(a) By sketching a suitable pair of graphs, show that the equation $2 + e^{-0.2x} = \ln(1 + x)$ has only one root. (b) Show by calculation that this root lies between 7 and 9. (c) Use the iterative formula $$ x_{n+1} = \exp(2 + e^{-0.2x_n}) - 1 $$ to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [$\exp(x)$ is an alternative notation for $e^x$.]

(a) The curves y = 2 + e^{-0.2x} and y = ln(1+x) intersect at exactly one point. · (b) f(7) = 0.1672 > 0 and f(9) = -0.1373 < 0, so the root lies between 7 and 9. · (c) 8.03

Let $f(x) = \frac{5x^2 + 8x + 5}{(1 + 2x)(2 + x^2)}$. (a) Express $f(x)$ in partial fractions. (b) Hence find the coefficient of $x^3$ in the expansion of $f(x)$.

(a) f(x) = 1/(1+2x) + (2x+3)/(2+x^2) · (b) -17/2

(a) Given that $z = 1 + yi$ and that $y$ is a real number, express $\frac{1}{z}$ in the form $a + bi$, where $a$ and $b$ are functions of $y$. (b) Show that $(a - \frac{1}{2})^2 + b^2 = \frac{1}{4}$, where $a$ and $b$ are the functions of $y$ found in part (a). (c) On a single Argand diagram, sketch the loci given by the equations $\text{Re}(z) = 1$ and $ z - \frac{1}{2} = \frac{1}{2}$, where $z$ is a complex number. (d) The complex number $z$ is such that $\text{Re}(z) =…

(a) a = 1/(1+y^2), b = -y/(1+y^2) · (b) (a - 1/2)^2 + b^2 = 1/4 · (c) Vertical line x = 1 and circle centre (1/2, 0) radius 1/2 · (d) Circle with centre 1/2 + 0i and radius 1/2

The position vector of point $A$ relative to the origin $O$ is $\vec{OA} = 8\mathbf{i} - 5\mathbf{j} + 6\mathbf{k}$. The line $l$ passes through $A$ and is parallel to the vector $2\mathbf{i} + \mathbf{j} + 4\mathbf{k}$. (a) State a vector equation for $l$. (b) The position vector of point $B$ relative to the origin $O$ is $\vec{OB} = -t\mathbf{i} + 4t\mathbf{j} + 3t\mathbf{k}$, where $t$ is a constant. The line $l$ also passes through $B$. Find the value of $t$. (c) The…

(a) r = 8i - 5j + 6k + λ(2i + j + 4k) · (b) t = -2 · (c) a = -2 or a = -86

9709/31 Oct/Nov 2024 — Mathematics Solved Past Paper

Q15MMediumAlgebra

The polynomial $4x^3 + ax^2 + 5x + b$ , where $a$ and $b$ are constants, is denoted by $p(x)$ . It is given that $(2x + 1)$ is a factor of $p(x)$ . When $p(x)$ is divided by $(x - 4)$ the remainder is equal to 3 times the remainder when $p(x)$ is divided by $(x - 2)$ .

Find the values of $a$ and $b$ .

DifficultyMedium

Worked solution

Approach

Use the factor theorem: since $(2x+1)$ is a factor, $p(-\frac{1}{2}) = 0$ . Use the remainder theorem: the remainder when dividing by $(x-k)$ is $p(k)$ , so the given condition gives $p(4) = 3p(2)$ . This yields two linear equations in $a$ and $b$ ; solve them simultaneously.

Working

Since $(2x+1)$ is a factor, $p(-\frac{1}{2}) = 0$ :

4\left(-\frac{1}{2}\right)^3 + a\left(-\frac{1}{2}\right)^2 + 5\left(-\frac{1}{2}\right) + b = 0

-\frac{1}{2} + \frac{a}{4} - \frac{5}{2} + b = 0

-3 + \frac{a}{4} + b = 0

So $\frac{a}{4} + b = 3$ , i.e. $a + 4b = 12$ .

By the remainder theorem, the remainder on division by $(x-4)$ is $p(4)$ , and on division by $(x-2)$ is $p(2)$ . Given $p(4) = 3p(2)$ :

p(4) = 4(4)^3 + a(4)^2 + 5(4) + b = 256 + 16a + 20 + b = 276 + 16a + b

p(2) = 4(2)^3 + a(2)^2 + 5(2) + b = 32 + 4a + 10 + b = 42 + 4a + b

276 + 16a + b = 3(42 + 4a + b)

276 + 16a + b = 126 + 12a + 3b

150 + 4a - 2b = 0

So $-2a + b = 75$ .

Now solve the system:

\begin{aligned} a + 4b &= 12 \\ -2a + b &= 75 \end{aligned}

From the second equation, $b = 75 + 2a$ . Substitute into the first:

a + 4(75 + 2a) = 12

a + 300 + 8a = 12

9a = -288

a = -32

Then $b = 75 + 2(-32) = 75 - 64 = 11$ .

Answer

a = -32, \quad b = 11

Final answer

a = -32, b = 11

Detailed explanation

Walkthrough

We are told that $(2x+1)$ is a factor of $p(x)$ . By the factor theorem, if $(2x+1)$ is a factor, then substituting the value of $x$ that makes $2x+1 = 0$ (i.e. $x = -\frac{1}{2}$ ) into $p(x)$ must give 0. This gives our first equation in $a$ and $b$ .

Then we use the remainder theorem: when a polynomial is divided by $(x-k)$ , the remainder is $p(k)$ . So the remainder when dividing by $(x-4)$ is $p(4)$ , and when dividing by $(x-2)$ is $p(2)$ . The problem tells us the first remainder is 3 times the second, so $p(4) = 3p(2)$ . This gives our second equation.

Finally we have two linear equations in $a$ and $b$ , which we solve by substitution (or elimination).

Key Takeaways

The factor theorem: if $(cx + d)$ is a factor of $p(x)$ , then $p(-d/c) = 0$ .
The remainder theorem: the remainder on division by $(x - k)$ is $p(k)$ .
Forming equations from given conditions and solving simultaneous linear equations.

Common Mistakes

Forgetting to use $x = -\frac{1}{2}$ rather than $x = \frac{1}{2}$ when $(2x+1)$ is the factor. The root is $x = -\frac{1}{2}$ .
Sign errors when substituting negative values into the polynomial.
Confusing which remainder is larger: the problem says the remainder on division by $(x-4)$ is 3 times the remainder on division by $(x-2)$ , so $p(4) = 3p(2)$ , not $3p(4) = p(2)$ . The mark scheme condones if 3 is on the wrong side, but it is better to get it right.
Arithmetic errors when simplifying $p(4)$ and $p(2)$ .
Not showing the substitution step — the mark scheme requires M1 for substituting $x = -\frac{1}{2}$ and equating to zero.

Things to Be Careful About

The factor $(2x+1)$ gives root $x = -\frac{1}{2}$ , not $x = 2$ or $x = 1$ .
When computing $p(4)$ : $4 \cdot 4^3 = 4 \cdot 64 = 256$ , $a \cdot 4^2 = 16a$ , $5 \cdot 4 = 20$ .
When computing $p(2)$ : $4 \cdot 2^3 = 4 \cdot 8 = 32$ , $a \cdot 2^2 = 4a$ , $5 \cdot 2 = 10$ .
The mark scheme allows equivalent forms of the equations, e.g. $\frac{a}{4} + b = 3$ or $a + 4b = 12$ ; and $-2a + b = 75$ or $b = 75 + 2a$ .
The final answer must give both $a$ and $b$ .

Techniques used

apply the factor theoremapply the remainder theoremform and solve a system of linear equations

Approach

Use integration by parts with $u = \ln 3x$ and $\frac{dv}{dx} = x^2$ , since differentiating the logarithm simplifies the integrand. Then evaluate the resulting definite integral between the limits and simplify using $\ln 9 = 2\ln 3$ .

Working

Let

I = \int_1^3 x^2 \ln 3x \, dx.

Choose

u = \ln 3x, \quad \frac{dv}{dx} = x^2.

Then

\frac{du}{dx} = \frac{1}{x}, \quad v = \frac{x^3}{3}.

Integration by parts gives

\begin{aligned} I &= \frac{x^3}{3}\ln 3x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx \\ &= \frac{x^3}{3}\ln 3x - \frac{1}{3}\int x^2 \, dx \\ &= \frac{x^3}{3}\ln 3x - \frac{1}{9}x^3. \end{aligned}

Evaluate from 1 to 3:

\begin{aligned} I &= \left[\frac{x^3}{3}\ln 3x - \frac{x^3}{9}\right]_1^3 \\ &= \left(9\ln 9 - 3\right) - \left(\frac{1}{3}\ln 3 - \frac{1}{9}\right) \\ &= 9\ln 9 - \frac{1}{3}\ln 3 - \frac{26}{9}. \end{aligned}

Since $\ln 9 = 2\ln 3$ ,

I = 18\ln 3 - \frac{1}{3}\ln 3 - \frac{26}{9} = \frac{53}{3}\ln 3 - \frac{26}{9}.

Answer

\frac{53}{3}\ln 3 - \frac{26}{9}

Final answer

53/3 ln 3 - 26/9

Detailed explanation

Walkthrough

This is a definite integral of a product, so integration by parts is the natural method. The product is $x^2 \ln 3x$ . We choose $u = \ln 3x$ because its derivative is $\frac{1}{x}$ , which lowers the power of $x$ when multiplied by $x^3$ . We choose $\frac{dv}{dx} = x^2$ , giving $v = \frac{x^3}{3}$ .

Applying the formula $\int u \frac{dv}{dx}\,dx = uv - \int v \frac{du}{dx}\,dx$ , the new integral is $\int \frac{x^3}{3} \cdot \frac{1}{x}\,dx = \frac{1}{3}\int x^2\,dx$ . This is now a simple power integral, so the antiderivative becomes $\frac{x^3}{3}\ln 3x - \frac{x^3}{9}$ .

Next, substitute the upper limit $x=3$ and the lower limit $x=1$ , and subtract the lower value from the upper value. At $x=3$ , the value is $9\ln 9 - 3$ ; at $x=1$ , it is $\frac{1}{3}\ln 3 - \frac{1}{9}$ . Subtracting gives $9\ln 9 - \frac{1}{3}\ln 3 - \frac{26}{9}$ .

Finally, use $\ln 9 = \ln 3^2 = 2\ln 3$ to combine the logarithmic terms: $18\ln 3 - \frac{1}{3}\ln 3 = \frac{53}{3}\ln 3$ . The exact value is $\frac{53}{3}\ln 3 - \frac{26}{9}$ .

