(a) Find the first three terms, in ascending powers of $x$, in the expansion of each of the following expressions. (i) $(2 - px)^5$ (ii) $\left(1 - \frac{1}{2}x\right)^4$ (b) Given that the coefficient of $x^2$ in the expansion of $(2 - px)^5 \left(1 - \frac{1}{2}x\right)^4$ is 93, find the possible values of the constant $p$.

(a)(i) 32 - 80px + 80p^2x^2 · (a)(ii) 1 - 2x + 3/2 x^2 · (b) p = -9/4 and p = 1/4

9709/11 May/June 2025 — Mathematics Solved Past Paper

Q14MMediumTrigonometryQuadratics

Solve the equation $6\sin\theta = 1 + \frac{2}{\sin\theta}$ for $-180^\circ < \theta < 180^\circ$ .

DifficultyMedium

Worked solution

Approach

The equation contains $\sin\theta$ in a denominator, so first multiply through by $\sin\theta$ (with $\sin\theta \neq 0$ ) to obtain a quadratic in $\sin\theta$ . Factorise the quadratic, solve for $\sin\theta$ , then find every angle in the interval $-180^\circ < \theta < 180^\circ$ that gives those sine values.

Working

Since $\frac{2}{\sin\theta}$ is defined, $\sin\theta \neq 0$ . Multiplying both sides by $\sin\theta$ :

6\sin^2\theta = \sin\theta + 2

Rearrange:

6\sin^2\theta - \sin\theta - 2 = 0

Factorise:

(3\sin\theta - 2)(2\sin\theta + 1) = 0

Hence

\sin\theta = \frac{2}{3} \quad \text{or} \quad \sin\theta = -\frac{1}{2}

For $\sin\theta = \frac{2}{3}$ :

\theta = \sin^{-1}\left(\frac{2}{3}\right) \approx 41.8^\circ

The other solution in the interval is

\theta = 180^\circ - 41.8^\circ = 138.2^\circ

For $\sin\theta = -\frac{1}{2}$ :

\theta = -30^\circ

The other solution in the interval is

\theta = -180^\circ + 30^\circ = -150^\circ

Answer

\theta = -150^\circ, -30^\circ, 41.8^\circ, 138.2^\circ

Final answer

-150°, -30°, 41.8°, 138.2°

Detailed explanation

Walkthrough

The equation has $\sin\theta$ in a denominator, so $\sin\theta$ cannot be zero. Multiplying every term by $\sin\theta$ clears the fraction and gives a quadratic in $\sin\theta$ :

6\sin^2\theta = \sin\theta + 2

Bring all terms to one side and factorise. The factorisation

(3\sin\theta - 2)(2\sin\theta + 1) = 0

gives two possible values for $\sin\theta$ . For each value, use inverse sine to find the principal angle, then use the symmetry of the sine graph to find every angle in $-180^\circ < \theta < 180^\circ$ . For $\sin\theta = \frac{2}{3}$ , the two angles are $41.8^\circ$ and $138.2^\circ$ ; for $\sin\theta = -\frac{1}{2}$ , they are $-30^\circ$ and $-150^\circ$ .

Key Takeaways

A trigonometric equation can often be reduced to a quadratic by treating $\sin\theta$ as the variable.
Factorising a quadratic in $\sin\theta$ gives candidate sine values; each sine value may give two angles within a $360^\circ$ interval.
Use inverse sine for the principal value and the symmetry of the sine graph to find all solutions in the required interval.

Common Mistakes

Multiplying by $\sin\theta$ without noting that $\sin\theta \neq 0$ ; although no zero solution arises here, the division by zero must be acknowledged.
Stopping at $\sin\theta = -\frac{1}{2}$ and only giving $-30^\circ$ , forgetting $-150^\circ$ .
Giving angles outside the required interval, such as $210^\circ$ or $330^\circ$ .
Mixing degrees and radians without converting; the question uses degrees, so answers should be in degrees.

Things to Be Careful About

The interval is $-180^\circ < \theta < 180^\circ$ , not $0^\circ$ to $360^\circ$ , so negative angles must be included.
For $\sin\theta = -\frac{1}{2}$ , the two solutions are $-30^\circ$ and $-150^\circ$ ; for $\sin\theta = \frac{2}{3}$ , the two are $41.8^\circ$ and $138.2^\circ$ .
The mark scheme allows answers within a tolerance (AWRT), so rounding to one decimal place is acceptable.
If radians were used, the equivalent answers would be $-\frac{5\pi}{6}$ , $-\frac{\pi}{6}$ , $0.730$ , $2.41$ ; but the question asks for degrees.

