e^x\right) &= 4e^{4x}-12e^{3x}+8e^{2x}-4e^{4x}+6e^{3x} \\ &= -6e^{3x}+8e^{2x} \\ &= 2e^{2x}\left(4-3e^x\right). \end{aligned}

Hence

f'(x) = \frac{2e^{2x}\left(4-3e^x\right)}{\left(e^{2x}-3e^x+2\right)^2}.

At a stationary point, $f'(x)=0$ . Since $2e^{2x}>0$ and the denominator is non-zero at the stationary point,

4-3e^x = 0 \quad \Rightarrow \quad e^x = \frac{4}{3}.

Therefore

x = \ln\frac{4}{3}.

Substitute $u=e^x=\frac{4}{3}$ into $f(x)$ :

\begin{aligned} y &= \frac{2u^2}{u^2-3u+2} \\ &= \frac{2\left(\frac{16}{9}\right)}{\frac{16}{9} - 4 + 2} \\ &= \frac{\frac{32}{9}}{-\frac{2}{9}} \\ &= -16. \end{aligned}

Answer

The stationary point is

\left(\ln\frac{4}{3},\ -16\right).

Final answer

x = ln(4/3), y = -16

Detailed explanation

Walkthrough

We are told that $f(x)$ is a quotient of two exponential expressions, so the quotient rule is the natural tool. Write the numerator as $N=2e^{2x}$ and the denominator as $D=e^{2x}-3e^x+2$ . Differentiate each separately: $N'=4e^{2x}$ and $D'=2e^{2x}-3e^x$ . Then substitute into the quotient rule formula. After expanding the numerator, the $e^{4x}$ terms cancel and the remaining terms factor as $2e^{2x}(4-3e^x)$ .

For a stationary point, set $f'(x)=0$ . A fraction is zero when its numerator is zero, provided the denominator is not also zero. Since $2e^{2x}$ is never zero, we need $4-3e^x=0$ , giving $e^x=\frac{4}{3}$ and hence $x=\ln\frac{4}{3}$ . Finally substitute this value back into $f(x)$ to get the $y$ -coordinate. Using $u=e^x=\frac{4}{3}$ makes the substitution tidy and gives $y=-16$ .

Key Takeaways

This question tests the quotient rule for a quotient of exponential functions, the fact that stationary points occur where the derivative is zero, and the use of natural logarithms to solve $e^x=k$ . It also reinforces that finding a stationary point requires both coordinates, not just the $x$ -value.

Common Mistakes

Forgetting to square the denominator in the quotient rule.
Expanding the numerator incorrectly, especially the signs when subtracting $ND'$ .
Setting the whole fraction equal to zero and incorrectly trying to solve with the denominator.
Forgetting that $e^{2x}>0$ for all real $x$ , so it cannot contribute a solution.
Finding only $x=\ln\frac{4}{3}$ and not substituting back to find $y=-16$ .

Things to Be Careful About

The question asks for $f'(x)$ , so the derivative must be given in a complete form, not just the numerator. The coordinates must be exact, so leave the answer as $x=\ln\frac{4}{3}$ rather than a decimal. Also check that the denominator is non-zero at the stationary point; here it equals $-\frac{2}{9}$ , so the point is valid.

Techniques used

apply the quotient rule to differentiate a quotient of exponentialsset the derivative equal to zero to locate stationary pointssolve an exponential equation using natural logarithmssubstitute back to find the y-coordinate

(b)

Use the substitution $u = e^x$ and partial fractions to find the exact value of $\int_{\ln 3}^{\ln 5} f(x) \, dx$ .

Give your answer in the form $\ln a$ , where $a$ is a rational number in its simplest form.

9M Similar

DifficultyMedium-Hard

Worked solution

Approach

Use the substitution $u=e^x$ to convert the integral into one in $u$ , decompose the integrand into partial fractions, integrate term by term, and evaluate at the new limits.

Working

Let $u=e^x$ . Then

\frac{du}{dx}=e^x \quad \Rightarrow \quad du = e^x\,dx.

Since $e^{2x}\,dx = e^x \cdot e^x\,dx = u\,du$ , the integral becomes

\int_{\ln 3}^{\ln 5} \frac{2e^{2x}}{e^{2x}-3e^x+2}\,dx = \int_{3}^{5} \frac{2u}{u^2-3u+2}\,du.