Key Takeaways

Integration by parts is used for products where one factor simplifies when differentiated.
Choosing $u$ as the logarithmic factor and $dv$ as the polynomial factor is usually effective.
Definite integrals require evaluating the antiderivative at both limits and subtracting correctly.
Logarithm laws such as $\ln a^n = n\ln a$ are often needed to simplify the final answer into the required form.

Common Mistakes

Forgetting the minus sign in the integration by parts formula.
Writing the derivative of $\ln 3x$ as $\frac{1}{3x}$ instead of $\frac{1}{x}$ .
Forgetting to subtract the value at the lower limit when evaluating a definite integral.
Leaving the answer as $9\ln 9 - \frac{1}{3}\ln 3 - \frac{26}{9}$ instead of simplifying $\ln 9$ to $2\ln 3$ .
Not showing the integration-by-parts step, which is required for the method mark.

Things to Be Careful About

The derivative of $\ln 3x$ is $\frac{1}{x}$ by the chain rule, not $\frac{1}{3x}$ .
When substituting limits, compute $F(3) - F(1)$ , including the sign of the $\frac{1}{9}$ term.
The final answer must be exact; decimals are not acceptable.
The required form is $a\ln b + c$ with $a$ and $c$ rational and $b$ an integer, so $a = \frac{53}{3}$ , $b = 3$ , and $c = -\frac{26}{9}$ .

Techniques used

choose u and dv for integration by partsevaluate definite integral using limitssimplify logarithmic terms with logarithm laws

Approach

Differentiate both sides of the equation implicitly with respect to $x$ . The left side requires the chain rule on $\ln(x+y)$ , and the right side requires the product rule on $3x^2y$ . Then substitute the point $(1,0)$ and solve for $\frac{dy}{dx}$ .

Working

Differentiate implicitly:

\begin{aligned} \frac{d}{dx}\left[\ln(x+y)\right] &= \frac{d}{dx}\left[3x^2y\right] \\ \frac{1 + \frac{dy}{dx}}{x+y} &= 6xy + 3x^2\frac{dy}{dx} \end{aligned}

Substitute $x=1$ and $y=0$ :

\frac{1 + \frac{dy}{dx}}{1+0} = 6(1)(0) + 3(1)^2\frac{dy}{dx}

So

1 + \frac{dy}{dx} = 3\frac{dy}{dx}

Solve:

1 = 2\frac{dy}{dx}

\frac{dy}{dx} = \frac{1}{2}

Answer

\frac{dy}{dx} = \frac{1}{2}

Final answer

dy/dx = 1/2

Detailed explanation

Walkthrough

We are asked for the gradient of a curve given by an equation linking $x$ and $y$ . Since $y$ is not written explicitly as a function of $x$ , we differentiate both sides with respect to $x$ implicitly.

Start with the left side, $\ln(x+y)$ . By the chain rule, the derivative of $\ln(u)$ is $\frac{1}{u}\frac{du}{dx}$ . Here $u = x+y$ , so $\frac{du}{dx} = 1 + \frac{dy}{dx}$ . Therefore the derivative is $\frac{1 + \frac{dy}{dx}}{x+y}$ .

Now differentiate the right side, $3x^2y$ . This is a product of $3x^2$ and $y$ . Using the product rule, $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$ , with $u=3x^2$ and $v=y$ , gives $3x^2\frac{dy}{dx} + 6xy$ .

Equate the two derivatives:

\frac{1 + \frac{dy}{dx}}{x+y} = 6xy + 3x^2\frac{dy}{dx}

Now substitute the point $(1,0)$ . Since $x+y = 1$ , $6xy = 0$ and $3x^2 = 3$ , the equation becomes:

1 + \frac{dy}{dx} = 3\frac{dy}{dx}

Rearranging gives $1 = 2\frac{dy}{dx}$ , so $\frac{dy}{dx} = \frac{1}{2}$ .

An alternative is to rewrite the equation as $x+y = e^{3x^2y}$ before differentiating. The derivative of the left side is still $1 + \frac{dy}{dx}$ , and the derivative of the right side is $(6xy + 3x^2\frac{dy}{dx})e^{3x^2y}$ . At $(1,0)$ , $e^{3x^2y}=e^0=1$ , giving the same equation and the same answer.

Key Takeaways

This question tests implicit differentiation. The key skills are:

Differentiating a composite function such as $\ln(x+y)$ using the chain rule.
Differentiating a product such as $3x^2y$ using the product rule, remembering that $y$ is a function of $x$ .
Substituting given coordinates into the differentiated equation before solving for $\frac{dy}{dx}$ .

Common Mistakes

Forgetting to multiply by $\frac{dy}{dx}$ when differentiating a term containing $y$ . For example, writing the derivative of $3x^2y$ as just $6xy$ misses the $3x^2\frac{dy}{dx}$ term.
Forgetting the factor $\frac{1}{x+y}$ when differentiating $\ln(x+y)$ , or omitting the $\frac{dy}{dx}$ inside the numerator.
Substituting $(1,0)$ before differentiating. This is invalid because the derivative must be found first.
Algebraic sign or rearrangement errors when solving $1 + y' = 3y'$ .

Things to Be Careful About

The point $(1,0)$ makes $x+y=1$ , so the denominator on the left is $1$ ; this simplifies the substitution but must still be written.
Since $e^{3x^2y} = e^0 = 1$ at the given point, the alternative method leads to the same simplified equation.
The gradient is a number at a point; do not leave $\frac{dy}{dx}$ as an expression involving $x$ and $y$ after substitution.
Marks are awarded for stating the correct derivative forms before substitution, so show both derivatives explicitly.

Techniques used

differentiate implicitlyapply chain rule to ln compositeapply product rulesubstitute coordinates and solve for dy/dx

Approach

Factor the difference of squares and use the Pythagorean identity $\sec^2 \theta - \tan^2 \theta = 1$ .

Working

\begin{aligned} \sec^4 \theta - \tan^4 \theta &= (\sec^2 \theta - \tan^2 \theta)(\sec^2 \theta + \tan^2 \theta) \\ &= 1 \cdot (\sec^2 \theta + \tan^2 \theta) \\ &= (1 + \tan^2 \theta) + \tan^2 \theta \\ &= 1 + 2\tan^2 \theta \end{aligned}

Answer

\sec^4 \theta - \tan^4 \theta \equiv 1 + 2\tan^2 \theta

Final answer

sec^4 θ - tan^4 θ ≡ 1 + 2 tan^2 θ

Detailed explanation

Walkthrough

We need to prove an identity. The left-hand side is a difference of squares: $\sec^4 \theta - \tan^4 \theta = (\sec^2 \theta)^2 - (\tan^2 \theta)^2$ . Factorising gives $(\sec^2 \theta - \tan^2 \theta)(\sec^2 \theta + \tan^2 \theta)$ .

The key Pythagorean identity for secant and tangent is $\sec^2 \theta - \tan^2 \theta = 1$ , so the first factor is simply 1. Also $\sec^2 \theta = 1 + \tan^2 \theta$ . Therefore the second factor becomes $(1 + \tan^2 \theta) + \tan^2 \theta = 1 + 2\tan^2 \theta$ . This is exactly the right-hand side.

Key Takeaways

The identity $\sec^2 \theta - \tan^2 \theta = 1$ is the secant-tangent form of Pythagoras. Recognising a difference of squares lets us factor the fourth powers and reduce the expression quickly.

Common Mistakes

Forgetting that $\sec^2 \theta - \tan^2 \theta = 1$ , not $\sec^2 \theta + \tan^2 \theta = 1$ .
Trying to expand $\sec^4 \theta$ directly without using the factorisation, which makes the proof longer and more error-prone.
Not showing the use of the identity, which is required for the method marks.

Things to Be Careful About

The identity is an identity, so it must hold for every value of $\theta$ where both sides are defined. The proof should use exact algebraic manipulation; no numerical substitution is sufficient.

Techniques used

factorise a difference of squaresapply the Pythagorean identity sec^2 theta - tan^2 theta = 1simplify to obtain the required identity

(b)

Hence or otherwise solve the equation $\sec^4 2\alpha - \tan^4 2\alpha = 2 \tan^2 2\alpha \sec^2 2\alpha$ for $0^\circ < \alpha < 180^\circ$ .

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

Use the identity from part (a) with $\theta = 2\alpha$ , replace $\sec^2 2\alpha$ by $1 + \tan^2 2\alpha$ , solve for $\tan 2\alpha$ , then list all solutions in the given interval.

Working

Let $t = \tan 2\alpha$ . From part (a),

\sec^4 2\alpha - \tan^4 2\alpha = 1 + 2t^2.

The equation becomes

1 + 2t^2 = 2t^2 \sec^2 2\alpha = 2t^2(1 + t^2).

Expanding and simplifying:

\begin{aligned} 1 + 2t^2 &= 2t^2 + 2t^4 \\ 1 &= 2t^4 \\ t^4 &= \frac12 \\ t^2 &= \frac{1}{\sqrt{2}} \\ t &= \pm \frac{1}{\sqrt[4]{2}}. \end{aligned}

Thus $\tan 2\alpha = \pm k$ , where $k = \frac{1}{\sqrt[4]{2}} \approx 0.8409$ , so $\arctan k \approx 40.06^\circ$ .

Since $0^\circ < \alpha < 180^\circ$ , we need $0^\circ < 2\alpha < 360^\circ$ . The tangent has period $180^\circ$ , so:

$\tan 2\alpha = +k$ : $2\alpha \approx 40.06^\circ$ or $220.06^\circ$ .
$\tan 2\alpha = -k$ : $2\alpha \approx 139.94^\circ$ or $319.94^\circ$ .

Dividing by 2 gives all solutions.

Answer

\alpha \approx 20.0^\circ, 70.0^\circ, 110.0^\circ, 160.0^\circ

Final answer

alpha ≈ 20.0°, 70.0°, 110.0°, 160.0°

Detailed explanation

Walkthrough

Use the result from part (a) with $\theta = 2\alpha$ . The left-hand side of the given equation is exactly $\sec^4 2\alpha - \tan^4 2\alpha$ , so it equals $1 + 2\tan^2 2\alpha$ .