Techniques used

multiply through by sin θ to clear the denominatorform a quadratic in sin θfactorise the quadraticsolve for θ in the given interval using inverse sine and sine symmetry

Approach

The gradient of the tangent at $A$ is the value of $\frac{dy}{dx}$ when $x = 4$ . Since the normal is perpendicular to the tangent, its gradient is the negative reciprocal of the tangent gradient.

Working

Substitute $x = 4$ into the given derivative:

\frac{dy}{dx} = 4(2(4) - 5)^3 - 9(4)^{\frac{1}{2}}

= 4(3)^3 - 9(2) = 4(27) - 18 = 108 - 18 = 90

The gradient of the normal is therefore:

m_{\text{normal}} = -\frac{1}{90}

Answer

-\frac{1}{90}

Final answer

-1/90

Detailed explanation

Walkthrough

The derivative $\frac{dy}{dx}$ gives the gradient of the tangent to the curve at any point. At point $A$ , the $x$ -coordinate is $4$ , so substitute $x = 4$ into the derivative. This gives the gradient of the tangent as $90$ . A normal is perpendicular to the tangent, so its gradient is the negative reciprocal, $-\frac{1}{90}$ .

Key Takeaways

The value of the derivative at a point is the gradient of the tangent there.
If two lines are perpendicular, the product of their gradients is $-1$ , so the normal gradient is the negative reciprocal of the tangent gradient.

Common Mistakes

Using the tangent gradient $90$ as the normal gradient instead of taking the negative reciprocal.
Substituting the $y$ -coordinate $-\frac{11}{2}$ into the derivative; the derivative is evaluated using $x = 4$ only.
Arithmetic errors such as evaluating $4^{\frac{1}{2}}$ as $4$ instead of $2$ .

Things to Be Careful About

The mark scheme requires substitution of $x = 4$ into $\frac{dy}{dx}$ ; substituting the point into the equation of the curve is not enough unless it leads to $-\frac{1}{90}$ .
Keep the negative sign in $-\frac{1}{90}$ ; the normal slopes in the opposite direction to the tangent.

Techniques used

substitute the x-coordinate into the derivativecalculate the gradient of the normal as the negative reciprocal

(b)

Find the equation of the curve.

4M Similar

DifficultyMedium

Worked solution

Approach

Integrate the given derivative term by term to recover $y$ , then use the point $A(4, -\frac{11}{2})$ to find the constant of integration.

Working

Integrate each term. For the first term, use the reverse chain rule:

\int 4(2x-5)^3\,dx = \frac{1}{2}(2x-5)^4

For the second term:

\int -9x^{\frac{1}{2}}\,dx = -9 \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = -6x^{\frac{3}{2}}

Hence

y = \frac{1}{2}(2x-5)^4 - 6x^{\frac{3}{2}} + c

Use $x = 4$ and $y = -\frac{11}{2}$ :

-\frac{11}{2} = \frac{1}{2}(2(4)-5)^4 - 6(4)^{\frac{3}{2}} + c

-\frac{11}{2} = \frac{1}{2}(3)^4 - 6(8) + c = \frac{81}{2} - 48 + c

-\frac{11}{2} = -\frac{15}{2} + c

c = 2

Therefore the equation of the curve is:

y = \frac{1}{2}(2x-5)^4 - 6x^{\frac{3}{2}} + 2

Answer

y = \frac{1}{2}(2x-5)^4 - 6x^{\frac{3}{2}} + 2

Final answer

y = (1/2)(2x - 5)^4 - 6x^(3/2) + 2

Detailed explanation

Walkthrough

To recover the equation of the curve from its derivative, integrate $\frac{dy}{dx}$ with respect to $x$ . The first term $4(2x-5)^3$ is a composite power, so use the reverse chain rule: the derivative of $(2x-5)^4$ is $8(2x-5)^3$ , so multiplying by $\frac{1}{2}$ gives $4(2x-5)^3$ . Hence its integral is $\frac{1}{2}(2x-5)^4$ . The second term $-9x^{\frac{1}{2}}$ is integrated by increasing the power by $1$ and dividing by the new power: $-9 \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = -6x^{\frac{3}{2}}$ . Add the constant of integration $c$ . Then substitute $x = 4$ and $y = -\frac{11}{2}$ into the integrated expression to find $c$ . The working gives $c = 2$ , so the equation of the curve is $y = \frac{1}{2}(2x-5)^4 - 6x^{\frac{3}{2}} + 2$ .