Factorise the denominator:

u^2-3u+2 = (u-1)(u-2).

Write

\frac{2u}{(u-1)(u-2)} = \frac{A}{u-1} + \frac{B}{u-2}.

Multiplying through by $(u-1)(u-2)$ :

2u = A(u-2)+B(u-1).

Set $u=1$ :

2 = -A \quad \Rightarrow \quad A = -2.

Set $u=2$ :

4 = B \quad \Rightarrow \quad B = 4.

Thus

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

Integrate:

\int \left(\frac{-2}{u-1} + \frac{4}{u-2}\right) du = -2\ln(u-1) + 4\ln(u-2) + C.

Evaluate from $u=3$ to $u=5$ :

\begin{aligned} \int_3^5 \frac{2u}{(u-1)(u-2)}\,du &= \left[-2\ln(u-1)+4\ln(u-2)\right]_3^5 \\ &= \left(-2\ln 4 + 4\ln 3\right) - \left(-2\ln 2 + 4\ln 1\right) \\ &= -2\ln 4 + 4\ln 3 + 2\ln 2. \end{aligned}

Simplify using $\ln 4 = 2\ln 2$ and $\ln 1 = 0$ :

\begin{aligned} -2\ln 4 + 4\ln 3 + 2\ln 2 &= -4\ln 2 + 4\ln 3 + 2\ln 2 \\ &= 4\ln 3 - 2\ln 2 \\ &= \ln\left(\frac{3^4}{2^2}\right) \\ &= \ln\frac{81}{4}. \end{aligned}

Answer

\int_{\ln 3}^{\ln 5} f(x)\,dx = \ln\frac{81}{4}.

Final answer

ln(81/4)

Detailed explanation

Walkthrough

Because the integrand contains $e^x$ and $e^{2x}$ , the substitution $u=e^x$ is natural. Since $\frac{du}{dx}=e^x$ , we have $du=e^x\,dx$ , so $e^{2x}\,dx = e^x \cdot e^x\,dx = u\,du$ . This turns the integral into $\int \frac{2u}{u^2-3u+2}\,du$ . Remember to change the limits: $x=\ln 3$ becomes $u=3$ , and $x=\ln 5$ becomes $u=5$ .

Next factor the denominator as $(u-1)(u-2)$ and write the integrand as $\frac{A}{u-1}+\frac{B}{u-2}$ . Clearing denominators gives $2u=A(u-2)+B(u-1)$ . Substituting $u=1$ and $u=2$ quickly gives $A=-2$ and $B=4$ . This is the partial fraction decomposition.

Integrating term by term, $\int \frac{-2}{u-1}\,du = -2\ln(u-1)$ and $\int \frac{4}{u-2}\,du = 4\ln(u-2)$ . Evaluating from $u=3$ to $u=5$ and using log laws gives $\ln\frac{81}{4}$ .

Key Takeaways

This question combines three core techniques: substitution to simplify an exponential integrand, partial fractions to split a rational function, and integration of reciprocal linear terms to logarithms. It also reviews the laws of logarithms needed to combine terms into a single logarithm.

Common Mistakes

Forgetting to change the limits when substituting $u=e^x$ .
Writing $du=dx$ instead of $du=e^x\,dx$ .
Failing to convert $e^{2x}\,dx$ into $u\,du$ , leaving an incorrect integrand.
Making sign errors when finding $A$ and $B$ in the partial fractions.
Combining logarithms incorrectly, e.g. writing $4\ln 3-2\ln 2$ as $\ln(3^4/2^2)$ incorrectly.

Things to Be Careful About

The final answer must be in the form $\ln a$ with $a$ rational in simplest form, so simplify $\ln\frac{81}{4}$ fully. Since the interval is $u=3$ to $u=5$ , both $u-1$ and $u-2$ are positive, so no absolute value signs are needed. The mark scheme requires the substitution step, the partial fraction form, and the use of limits to be clearly shown.

Techniques used

use an exponential substitution to change variablesdecompose a rational integrand into partial fractionsintegrate reciprocal linear terms to logarithmsevaluate a definite integral and combine logarithms

Mathematics 9709/32 — October/November 2024

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working