The right-hand side is $2\tan^2 2\alpha \sec^2 2\alpha$ . Replace $\sec^2 2\alpha$ with $1 + \tan^2 2\alpha$ . Let $t = \tan 2\alpha$ ; then the equation becomes $1 + 2t^2 = 2t^2(1 + t^2)$ . Expanding gives $1 + 2t^2 = 2t^2 + 2t^4$ , so $1 = 2t^4$ . Thus $t^4 = \frac12$ , $t^2 = \frac{1}{\sqrt{2}}$ , and $t = \pm \frac{1}{\sqrt[4]{2}}$ .

Now solve $\tan 2\alpha = \pm k$ with $k = \frac{1}{\sqrt[4]{2}}$ . Because $0^\circ < \alpha < 180^\circ$ , the angle $2\alpha$ lies in $(0^\circ, 360^\circ)$ . Tangent is positive in the first and third quadrants and negative in the second and fourth. With $\arctan k \approx 40.06^\circ$ , the four possible values of $2\alpha$ are approximately $40.06^\circ$ , $180^\circ - 40.06^\circ = 139.94^\circ$ , $180^\circ + 40.06^\circ = 220.06^\circ$ , and $360^\circ - 40.06^\circ = 319.94^\circ$ . Dividing by 2 gives $\alpha \approx 20.0^\circ, 70.0^\circ, 110.0^\circ, 160.0^\circ$ .

Key Takeaways

A "hence" instruction usually means the previous identity should be substituted directly. Reducing a trigonometric equation to an equation in $\tan$ often makes it algebraic. When solving $\tan x = c$ over a full circle, remember tangent has period $180^\circ$ and there are two solutions in any $360^\circ$ interval.

Common Mistakes

Forgetting to use the identity from part (a), or using it incorrectly.
Replacing $\sec^2 2\alpha$ by $1 - \tan^2 2\alpha$ (wrong sign).
Taking only the positive square root and losing the negative solutions.
Solving only for $2\alpha$ and forgetting to divide by 2 to get $\alpha$ .
Stopping at the acute solution and missing the other three solutions in the range.
Including $\alpha = 0^\circ$ or $180^\circ$ , which are excluded by the open interval.

Things to Be Careful About

The domain is for $\alpha$ , not for $2\alpha$ ; this is why there are four solutions. The value $k = \frac{1}{\sqrt[4]{2}}$ is exact; the angles are approximate. Use degree mode on the calculator. The mark scheme requires all four solutions and no others in the range.

Techniques used

apply the identity from part (a)substitute sec^2 = 1 + tan^2solve the resulting quartic equation for tan 2alphafind all angles in the given interval using tangent periodicity

Q5MediumLogarithmic and Exponential FunctionsNumerical Solution of Equations

(a)

By sketching a suitable pair of graphs, show that the equation $2 + e^{-0.2x} = \ln(1 + x)$ has only one root.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Split the equation $2 + e^{-0.2x} = \ln(1+x)$ into two separate functions and sketch both on the same axes. The roots of the original equation correspond to the $x$ -coordinates of the intersection points of the two graphs.

Working

Consider the two curves:

y_1 = 2 + e^{-0.2x} \quad \text{and} \quad y_2 = \ln(1 + x)

Graph of $y_1 = 2 + e^{-0.2x}$ :

This is an exponential decay curve shifted vertically upwards by 2.
$y$ -intercept: at $x = 0$ , $y_1 = 2 + e^0 = 2 + 1 = 3$ , so the curve passes through $(0, 3)$ .
As $x \to \infty$ , $e^{-0.2x} \to 0$ , so $y_1 \to 2$ . There is a horizontal asymptote at $y = 2$ .
The curve is strictly decreasing and concave up.

Graph of $y_2 = \ln(1 + x)$ :

This is a logarithmic curve shifted left by 1.
$y$ -intercept: at $x = 0$ , $y_2 = \ln(1) = 0$ , so the curve passes through the origin $(0, 0)$ .
The domain is $x > -1$ .
As $x \to \infty$ , $y_2 \to \infty$ with decreasing gradient (concave down).
The curve is strictly increasing.

Since $y_1$ is strictly decreasing from 3 towards 2, and $y_2$ is strictly increasing from 0 towards $\infty$ , the two curves must intersect exactly once. The intersection occurs in the first quadrant where $y_1 > 2$ and $y_2 > 0$ .

Answer

The two curves intersect at exactly one point, so the equation $2 + e^{-0.2x} = \ln(1+x)$ has only one root.

Final answer

The curves y = 2 + e^{-0.2x} and y = ln(1+x) intersect at exactly one point.

Detailed explanation

Walkthrough

We are asked to show that the equation $2 + e^{-0.2x} = \ln(1+x)$ has only one root. The most natural approach is to split this into two separate functions and sketch them on the same set of axes.

Step 1: Define the two curves. Let $y_1 = 2 + e^{-0.2x}$ and $y_2 = \ln(1+x)$ . Any $x$ -value where these two curves meet is a root of the original equation.

Step 2: Sketch $y_1 = 2 + e^{-0.2x}$ . This is a transformed exponential function. The base function $e^{-0.2x}$ is an exponential decay (since the coefficient of $x$ is negative). Adding 2 shifts the entire graph up by 2 units. At $x = 0$ , we get $y_1 = 2 + 1 = 3$ . As $x$ grows large, $e^{-0.2x}$ approaches 0, so $y_1$ approaches 2 from above. The curve is always decreasing and always concave up.

Step 3: Sketch $y_2 = \ln(1+x)$ . This is a logarithmic function. At $x = 0$ , $y_2 = \ln(1) = 0$ . The function is defined for $x > -1$ . As $x$ increases, $y_2$ increases without bound, but the rate of increase slows down (concave down). The curve is always increasing.

Step 4: Argue uniqueness of intersection. Since $y_1$ is strictly decreasing and $y_2$ is strictly increasing, they can intersect at most once. Since $y_1(0) = 3 > 0 = y_2(0)$ and $y_1$ approaches 2 while $y_2$ grows without bound, there must be a point where $y_2$ overtakes $y_1$ . Therefore, they intersect exactly once.

Key Takeaways

Splitting an equation $f(x) = g(x)$ into two curves $y = f(x)$ and $y = g(x)$ and sketching them is a powerful way to visualise roots.
The monotonicity of the two curves (one increasing, one decreasing) guarantees at most one intersection.
A sign change or crossing of values at two points guarantees at least one intersection.

Common Mistakes

Sketching $e^{-0.2x}$ without the vertical shift of +2, giving a curve that approaches 0 instead of 2.
Forgetting the $y$ -intercept of $y = \ln(1+x)$ is at $(0, 0)$ , not at $(-1, 0)$ .
Claiming the curves do not intersect without checking values at specific points.
Not justifying why there is only one intersection — merely sketching is insufficient.

Things to Be Careful About

The horizontal asymptote $y = 2$ need not be drawn explicitly, but the sketch should show the curve approaching a value near 2.
The scale on the axes does not need to be marked, but the key points $(0, 3)$ and $(0, 0)$ should be approximately correct.
The justification for a single root must reference the monotonic behaviour of both curves.

Techniques used

sketch exponential decay curvesketch logarithmic curveidentify intersection pointargue uniqueness of root

(b)

Show by calculation that this root lies between 7 and 9.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Define $f(x) = 2 + e^{-0.2x} - \ln(1+x)$ . If $f(7)$ and $f(9)$ have opposite signs, then by the intermediate value theorem, there is a root between $x = 7$ and $x = 9$ .

Working

Calculate $f(7)$ :

f(7) = 2 + e^{-0.2 \times 7} - \ln(1 + 7) = 2 + e^{-1.4} - \ln 8

e^{-1.4} \approx 0.2466

\ln 8 \approx 2.0794

f(7) = 2 + 0.2466 - 2.0794 = 0.1672

So $f(7) > 0$ .

Calculate $f(9)$ :

f(9) = 2 + e^{-0.2 \times 9} - \ln(1 + 9) = 2 + e^{-1.8} - \ln 10

e^{-1.8} \approx 0.1653

\ln 10 \approx 2.3026

f(9) = 2 + 0.1653 - 2.3026 = -0.1373

So $f(9) < 0$ .

Since $f(7) = 0.1672 > 0$ and $f(9) = -0.1373 < 0$ , there is a sign change in the interval $(7, 9)$ . Since $f(x)$ is continuous for $x > -1$ , by the intermediate value theorem, there is at least one root in $(7, 9)$ .

Answer

The root lies between $x = 7$ and $x = 9$ .

Final answer

f(7) = 0.1672 > 0 and f(9) = -0.1373 < 0, so the root lies between 7 and 9.

Detailed explanation

Walkthrough

We need to show that the root of $2 + e^{-0.2x} = \ln(1+x)$ lies between 7 and 9. The standard technique is to define a single function $f(x) = 2 + e^{-0.2x} - \ln(1+x)$ and evaluate it at the two boundary values.

Step 1: Evaluate at x = 7.

$f(7) = 2 + e^{-1.4} - \ln 8$

$e^{-1.4} \approx 0.2466$ and $\ln 8 \approx 2.0794$ , so $f(7) \approx 2 + 0.2466 - 2.0794 = 0.1672$ .

Since $f(7) > 0$ , the left-hand side $2 + e^{-0.2x}$ is greater than the right-hand side $\ln(1+x)$ at $x = 7$ .

Step 2: Evaluate at x = 9.

$f(9) = 2 + e^{-1.8} - \ln 10$

$e^{-1.8} \approx 0.1653$ and $\ln 10 \approx 2.3026$ , so $f(9) \approx 2 + 0.1653 - 2.3026 = -0.1373$ .

Since $f(9) < 0$ , the left-hand side is now less than the right-hand side at $x = 9$ .

Step 3: Conclude.

Since $f(x)$ is continuous on $(7, 9)$ and changes sign from positive to negative, there must be at least one value of $x$ in $(7, 9)$ where $f(x) = 0$ , i.e., a root of the original equation.

Key Takeaways

Defining $f(x) = \text{LHS} - \text{RHS}$ and checking for a sign change is the standard method for locating roots in an interval.
The intermediate value theorem guarantees a root exists when a continuous function changes sign.
Calculations should be done to at least 3 decimal places to ensure the sign is unambiguous.

Common Mistakes

Calculating $e^{-1.4}$ incorrectly (e.g., using $e^{1.4}$ instead of $e^{-1.4}$ ).
Forgetting to include the +2 in the expression.
Not clearly stating that $f(7)$ and $f(9)$ have opposite signs.
Not mentioning continuity of $f(x)$ , though this is often implied.