Key Takeaways

Integration is the reverse of differentiation, and every indefinite integral must include a constant of integration.
A point on the curve provides the initial condition needed to determine that constant.
Composite functions of the form $(ax+b)^n$ are integrated by reversing the chain rule.

Common Mistakes

Forgetting to add $+c$ before using the point.
Getting the coefficient wrong when integrating $4(2x-5)^3$ ; the factor $2$ from differentiating $2x-5$ must be accounted for.
Miscomputing $4^{\frac{3}{2}}$ : it equals $8$ , not $16$ .
Not simplifying the fraction $-\frac{11}{2} = -\frac{15}{2} + c$ correctly.

Things to Be Careful About

The mark scheme accepts unsimplified integrated expressions for the first two marks, but the final answer must be simplified and include the constant.
Fractions must be simplified; for example $c = \frac{4}{2}$ should be written as $2$ .
The final answer can be written as $f(x)$ instead of $y$ and is still accepted.

Techniques used

integrate a composite linear powerintegrate a fractional power termsubstitute the given point to determine the constant of integration

Q3MediumSeries

The third term of a geometric progression is 18 and the sum of the first three terms is 26. It is given that the common ratio is negative.

(a)

Find the tenth term of the progression. Give your answer correct to 3 significant figures.

5M Similar

DifficultyMedium

Worked solution

Approach

Let the first term be $a$ and the common ratio be $r$ . Use the third term and the sum of the first three terms to write two equations, eliminate $a$ , solve the resulting quadratic in $r$ , and select the negative root. Then compute the tenth term using $ar^{n-1}$ .

Working

The third term is

ar^2 = 18

The sum of the first three terms is

a + ar + ar^2 = 26

Since $ar^2 = 18$ , this becomes

a + ar + 18 = 26

so $a(1+r) = 8$ . Substitute $a = \dfrac{18}{r^2}$ :

\frac{18}{r^2} + \frac{18}{r} = 8

Multiply through by $r^2$ :

8r^2 - 18r - 18 = 0

Factorise:

(4r + 3)(r - 3) = 0

So $r = -\dfrac{3}{4}$ or $r = 3$ . Since the common ratio is negative, $r = -\dfrac{3}{4}$ .

Now

a = \frac{18}{r^2} = \frac{18}{\left(-\frac{3}{4}\right)^2} = 32

The tenth term is

ar^{9} = 32\left(-\frac{3}{4}\right)^9 = -2.400

Correct to 3 significant figures:

-2.40

Answer

-2.40

Final answer

-2.40

Detailed explanation

Walkthrough

Write the progression as $a, ar, ar^2, \ldots$ . The third term is $ar^2 = 18$ , and the sum of the first three terms is $a + ar + ar^2 = 26$ . Since $ar^2 = 18$ , the sum equation becomes $a + ar + 18 = 26$ , so $a(1+r) = 8$ .

To remove $a$ , use $a = \dfrac{18}{r^2}$ from the third term. Substituting into $a(1+r) = 8$ gives

\frac{18}{r^2} + \frac{18}{r} = 8.

Multiplying by $r^2$ gives $18 + 18r = 8r^2$ , or $8r^2 - 18r - 18 = 0$ . Factorising,

(4r+3)(r-3) = 0,

so $r = -\dfrac{3}{4}$ or $r = 3$ . The question tells us the common ratio is negative, so $r = -\dfrac{3}{4}$ .

Then $a = \dfrac{18}{r^2} = 32$ . The $n$ th term of a GP is $ar^{n-1}$ , so the tenth term is

32\left(-\frac{3}{4}\right)^9 = -2.400,

which is $-2.40$ to 3 significant figures.