Things to Be Careful About

Ensure all values are calculated to sufficient precision (at least 3 decimal places) so the signs are unambiguous.
The mark scheme also accepts comparing the two sides directly: at $x=7$ , $2 + e^{-1.4} \approx 2.2466 > 2.0794 \approx \ln 8$ , and at $x=9$ , $2 + e^{-1.8} \approx 2.1653 < 2.3026 \approx \ln 10$ .
Make sure to state the conclusion clearly: the root lies between 7 and 9.

Techniques used

evaluate function at boundary valuescheck sign changeapply intermediate value theorem

(c)

Use the iterative formula

x_{n+1} = \exp(2 + e^{-0.2x_n}) - 1

to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

[ $\exp(x)$ is an alternative notation for $e^x$ .]

3M Similar

DifficultyMedium

Worked solution

Approach

Use the given iterative formula $x_{n+1} = \exp(2 + e^{-0.2x_n}) - 1$ starting with an initial value in the interval $(7, 9)$ . Continue iterating until successive values agree to 2 decimal places. Show all iterations to 4 decimal places.

Working

Choose $x_1 = 8$ as the initial value (any value in $(7, 9)$ is acceptable).

Iteration 1:

x_2 = \exp(2 + e^{-0.2 \times 8}) - 1 = \exp(2 + e^{-1.6}) - 1

e^{-1.6} \approx 0.2019

x_2 = \exp(2.2019) - 1 \approx 9.0421 - 1 = 8.0421

Iteration 2:

x_3 = \exp(2 + e^{-0.2 \times 8.0421}) - 1 = \exp(2 + e^{-1.6084}) - 1

e^{-1.6084} \approx 0.1999

x_3 = \exp(2.1999) - 1 \approx 9.0268 - 1 = 8.0268

Iteration 3:

x_4 = \exp(2 + e^{-0.2 \times 8.0268}) - 1 = \exp(2 + e^{-1.6054}) - 1

e^{-1.6054} \approx 0.2005

x_4 = \exp(2.2005) - 1 \approx 9.0324 - 1 = 8.0324

Iteration 4:

x_5 = \exp(2 + e^{-0.2 \times 8.0324}) - 1 = \exp(2 + e^{-1.6065}) - 1

e^{-1.6065} \approx 0.2002

x_5 = \exp(2.2002) - 1 \approx 9.0303 - 1 = 8.0303

Iteration 5:

x_6 = \exp(2 + e^{-0.2 \times 8.0303}) - 1 = \exp(2 + e^{-1.6061}) - 1

e^{-1.6061} \approx 0.2003

x_6 = \exp(2.2003) - 1 \approx 9.0309 - 1 = 8.0309

Iteration 6:

x_7 = \exp(2 + e^{-0.2 \times 8.0309}) - 1 = \exp(2 + e^{-1.6062}) - 1

e^{-1.6062} \approx 0.2003

x_7 = \exp(2.2003) - 1 \approx 9.0310 - 1 = 8.0310

The values are converging. We have $x_5 = 8.0303$ and $x_6 = 8.0309$ , which both round to $8.03$ to 2 decimal places. To confirm, we check the sign change in $(8.025, 8.035)$ :

f(8.025) = 2 + e^{-1.605} - \ln(9.025) \approx 2.2006 - 2.1999 = 0.0007 > 0

f(8.035) = 2 + e^{-1.607} - \ln(9.035) \approx 2.1997 - 2.2010 = -0.0013 < 0

Since $f(8.025) > 0$ and $f(8.035) < 0$ , the root lies in $(8.025, 8.035)$ , confirming it is $8.03$ to 2 decimal places.

Answer

The root is $8.03$ correct to 2 decimal places.

Final answer

8.03

Detailed explanation

Walkthrough

We are given the iterative formula $x_{n+1} = \exp(2 + e^{-0.2x_n}) - 1$ and asked to find the root to 2 decimal places.

Step 1: Choose an initial value. From part (b), we know the root lies between 7 and 9. We choose $x_1 = 8$ as a convenient starting point.

Step 2: Apply the iterative formula repeatedly.

Substitute $x_1 = 8$ :
$x_2 = \exp(2 + e^{-1.6}) - 1 = \exp(2.2019) - 1 \approx 8.0421$

Substitute $x_2 = 8.0421$ :
$x_3 = \exp(2 + e^{-1.6084}) - 1 = \exp(2.1999) - 1 \approx 8.0268$

Substitute $x_3 = 8.0268$ :
$x_4 = \exp(2 + e^{-1.6054}) - 1 = \exp(2.2005) - 1 \approx 8.0324$

Continue this process. The values oscillate around the true root, getting closer each time:
$x_5 \approx 8.0303$ , $x_6 \approx 8.0309$ , $x_7 \approx 8.0310$ .

Step 3: Determine convergence. The values $x_5 = 8.0303$ and $x_6 = 8.0309$ both round to $8.03$ to 2 decimal places. To be rigorous, we verify that the root lies in the interval $(8.025, 8.035)$ by checking the sign of $f(x) = 2 + e^{-0.2x} - \ln(1+x)$ at the endpoints. Since $f(8.025) > 0$ and $f(8.035) < 0$ , the root is indeed $8.03$ to 2 decimal places.

Key Takeaways

Iterative formulas converge to a root when $|F'(x)| < 1$ near the root. Here, the formula is rearranged from the original equation.
Always show sufficient iterations to demonstrate convergence, not just the final answer.
To confirm a value to $n$ decimal places, either show successive iterations agree to $n$ decimal places, or show a sign change in the interval $(\text{value} - 0.005, \text{value} + 0.005)$ .

Common Mistakes

Making arithmetic errors in the exponential calculations.
Not showing enough iterations to justify the final answer.
Rounding intermediate values too early, which can cause the iteration to diverge or give incorrect results.
Forgetting to check convergence properly — just stating the last value is not enough.

Things to Be Careful About

Keep all intermediate values to at least 4 decimal places as specified.
The mark scheme accepts various starting values (7, 8, or 9) and shows different sequences, all converging to 8.03.
The final answer must be justified either by showing convergence of iterations or by a sign change in the appropriate interval.
$\exp(x)$ is the same as $e^x$ ; do not confuse $\exp(2 + e^{-0.2x})$ with $e^{2 + e^{-0.2x}}$ — they are the same thing, but be careful with the notation.

Techniques used

apply iterative formula repeatedlycheck convergence to required precisionverify sign change in final interval

Q6MediumDifferentiationTrigonometryIntegration

Fig. 1

The diagram shows the curve $y = \sin 2x(1 + \sin 2x)$ , for $0 \le x \le \frac{3}{4}\pi$ , and its minimum point $M$ . The shaded region bounded by the curve that lies above the $x$ -axis and the $x$ -axis itself is denoted by $R$ .

(a)

Given that the $x$ -coordinate of $M$ lies in the interval $\frac{1}{2}\pi < x < \frac{3}{4}\pi$ , find the exact coordinates of $M$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

To find the minimum point $M$ , differentiate $y$ with respect to $x$ using the product rule, set $\frac{dy}{dx} = 0$ , and solve for $x$ within the given interval $\frac{1}{2}\pi < x < \frac{3}{4}\pi$ . Then substitute $x$ back into the original equation to find $y$ .

Working

Given $y = \sin 2x(1 + \sin 2x)$ .

Differentiate using the product rule:

\begin{aligned} \frac{dy}{dx} &= (2\cos 2x)(1 + \sin 2x) + (\sin 2x)(2\cos 2x) \\ &= 2\cos 2x + 2\sin 2x \cos 2x + 2\sin 2x \cos 2x \\ &= 2\cos 2x + 4\sin 2x \cos 2x \\ &= 2\cos 2x(1 + 2\sin 2x) \end{aligned}

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

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

This gives two cases:

$\cos 2x = 0 \Rightarrow 2x = \frac{\pi}{2}, \frac{3\pi}{2}, \dots \Rightarrow x = \frac{\pi}{4}, \frac{3\pi}{4}, \dots$
$1 + 2\sin 2x = 0 \Rightarrow \sin 2x = -\frac{1}{2}$

We are given that $\frac{1}{2}\pi < x < \frac{3}{4}\pi$ , which means $\pi < 2x < \frac{3}{2}\pi$ .

For $\cos 2x = 0$ in this range, $2x = \frac{3\pi}{2} \Rightarrow x = \frac{3\pi}{4}$ . At this point, $y = \sin(\frac{3\pi}{2})(1 + \sin(\frac{3\pi}{2})) = (-1)(1 - 1) = 0$ . This is an x-intercept, not the minimum point $M$ described as being strictly inside the interval.
For $\sin 2x = -\frac{1}{2}$ in the range $\pi < 2x < \frac{3\pi}{2}$ (third quadrant), the reference angle is $\frac{\pi}{6}$ , so $2x = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \Rightarrow x = \frac{7\pi}{12}$ .

Substitute $x = \frac{7\pi}{12}$ into the original equation:

\sin 2x = \sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}

y = \left(-\frac{1}{2}\right)\left(1 + \left(-\frac{1}{2}\right)\right) = \left(-\frac{1}{2}\right)\left(\frac{1}{2}\right) = -\frac{1}{4}

Thus, the coordinates of $M$ are $\left(\frac{7\pi}{12}, -\frac{1}{4}\right)$ .

Answer

M = \left(\frac{7\pi}{12}, -\frac{1}{4}\right)

Final answer

M = (7π/12, -1/4)

Detailed explanation

Walkthrough

First, we differentiate the function $y = \sin 2x(1 + \sin 2x)$ using the product rule. Let $u = \sin 2x$ and $v = 1 + \sin 2x$ . Then $\frac{du}{dx} = 2\cos 2x$ and $\frac{dv}{dx} = 2\cos 2x$ . Applying $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ , we get $2\sin 2x \cos 2x + 2\cos 2x(1 + \sin 2x)$ , which simplifies to $2\cos 2x(1 + 2\sin 2x)$ .

Next, we set the derivative to zero to find stationary points. This gives $\cos 2x = 0$ or $\sin 2x = -1/2$ . We restrict our attention to the interval $\frac{1}{2}\pi < x < \frac{3}{4}\pi$ , meaning $\pi < 2x < \frac{3\pi}{2}$ . In this third-quadrant range, $\cos 2x = 0$ at $2x = \frac{3\pi}{2}$ (giving $x = \frac{3\pi}{4}$ ), but this yields $y=0$ , which is an intercept. The equation $\sin 2x = -1/2$ has the solution $2x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$ , so $x = \frac{7\pi}{12}$ . Substituting this back into $y = \sin 2x(1 + \sin 2x)$ with $\sin 2x = -1/2$ gives $y = (-1/2)(1/2) = -1/4$ .

Key Takeaways

The product rule is essential for differentiating products of trigonometric functions.
When finding stationary points in a restricted interval, always check which solutions fall within the specified range and evaluate the function to distinguish between maxima, minima, and intercepts.

Common Mistakes

Forgetting the chain rule when differentiating $\sin 2x$ (must multiply by 2).
Missing the solution $\sin 2x = -1/2$ and only considering $\cos 2x = 0$ .
Using degrees instead of radians when the question asks for exact answers in terms of $\pi$ .

Things to Be Careful About

The interval for $x$ is $\frac{1}{2}\pi < x < \frac{3}{4}\pi$ , so $2x$ is in $\pi < 2x < \frac{3\pi}{2}$ . Be careful to find the correct angle in this range for $\sin 2x = -1/2$ .
The question asks for exact coordinates, so do not give decimal approximations.

Techniques used

apply product rule to differentiatesolve trigonometric equation for minimum point

(b)

Find the exact area of the region $R$ .

4M Similar

DifficultyMedium

Worked solution

Approach

The region $R$ is bounded by the curve and the x-axis where $y \ge 0$ . From the diagram and the function $y = \sin 2x(1 + \sin 2x)$ , $y = 0$ at $x = 0$ and $x = \frac{\pi}{2}$ (since $\sin 2x = 0$ at $2x = \pi$ ). Between $0$ and $\frac{\pi}{2}$ , $\sin 2x > 0$ , so $y > 0$ . The area is given by $\int_0^{\pi/2} y \, dx$ . We expand the integrand and use the double angle identity $\sin^2 2x = \frac{1 - \cos 4x}{2}$ to integrate.

Working

The area of region $R$ is:

A = \int_0^{\pi/2} \sin 2x(1 + \sin 2x) \, dx

Expand the integrand:

A = \int_0^{\pi/2} (\sin 2x + \sin^2 2x) \, dx

Use the identity $\sin^2 2x = \frac{1 - \cos 4x}{2}$ :

A = \int_0^{\pi/2} \left(\sin 2x + \frac{1}{2} - \frac{1}{2}\cos 4x\right) \, dx

Integrate term by term:

A = \left[-\frac{1}{2}\cos 2x + \frac{1}{2}x - \frac{1}{8}\sin 4x\right]_0^{\pi/2}

Evaluate at the upper limit $x = \frac{\pi}{2}$ :

-\frac{1}{2}\cos(\pi) + \frac{1}{2}\left(\frac{\pi}{2}\right) - \frac{1}{8}\sin(2\pi) = -\frac{1}{2}(-1) + \frac{\pi}{4} - 0 = \frac{1}{2} + \frac{\pi}{4}

Evaluate at the lower limit $x = 0$ :

-\frac{1}{2}\cos(0) + 0 - \frac{1}{8}\sin(0) = -\frac{1}{2}(1) + 0 - 0 = -\frac{1}{2}

Subtract the lower limit value from the upper limit value:

A = \left(\frac{1}{2} + \frac{\pi}{4}\right) - \left(-\frac{1}{2}\right) = \frac{1}{2} + \frac{\pi}{4} + \frac{1}{2} = 1 + \frac{\pi}{4}

Answer

\text{Area of } R = \frac{\pi}{4} + 1

Final answer

π/4 + 1

Detailed explanation

Walkthrough

Region $R$ is the area under the first hump of the curve, above the x-axis. The curve intersects the x-axis when $\sin 2x = 0$ (at $x=0, \pi/2$ ) or $\sin 2x = -1$ (at $x=3\pi/4$ ). Between $x=0$ and $x=\pi/2$ , $\sin 2x$ is positive, so $y$ is positive. Thus, the limits of integration are $0$ and $\pi/2$ .

We integrate $y = \sin 2x + \sin^2 2x$ . To integrate $\sin^2 2x$ , we use the double angle identity $\cos 4x = 1 - 2\sin^2 2x$ , which rearranges to $\sin^2 2x = \frac{1 - \cos 4x}{2}$ . This transforms the integral into $\int (\sin 2x + \frac{1}{2} - \frac{1}{2}\cos 4x) dx$ .

Integrating term by term gives $-\frac{1}{2}\cos 2x + \frac{1}{2}x - \frac{1}{8}\sin 4x$ . Evaluating this from $0$ to $\pi/2$ : at $\pi/2$ , we get $\frac{1}{2} + \frac{\pi}{4}$ ; at $0$ , we get $-\frac{1}{2}$ . The difference is $1 + \frac{\pi}{4}$ .

Key Takeaways

Always determine the correct limits of integration by finding where the curve intersects the bounding line (here, the x-axis).
Trigonometric identities like $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$ are crucial for integrating squared trigonometric functions.
Be careful with signs when evaluating definite integrals, especially with negative values at the lower limit.

Common Mistakes

Using incorrect limits (e.g., integrating up to $3\pi/4$ instead of $\pi/2$ , which would include the negative area below the x-axis).
Forgetting to halve the coefficient when integrating $\cos 4x$ (must divide by 4).
Sign errors when evaluating $\cos(\pi) = -1$ .

Things to Be Careful About

The question asks for the area of the region bounded by the curve that lies above the x-axis. This restricts the limits to $0 \le x \le \pi/2$ .
Ensure all answers are exact; do not use decimal approximations for $\pi$ .

Techniques used

expand integrand using double angle identityevaluate definite integral for area under curve

Q7MediumAlgebra

Let $f(x) = \frac{5x^2 + 8x + 5}{(1 + 2x)(2 + x^2)}$ .

(a)

Express $f(x)$ in partial fractions.

5M Similar

DifficultyMedium

Worked solution

Approach

Since the denominator contains the irreducible quadratic $2+x^2$ , use the partial fraction form

\frac{A}{1+2x}+\frac{Bx+C}{2+x^2}.

Clear denominators, expand, and equate coefficients of $x^2$ , $x$ and the constant term.

Working

\frac{5x^2+8x+5}{(1+2x)(2+x^2)} = \frac{A}{1+2x}+\frac{Bx+C}{2+x^2}

Multiplying by $(1+2x)(2+x^2)$ :

5x^2+8x+5 = A(2+x^2)+(Bx+C)(1+2x)

Expanding the right-hand side:

5x^2+8x+5 = (A+2B)x^2 + (B+2C)x + (2A+C)

Equating coefficients:

A+2B=5,\quad B+2C=8,\quad 2A+C=5

Solving these equations gives:

A=1,\quad B=2,\quad C=3

Answer

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

Final answer

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

Detailed explanation

Walkthrough

We want to write the fraction as a sum of simpler fractions. The denominator has a linear factor $(1+2x)$ and a quadratic factor $(2+x^2)$ . Because $2+x^2$ cannot be factored into real linear factors, we must use a numerator of the form $Bx+C$ over it; if we used a constant, we could not match all the terms in the numerator.

Set

\frac{5x^2+8x+5}{(1+2x)(2+x^2)} = \frac{A}{1+2x}+\frac{Bx+C}{2+x^2}.

Multiply both sides by the denominator to remove fractions. Then expand the right-hand side and group by powers of $x$ . Comparing the coefficients of $x^2$ , $x$ and the constant term gives three equations. Solving them gives $A=1$ , $B=2$ , $C=3$ , so the partial fraction decomposition is complete.

Key Takeaways

A quadratic denominator that cannot be factored over the reals requires a linear numerator $Bx+C$ in partial fractions.
Clearing denominators and equating coefficients is a reliable method for finding the constants.
Always check that the number of equations matches the number of unknown constants.

Common Mistakes

Using $\frac{A}{1+2x}+\frac{B}{2+x^2}$ instead of $\frac{Bx+C}{2+x^2}$ ; this cannot represent the $x$ term in the numerator.
Errors in expanding $(Bx+C)(1+2x)$ , especially the $2Bx^2$ and $2Cx$ terms.
Solving the simultaneous equations incorrectly because of sign errors.

Things to Be Careful About

The linear factor is $1+2x$ , not $x+1$ ; the constant term in the denominator is 1.
When equating coefficients, include the constant term from $A(2+x^2)$ , which is $2A$ .
There is no need to factor $2+x^2$ ; it is already irreducible over the reals.

Techniques used

clear denominatorsequate coefficientssolve simultaneous equations

(b)

Hence find the coefficient of $x^3$ in the expansion of $f(x)$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use the partial fraction result from part (a). Expand each term using the binomial theorem for negative indices, then collect only the coefficient of $x^3$ .

Working

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

First term:

(1+2x)^{-1} = 1 - 2x + 4x^2 - 8x^3 + \cdots

so the coefficient of $x^3$ from this term is $-8$ .

Second term:

\frac{1}{2+x^2} = \frac12\left(1+\frac{x^2}{2}\right)^{-1} = \frac12\left(1 - \frac{x^2}{2} + \cdots\right)

Therefore

\frac{2x+3}{2+x^2} = (2x+3)\cdot\frac12\left(1 - \frac{x^2}{2} + \cdots\right)

The only contribution to $x^3$ is

2x \cdot \frac12 \cdot \left(-\frac{x^2}{2}\right) = -\frac12 x^3.

Total coefficient of $x^3$ :

-8 - \frac12 = -\frac{17}{2}.

Answer

-\frac{17}{2}

Final answer

-17/2

Detailed explanation

Walkthrough

After part (a), write

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

For the first term, use the binomial expansion for a negative index:

(1+u)^{-1}=1-u+u^2-u^3+\cdots

with $u=2x$ . The $x^3$ term is $(-2x)^3=-8x^3$ , so its coefficient is $-8$ .

For the second term, factor out $\frac12$ :

\frac{1}{2+x^2}=\frac12\left(1+\frac{x^2}{2}\right)^{-1}.

Expanding only as far as needed:

\frac12\left(1-\frac{x^2}{2}+\cdots\right).

Multiplying by $2x+3$ , the $x^3$ term can only come from $2x$ times the $x^2$ term, because $3$ times the expansion gives only even powers. This contribution is

2x\cdot\frac12\cdot\left(-\frac{x^2}{2}\right)=-\frac12x^3.

Adding the two coefficients gives $-8-\frac12=-\frac{17}{2}$ .

Key Takeaways

Binomial expansions with negative indices are infinite series; we only need the terms up to the required power.
When multiplying series, identify which product of terms can produce the required power.
Partial fractions turn a complicated expansion into a sum of simpler binomial expansions.

Common Mistakes

Forgetting the factor $\frac12$ when expanding $(2+x^2)^{-1}$ .
Missing the minus sign in the $x^2$ term of $(1+x^2/2)^{-1}$ .
Using only the first term $-8$ and forgetting the contribution from the second partial fraction.
Trying to expand $(2+x^2)^{-1}$ without first factoring out $2$ .

Things to Be Careful About

The expansion of $(1+2x)^{-1}$ is valid for $|2x|<1$ , and $(1+x^2/2)^{-1}$ for $|x^2/2|<1$ ; the coefficient of $x^3$ is still found by the same series.
The coefficient is requested, not the full term; the final answer is $-\frac{17}{2}$ .
If you use the binomial coefficient formula, remember the signs alternate: $(-1)(-2)(-3)/3! = -1$ .

Techniques used

expand (1+2x)^-1 using the binomial theoremexpand (1+x^2/2)^-1 using the binomial theoremextract the coefficient of x^3

Q8MediumComplex Numbers

(a)

Given that $z = 1 + yi$ and that $y$ is a real number, express $\frac{1}{z}$ in the form $a + bi$ , where $a$ and $b$ are functions of $y$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

To express $\frac{1}{z}$ in the form $a + bi$ , multiply the numerator and denominator by the complex conjugate of the denominator.

Working

\frac{1}{z} = \frac{1}{1 + yi}

Multiply numerator and denominator by the conjugate $1 - yi$ :

\frac{1}{1 + yi} \times \frac{1 - yi}{1 - yi} = \frac{1 - yi}{(1 + yi)(1 - yi)}

Expand the denominator using the difference of two squares:

(1 + yi)(1 - yi) = 1 - (yi)^2 = 1 - y^2 i^2 = 1 + y^2

So:

\frac{1}{z} = \frac{1 - yi}{1 + y^2}

Separate into real and imaginary parts:

\frac{1}{z} = \frac{1}{1 + y^2} + \frac{-y}{1 + y^2}i

Therefore:

a = \frac{1}{1 + y^2}, \quad b = \frac{-y}{1 + y^2}

Answer

\frac{1}{z} = \frac{1}{1 + y^2} - \frac{y}{1 + y^2}i

Final answer

a = 1/(1+y^2), b = -y/(1+y^2)

Detailed explanation

Walkthrough

We are given $z = 1 + yi$ and need to express $\frac{1}{z}$ in Cartesian form $a + bi$ . The standard technique is to multiply numerator and denominator by the complex conjugate of the denominator. The conjugate of $1 + yi$ is $1 - yi$ . Multiplying gives $\frac{1 - yi}{(1+yi)(1-yi)}$ . The denominator simplifies via the difference of two squares: $(1+yi)(1-yi) = 1 - y^2i^2 = 1 + y^2$ since $i^2 = -1$ . Dividing both terms in the numerator by the real denominator gives the real part $a = \frac{1}{1+y^2}$ and the imaginary part $b = \frac{-y}{1+y^2}$ .

Key Takeaways

To find $\frac{1}{z}$ for $z = x + yi$ , multiply by the conjugate $x - yi$ over itself.
The denominator becomes $x^2 + y^2$ , which is always real and positive (for $z \neq 0$ ).
Always separate the result into real and imaginary components explicitly.

Common Mistakes

Forgetting to multiply both numerator and denominator by the conjugate.
Writing $i^2 = 1$ instead of $i^2 = -1$ , which gives the wrong denominator.
Not clearly identifying which part is $a$ and which is $b$ .

Things to Be Careful About

The question asks for $a$ and $b$ as functions of $y$ , so leave them in terms of $y$ .
The imaginary part $b$ is $\frac{-y}{1+y^2}$ , which is negative when $y > 0$ . Do not drop the minus sign.

Techniques used

multiply by complex conjugateseparate into real and imaginary parts

(b)

Show that $(a - \frac{1}{2})^2 + b^2 = \frac{1}{4}$ , where $a$ and $b$ are the functions of $y$ found in part (a).

3M Similar

DifficultyMedium

Worked solution

Approach

Substitute the expressions for $a$ and $b$ from part (a) into $(a - \frac{1}{2})^2 + b^2$ , expand, and simplify to show the result equals $\frac{1}{4}$ .

Working

From part (a):

a = \frac{1}{1 + y^2}, \quad b = \frac{-y}{1 + y^2}

Substitute into $(a - \frac{1}{2})^2 + b^2$ :

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

Expand the first bracket using $(p - q)^2 = p^2 - 2pq + q^2$ :

\frac{1}{(1 + y^2)^2} - 2 \cdot \frac{1}{1 + y^2} \cdot \frac{1}{2} + \frac{1}{4} + \frac{y^2}{(1 + y^2)^2}

Simplify the middle term:

\frac{1}{(1 + y^2)^2} - \frac{1}{1 + y^2} + \frac{1}{4} + \frac{y^2}{(1 + y^2)^2}

Combine the first and last terms over the common denominator $(1 + y^2)^2$ :

\frac{1 + y^2}{(1 + y^2)^2} - \frac{1}{1 + y^2} + \frac{1}{4}

Simplify $\frac{1 + y^2}{(1 + y^2)^2} = \frac{1}{1 + y^2}$ :

\frac{1}{1 + y^2} - \frac{1}{1 + y^2} + \frac{1}{4}

The first two terms cancel:

\frac{1}{4}

Answer

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

Final answer

(a - 1/2)^2 + b^2 = 1/4

Detailed explanation

Walkthrough

We substitute $a = \frac{1}{1+y^2}$ and $b = \frac{-y}{1+y^2}$ into $(a - \frac{1}{2})^2 + b^2$ . First, expand $(a - \frac{1}{2})^2$ as $a^2 - a + \frac{1}{4}$ , giving $\frac{1}{(1+y^2)^2} - \frac{1}{1+y^2} + \frac{1}{4}$ . Then $b^2 = \frac{y^2}{(1+y^2)^2}$ . Adding the two squared terms: $\frac{1}{(1+y^2)^2} + \frac{y^2}{(1+y^2)^2} = \frac{1+y^2}{(1+y^2)^2} = \frac{1}{1+y^2}$ . This exactly cancels the $-\frac{1}{1+y^2}$ term, leaving only $\frac{1}{4}$ .

Key Takeaways

When proving an identity involving complex number components, substitute and expand carefully.
Combining fractions with the same denominator is a key simplification step.
The result $(a - \frac{1}{2})^2 + b^2 = \frac{1}{4}$ is the equation of a circle, which connects to part (c).

Common Mistakes

Expanding $(a - \frac{1}{2})^2$ incorrectly (e.g., forgetting the $-2pq$ term or getting the sign wrong).
Not combining $\frac{1}{(1+y^2)^2} + \frac{y^2}{(1+y^2)^2}$ into $\frac{1+y^2}{(1+y^2)^2}$ before cancelling.
Algebraic errors when simplifying $\frac{1+y^2}{(1+y^2)^2}$ to $\frac{1}{1+y^2}$ .

Things to Be Careful About

The question says "show that", so you must show full working — unsupported answers earn no marks.
Follow their answer from part (a); if part (a) was wrong, part (b) can still be awarded method marks for correct substitution.
The final answer must be $\frac{1}{4}$ , not an expression in $y$ .

Techniques used

substitute expressions for a and bexpand binomial squaressimplify algebraic fraction

(c)

On a single Argand diagram, sketch the loci given by the equations $\text{Re}(z) = 1$ and $|z - \frac{1}{2}| = \frac{1}{2}$ , where $z$ is a complex number.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Interpret each equation geometrically on the Argand diagram, then sketch both loci.

Working

Locus 1: $\text{Re}(z) = 1$

Let $z = x + yi$ . Then $\text{Re}(z) = x = 1$ . This is a vertical straight line passing through the point $1 + 0i$ on the real axis, parallel to the imaginary axis.

Locus 2: $|z - \frac{1}{2}| = \frac{1}{2}$

Let $z = x + yi$ . Then:

\left|\left(x - \frac{1}{2}\right) + yi\right| = \frac{1}{2}

\sqrt{\left(x - \frac{1}{2}\right)^2 + y^2} = \frac{1}{2}

Squaring both sides:

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

This is the equation of a circle with centre $\left(\frac{1}{2}, 0\right)$ , i.e., $\frac{1}{2} + 0i$ , and radius $\frac{1}{2}$ .

The circle passes through the origin $(0, 0)$ and the point $(1, 0)$ , and intersects the imaginary axis at $\frac{1}{2}i$ and $-\frac{1}{2}i$ .

Answer

The locus $\text{Re}(z) = 1$ is a vertical line through $x = 1$ . The locus $|z - \frac{1}{2}| = \frac{1}{2}$ is a circle with centre $\frac{1}{2} + 0i$ and radius $\frac{1}{2}$ .

\text{Line: } x = 1, \quad \text{Circle: centre } \left(\tfrac{1}{2}, 0\right), \text{ radius } \tfrac{1}{2}

Final answer

Vertical line x = 1 and circle centre (1/2, 0) radius 1/2

Detailed explanation

Walkthrough

We need to sketch two loci on an Argand diagram.

For $\text{Re}(z) = 1$ : If $z = x + yi$ , then $\text{Re}(z) = x$ . Setting $x = 1$ gives a vertical line through the point $(1, 0)$ on the real axis. This line is parallel to the imaginary (y) axis.

For $|z - \frac{1}{2}| = \frac{1}{2}$ : The expression $|z - z_0|$ represents the distance from $z$ to the point $z_0$ in the Argand diagram. So $|z - \frac{1}{2}| = \frac{1}{2}$ is the set of all points at distance $\frac{1}{2}$ from $\frac{1}{2} + 0i$ . This is a circle with centre $\frac{1}{2} + 0i$ and radius $\frac{1}{2}$ .

Key points on the circle:

Centre: $(\frac{1}{2}, 0)$
Passes through origin: $|0 - \frac{1}{2}| = \frac{1}{2}$ ✓
Passes through $(1, 0)$ : $|1 - \frac{1}{2}| = \frac{1}{2}$ ✓
Intersects imaginary axis at $\pm\frac{1}{2}i$ : $|\pm\frac{1}{2}i - \frac{1}{2}| = \sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{1}{\sqrt{2}} \neq \frac{1}{2}$ — actually these are NOT on the circle. The circle intersects the imaginary axis where $x = 0$ : $(0 - \frac{1}{2})^2 + y^2 = \frac{1}{4}$ , so $y^2 = 0$ , meaning it only touches the imaginary axis at the origin.

Key Takeaways

$\text{Re}(z) = k$ is always a vertical line $x = k$ .
$|z - z_0| = r$ is always a circle with centre $z_0$ and radius $r$ .
The Argand diagram uses the real axis as the horizontal axis and the imaginary axis as the vertical axis.

Common Mistakes

Drawing the circle with centre at the origin instead of at $\frac{1}{2}$ .
Drawing the vertical line at $x = 0$ (the imaginary axis) instead of $x = 1$ .
Forgetting to label axes as $\text{Re}(z)$ and $\text{Im}(z)$ .

Things to Be Careful About

The circle $|z - \frac{1}{2}| = \frac{1}{2}$ passes through the origin and $(1, 0)$ , and is entirely in the right half-plane ( $x \geq 0$ ).
The vertical line $\text{Re}(z) = 1$ is tangent to the circle at the point $(1, 0)$ .
The mark scheme awards marks for: correct vertical line through $1+0i$ , correct circle centre at $\frac{1}{2}+0i$ , and correct radius $\frac{1}{2}$ with centre not at the origin.

Techniques used

interpret real part condition as vertical lineinterpret modulus condition as circlesketch loci on Argand diagram

(d)

The complex number $z$ is such that $\text{Re}(z) = 1$ . Use your answer to part (b) to give a geometrical description of the locus of $\frac{1}{z}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the result from part (b) and the condition $\text{Re}(z) = 1$ to describe the locus of $\frac{1}{z}$ geometrically.

Working

From part (a), when $z = 1 + yi$ (i.e., $\text{Re}(z) = 1$ with $y$ real), we have:

\frac{1}{z} = a + bi \quad \text{where} \quad a = \frac{1}{1 + y^2}, \quad b = \frac{-y}{1 + y^2}

From part (b), we showed that:

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

This is the equation of a circle in the $a$ - $b$ plane (which is the Argand diagram for $\frac{1}{z}$ ) with:

Centre: $\left(\frac{1}{2}, 0\right)$ , i.e., $\frac{1}{2} + 0i$
Radius: $\frac{1}{2}$

As $y$ varies over all real numbers, the point $(a, b)$ traces out this circle. Note that when $y = 0$ , $\frac{1}{z} = 1$ , which is the point $(1, 0)$ on the circle. As $y \to \pm\infty$ , $\frac{1}{z} \to 0$ , so the origin is approached but never reached (since $y$ is finite).

Answer

The locus of $\frac{1}{z}$ is a circle with centre $\frac{1}{2} + 0i$ and radius $\frac{1}{2}$ .

\text{Circle: centre } \frac{1}{2} + 0i, \text{ radius } \frac{1}{2}

Final answer

Circle with centre 1/2 + 0i and radius 1/2

Detailed explanation

Walkthrough

We are told that $\text{Re}(z) = 1$ , so $z = 1 + yi$ for some real number $y$ . From part (a), $\frac{1}{z} = a + bi$ where $a = \frac{1}{1+y^2}$ and $b = \frac{-y}{1+y^2}$ . From part (b), these satisfy $(a - \frac{1}{2})^2 + b^2 = \frac{1}{4}$ .

The equation $(a - \frac{1}{2})^2 + b^2 = \frac{1}{4}$ is the standard form of a circle equation $(x - h)^2 + (y - k)^2 = r^2$ with centre $(h, k) = (\frac{1}{2}, 0)$ and radius $r = \frac{1}{2}$ .

In the Argand diagram for $\frac{1}{z}$ (where the horizontal axis is $a = \text{Re}(\frac{1}{z})$ and the vertical axis is $b = \text{Im}(\frac{1}{z})$ ), this describes a circle centred at $\frac{1}{2} + 0i$ with radius $\frac{1}{2}$ .

Note: The origin $(0, 0)$ is on this circle (when $a = 0, b = 0$ : $(0 - \frac{1}{2})^2 + 0^2 = \frac{1}{4}$ ✓), but $\frac{1}{z}$ never actually equals $0$ for any finite $y$ . The mark scheme concedes inclusion of the origin.

Key Takeaways

Part (b) establishes that the transformation $z \mapsto \frac{1}{z}$ maps the vertical line $\text{Re}(z) = 1$ to a circle.
This is an example of a Möbius transformation mapping lines to circles (or lines).
The equation $(a - \frac{1}{2})^2 + b^2 = \frac{1}{4}$ directly gives the centre and radius.

Common Mistakes

Describing the locus as a line instead of a circle.
Getting the centre or radius wrong from the equation $(a - \frac{1}{2})^2 + b^2 = \frac{1}{4}$ .
Forgetting to mention that this is the locus of $\frac{1}{z}$ , not of $z$ .

Things to Be Careful About

The question asks for a geometrical description, so state centre and radius clearly.
The origin lies on the circle but is not actually attained by $\frac{1}{z}$ for any finite $y$ ; however, the mark scheme concedes inclusion of the origin.
Reference the result from part (b) explicitly to show the connection.

Techniques used

use result from part (b)recognise circle equationdescribe locus geometrically

Q9Medium-HardVectors

The position vector of point $A$ relative to the origin $O$ is $\vec{OA} = 8\mathbf{i} - 5\mathbf{j} + 6\mathbf{k}$ .
The line $l$ passes through $A$ and is parallel to the vector $2\mathbf{i} + \mathbf{j} + 4\mathbf{k}$ .

(a)

State a vector equation for $l$ .

2M Similar

DifficultyEasy

Worked solution

Approach

A line through a point with position vector $\mathbf{a}$ and parallel to direction vector $\mathbf{d}$ has equation $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$ .

Working

Here $\mathbf{a} = 8\mathbf{i} - 5\mathbf{j} + 6\mathbf{k}$ and $\mathbf{d} = 2\mathbf{i} + \mathbf{j} + 4\mathbf{k}$ . Therefore

\mathbf{r} = 8\mathbf{i} - 5\mathbf{j} + 6\mathbf{k} + \lambda(2\mathbf{i} + \mathbf{j} + 4\mathbf{k}), \quad \lambda \in \mathbb{R}

Answer

\mathbf{r} = 8\mathbf{i} - 5\mathbf{j} + 6\mathbf{k} + \lambda(2\mathbf{i} + \mathbf{j} + 4\mathbf{k})

Final answer

r = 8i - 5j + 6k + λ(2i + j + 4k)

Detailed explanation

Walkthrough

A vector equation of a line needs two pieces of information: a point on the line and a direction parallel to the line. Here the point is $A$ with position vector $8\mathbf{i} - 5\mathbf{j} + 6\mathbf{k}$ , and the direction is the given vector $2\mathbf{i} + \mathbf{j} + 4\mathbf{k}$ . The standard form is $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$ , where $\lambda$ is a scalar parameter. Substituting the given vectors gives the equation directly.

Key Takeaways

The vector equation of a line is $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$ . It is not unique: any point on the line and any non-zero scalar multiple of the direction vector give an equivalent equation.

Common Mistakes

Using the direction vector as the position vector, or omitting the parameter $\lambda$ .
Writing the equation as $\mathbf{r} = \lambda \mathbf{a} + \mathbf{d}$ instead of $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$ .

Things to Be Careful About

The parameter is often denoted $\lambda$ or $\mu$ ; any letter is acceptable. The direction vector must be parallel to the line, not perpendicular.

Techniques used

identify a point on the line and a direction vectorwrite the vector equation in the form r = a + tb

(b)

The position vector of point $B$ relative to the origin $O$ is $\vec{OB} = -t\mathbf{i} + 4t\mathbf{j} + 3t\mathbf{k}$ , where $t$ is a constant. The line $l$ also passes through $B$ .

Find the value of $t$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $B$ lies on $l$ , its position vector must equal the vector equation of $l$ for some value of $\lambda$ . Equate components and solve for $\lambda$ and $t$ .

Working

Let $l$ be $\mathbf{r} = 8\mathbf{i} - 5\mathbf{j} + 6\mathbf{k} + \lambda(2\mathbf{i} + \mathbf{j} + 4\mathbf{k})$ . For $B$ :

8 + 2\lambda = -t

-5 + \lambda = 4t

6 + 4\lambda = 3t

From the first equation, $t = -8 - 2\lambda$ . Substitute into the second:

-5 + \lambda = 4(-8 - 2\lambda) = -32 - 8\lambda

9\lambda = -27

\lambda = -3

Then $t = -8 - 2(-3) = -2$ . Check the third equation: $6 + 4(-3) = -6$ and $3(-2) = -6$ .

Answer

t = -2

Final answer

t = -2

Detailed explanation

Walkthrough

We know that $B$ lies on the line $l$ . This means there is one value of the parameter $\lambda$ for which the position vector of $B$ is exactly the same as the position vector given by the equation of $l$ . Write the line equation in components: $x = 8 + 2\lambda$ , $y = -5 + \lambda$ , $z = 6 + 4\lambda$ . These must equal $-t$ , $4t$ and $3t$ respectively. Use the first equation to express $t$ in terms of $\lambda$ , substitute into the second to find $\lambda$ , then find $t$ . Finally check that the third component is also satisfied.

Key Takeaways

A point lies on a line if its position vector satisfies the line equation for some parameter value. Equating components gives simultaneous equations that can be solved for the unknown parameter and any other unknown constant.

Common Mistakes

Using different parameters for different components. The same $\lambda$ must work for all three components.
Confusing the constant $t$ in the coordinates of $B$ with the parameter $\lambda$ of the line.
Sign errors when substituting $t = -8 - 2\lambda$ into $4t$ .
Forgetting to check the third component after solving two equations.

Things to Be Careful About

If the third component did not match, the point would not lie on the line, so the check is important. Here it does match, confirming $t = -2$ .

Techniques used

substitute the coordinates of B into the line equationequate components and solve simultaneous equationscheck the remaining component

(c)

The line $m$ has vector equation $\mathbf{r} = 5\mathbf{i} - \mathbf{j} + 2\mathbf{k} + \mu(a\mathbf{i} - \mathbf{j} + 3\mathbf{k})$ . The acute angle between the directions of $l$ and $m$ is $\theta$ , where $\cos \theta = \frac{1}{\sqrt{6}}$ .

Find the possible values of $a$ .

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

The direction of $l$ is $\mathbf{d}_1 = 2\mathbf{i} + \mathbf{j} + 4\mathbf{k}$ and the direction of $m$ is $\mathbf{d}_2 = a\mathbf{i} - \mathbf{j} + 3\mathbf{k}$ . Use the scalar product formula

\cos \phi = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1||\mathbf{d}_2|}

Since $\theta$ is the acute angle between the lines, use $\pm \frac{1}{\sqrt{6}}$ and solve for $a$ .

Working

\mathbf{d}_1 \cdot \mathbf{d}_2 = 2a + 1(-1) + 4(3) = 2a + 11

|\mathbf{d}_1| = \sqrt{2^2 + 1^2 + 4^2} = \sqrt{21}

|\mathbf{d}_2| = \sqrt{a^2 + (-1)^2 + 3^2} = \sqrt{a^2 + 10}

So

\frac{2a + 11}{\sqrt{21}\sqrt{a^2 + 10}} = \pm \frac{1}{\sqrt{6}}

Squaring both sides:

6(2a + 11)^2 = 21(a^2 + 10)

Divide by 3:

2(2a + 11)^2 = 7(a^2 + 10)

8a^2 + 88a + 242 = 7a^2 + 70

a^2 + 88a + 172 = 0

(a + 2)(a + 86) = 0

So $a = -2$ or $a = -86$ .

Answer

a = -2 \quad \text{or} \quad a = -86

Final answer

a = -2 or a = -86

Detailed explanation

Walkthrough

The direction of a line given by $\mathbf{r} = \mathbf{p} + \mu \mathbf{d}$ is the vector $\mathbf{d}$ . For $l$ the direction vector is $2\mathbf{i} + \mathbf{j} + 4\mathbf{k}$ ; for $m$ it is $a\mathbf{i} - \mathbf{j} + 3\mathbf{k}$ . The cosine of the angle between two direction vectors is their scalar product divided by the product of their magnitudes. Compute the scalar product: $2a + 11$ . Compute the magnitudes: $\sqrt{21}$ and $\sqrt{a^2 + 10}$ . The angle between the lines is acute, so its cosine is positive, but the scalar product of the chosen direction vectors may be positive or negative depending on the orientation. Therefore we set the expression equal to $\pm \frac{1}{\sqrt{6}}$ . Squaring removes the sign and gives a quadratic equation in $a$ . Solving it gives the two possible values.

Key Takeaways

The scalar product formula $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos \theta$ connects vectors to angles. For the acute angle between lines, use the absolute value of the cosine, which is why the $\pm$ sign appears. Solving the resulting equation often leads to a quadratic.

Common Mistakes

Forgetting the $\pm$ sign, which would lose one solution.
Sign errors in the scalar product, especially the $-1$ from $-\mathbf{j}$ .
Writing the magnitude of $\mathbf{d}_2$ as $\sqrt{a^2 + 1 + 9}$ incorrectly; it is $\sqrt{a^2 + 10}$ .
Squaring incorrectly: remember to square the whole numerator and denominator.

Things to Be Careful About

Both $a = -2$ and $a = -86$ are valid. For $a = -2$ the scalar product is positive; for $a = -86$ it is negative, but the acute angle between the two lines is the same because a direction vector can be reversed. Always check that the values satisfy the original equation after squaring.

Techniques used

extract direction vectors from the line equationscompute the scalar product of the direction vectorscompute the magnitudes of the direction vectorsset the cosine formula equal to the given acute angle valuesolve the resulting quadratic equation

Q10MediumDifferential Equations

Fig. 2

A large cylindrical tank is used to store water. The base of the tank is a circle of radius 4 metres. At time $t$ minutes, the depth of the water in the tank is $h$ metres. There is a tap at the bottom of the tank. When the tap is open, water flows out of the tank at a rate proportional to the square root of the volume of water in the tank.

(a)

Show that $\frac{dh}{dt} = -\lambda \sqrt{h}$ , where $\lambda$ is a positive constant.

4M Similar

DifficultyMedium

Worked solution

Approach

Relate the volume of water $V$ to the depth $h$ using the formula for the volume of a cylinder. Use the given proportionality for the rate of change of volume and the chain rule to find $\frac{dh}{dt}$ .

Working

The volume of water in the cylindrical tank with radius $r=4$ and depth $h$ is:

V = \pi r^2 h = 16\pi h

Differentiating with respect to $h$ :

\frac{dV}{dh} = 16\pi

The rate at which water flows out is proportional to $\sqrt{V}$ . Since water is leaving the tank, $\frac{dV}{dt}$ is negative. Let $k$ be a positive constant of proportionality:

\frac{dV}{dt} = -k\sqrt{V}

Substitute $V = 16\pi h$ :

\frac{dV}{dt} = -k\sqrt{16\pi h} = -4k\sqrt{\pi}\sqrt{h}

Using the chain rule, $\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}$ :

16\pi \frac{dh}{dt} = -4k\sqrt{\pi}\sqrt{h}

Solve for $\frac{dh}{dt}$ :

\frac{dh}{dt} = \frac{-4k\sqrt{\pi}}{16\pi} \sqrt{h} = -\frac{k}{4\sqrt{\pi}} \sqrt{h}

Let $\lambda = \frac{k}{4\sqrt{\pi}}$ . Since $k$ and $\pi$ are positive constants, $\lambda$ is a positive constant.

\frac{dh}{dt} = -\lambda \sqrt{h}

Answer

Shown as required.

Final answer

dh/dt = -lambda sqrt(h)

Detailed explanation

Walkthrough

First, we express the volume of water $V$ in terms of the depth $h$ . For a cylinder of radius 4, $V = \pi(4)^2 h = 16\pi h$ . Differentiating this gives $dV/dh = 16\pi$ .

Next, we translate the word statement 'water flows out at a rate proportional to the square root of the volume' into a mathematical equation. Since the volume is decreasing, $dV/dt = -k\sqrt{V}$ for some $k > 0$ . Substituting $V = 16\pi h$ gives $dV/dt = -k\sqrt{16\pi h} = -4k\sqrt{\pi}\sqrt{h}$ .

We then use the chain rule $dV/dt = (dV/dh)(dh/dt)$ to link the rate of change of volume to the rate of change of depth. Equating the two expressions for $dV/dt$ allows us to solve for $dh/dt$ . The constant coefficient $k/(4\sqrt{\pi})$ is renamed $\lambda$ to match the required form.

Key Takeaways

Volume of a cylinder is $V = \pi r^2 h$ .
Related rates problems often require the chain rule to connect variables.
Proportionality in a decreasing quantity implies a negative constant of proportionality.

Common Mistakes

Forgetting the negative sign in $dV/dt = -k\sqrt{V}$ , which is crucial because the water level is dropping.
Incorrectly applying the chain rule or forgetting to multiply by $dV/dh$ .
Not simplifying $\sqrt{16\pi}$ correctly to $4\sqrt{\pi}$ .

Things to Be Careful About

Ensure $\lambda$ is shown to be positive since $k$ and $\pi$ are positive.
The question asks to 'show', so the final line must match the required expression exactly.

Techniques used

express volume in terms of depthapply chain rule for related ratessubstitute proportionality condition

(b)

At time $t = 0$ the tap is opened. It is given that $h = 4$ when $t = 0$ and that $h = 2.25$ when $t = 20$ .

Solve the differential equation to obtain an expression for $t$ in terms of $h$ , and hence find the time taken to empty the tank.

6M Similar

DifficultyMedium

Worked solution

Approach

Separate the variables in the differential equation $\frac{dh}{dt} = -\lambda \sqrt{h}$ and integrate. Use the given initial conditions ( $t=0, h=4$ and $t=20, h=2.25$ ) to find the constants $\lambda$ and $C$ . Finally, set $h=0$ to find the time to empty the tank.

Working

Separate variables:

\frac{1}{\sqrt{h}} \, dh = -\lambda \, dt

h^{-1/2} \, dh = -\lambda \, dt

Integrate both sides:

\int h^{-1/2} \, dh = \int -\lambda \, dt

2h^{1/2} = -\lambda t + C

2\sqrt{h} = -\lambda t + C

Apply the first boundary condition: when $t=0$ , $h=4$ :

2\sqrt{4} = -\lambda(0) + C \implies 4 = C

Substitute $C=4$ into the equation:

2\sqrt{h} = -\lambda t + 4

Apply the second boundary condition: when $t=20$ , $h=2.25$ :

2\sqrt{2.25} = -\lambda(20) + 4

2(1.5) = -20\lambda + 4

3 = -20\lambda + 4

20\lambda = 1 \implies \lambda = \frac{1}{20}

Substitute $\lambda = 1/20$ back into the equation:

2\sqrt{h} = -\frac{1}{20}t + 4

Rearrange to obtain $t$ in terms of $h$ :

\frac{t}{20} = 4 - 2\sqrt{h}

t = 80 - 40\sqrt{h}

To find the time taken to empty the tank, set $h=0$ :

t = 80 - 40\sqrt{0} = 80

Answer

t = 80 - 40\sqrt{h}

Time to empty the tank is 80 minutes.

Final answer

t = 80 - 40sqrt(h); 80 minutes

Detailed explanation

Walkthrough

We start with the differential equation $dh/dt = -\lambda\sqrt{h}$ . Separating variables gives $h^{-1/2} dh = -\lambda dt$ . Integrating both sides yields $2\sqrt{h} = -\lambda t + C$ .

We use the initial condition $h=4$ when $t=0$ to find $C$ : $2\sqrt{4} = C \implies C=4$ .

We then use the second condition $h=2.25$ when $t=20$ to find $\lambda$ : $2\sqrt{2.25} = -20\lambda + 4 \implies 3 = -20\lambda + 4 \implies \lambda = 1/20$ .

Substituting these constants back gives $2\sqrt{h} = -t/20 + 4$ . Rearranging for $t$ gives $t = 80 - 40\sqrt{h}$ .

Finally, the tank is empty when $h=0$ . Substituting $h=0$ gives $t=80$ minutes.

Key Takeaways

Separation of variables is a standard technique for first-order ODEs of the form $dy/dx = f(y)g(x)$ .
Integration of $y^n$ gives $y^{n+1}/(n+1)$ ; here $n=-1/2$ .
Boundary conditions are used to determine constants of integration.
Physical interpretation: 'empty tank' means depth $h=0$ .

Common Mistakes

Forgetting the constant of integration $C$ .
Sign errors when rearranging the equation for $t$ .
Calculating $\sqrt{2.25}$ incorrectly (it is 1.5).
Forgetting to set $h=0$ at the end to find the emptying time.

Things to Be Careful About

Ensure the final expression for $t$ is explicitly solved for $t$ .
Check that $\lambda$ is positive as derived in part (a).
Units are in minutes, so the final answer is 80 minutes.

Techniques used

separate variablesintegrate power functionapply initial conditions to find constantssolve linear equation for time

Mathematics 9709/31 — October/November 2024

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