Key Takeaways

A geometric progression is fully described by its first term $a$ and common ratio $r$ .
The third term is $ar^2$ , and the sum of the first three terms is $a+ar+ar^2$ .
Eliminating one variable from two equations can reduce the problem to a quadratic.
The sign of the common ratio is essential when choosing between two possible roots.

Common Mistakes

Using $r^{10}$ instead of $r^9$ for the tenth term: the $n$ th term is $ar^{n-1}$ .
Choosing $r=3$ and ignoring the condition that the common ratio is negative.
Sign errors when rearranging $18 + 18r = 8r^2$ ; the correct quadratic is $8r^2 - 18r - 18 = 0$ .
Not rounding the final answer to 3 significant figures.

Things to Be Careful About

The tenth term is negative because $r$ is negative and the exponent $9$ is odd.
The exact value $r = -\dfrac{3}{4}$ is also accepted as $-0.75$ .
The final answer should be written as $-2.40$ to show 3 significant figures, not $-2.4$ .

Techniques used

write equations for the third term and the sum of the first three termseliminate the first term to form a quadratic in the common ratiosolve the quadratic and select the negative rootsubstitute into the nth term formula

(b)

Find the exact value of the sum to infinity of the progression.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $|r| < 1$ , the sum to infinity exists. Use the formula $S_\infty = \dfrac{a}{1-r}$ with the values found in part (a).

Working

With $a = 32$ and $r = -\dfrac{3}{4}$ :

S_\infty = \frac{32}{1 - \left(-\frac{3}{4}\right)}

Simplify:

S_\infty = \frac{32}{\frac{7}{4}} = 32 \times \frac{4}{7} = \frac{128}{7}

Answer

\frac{128}{7}

Final answer

128/7

Detailed explanation

Walkthrough

The sum to infinity of a geometric progression exists only when $|r| < 1$ . Here $r = -\dfrac{3}{4}$ , so $|r| = \dfrac{3}{4} < 1$ and the formula applies. Using $a = 32$ and $r = -\dfrac{3}{4}$ in $S_\infty = \dfrac{a}{1-r}$ :

S_\infty = \frac{32}{1 - \left(-\frac{3}{4}\right)} = \frac{32}{\frac{7}{4}} = 32 \times \frac{4}{7} = \frac{128}{7}.

This is the exact value; no decimal approximation is required.

Key Takeaways

The sum to infinity formula is $S_\infty = \dfrac{a}{1-r}$ and is valid only when $|r| < 1$ .
A negative common ratio still gives a finite sum if its absolute value is less than 1.
Exact answers should be left as fractions when the question asks for the exact value.

Common Mistakes

Using $1+r$ in the denominator instead of $1-r$ .
Forgetting to check the convergence condition $|r| < 1$ .
Giving a decimal approximation such as $18.2857$ instead of the exact fraction $\dfrac{128}{7}$ .

Things to Be Careful About

The mark scheme follows through on the values of $a$ and $r$ from part (a), provided $-1 < r < 1$ .
Since the question asks for the exact value, leave the answer as $\dfrac{128}{7}$ .
The denominator is $1 - r$ ; because $r$ is negative, $1 - r = 1 - \left(-\dfrac{3}{4}\right) = \dfrac{7}{4}$ .

Techniques used

check the convergence condition |r| < 1apply the sum to infinity formulasimplify the resulting fraction

Q45MMediumIntegration

Fig. 4.1

The diagram shows the curve with equation $y = 5x^{\frac{3}{2}} - 20x$ and the line with equation $y = x - 16$ . The $x$ -coordinates of the points of intersection of the curve and line are 1 and 16.

Find the area of the shaded region between the curve and the line.

DifficultyMedium

Worked solution

Approach

The area between two curves $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$ , where $f(x) \geq g(x)$ on $[a, b]$ , is given by:

\int_a^b [f(x) - g(x)] \, dx

From the diagram , the line $y = x - 16$ lies above the curve $y = 5x^{3/2} - 20x$ between the intersection points $x = 1$ and $x = 16$ . We integrate the difference (line $-$ curve) with respect to $x$ .

Working

Set up the integral for the shaded area $A$ :

A = \int_1^{16} \left[ (x - 16) - \left( 5x^{\frac{3}{2}} - 20x \right) \right] \, dx

Simplify the integrand:

\begin{aligned} (x - 16) - \left( 5x^{\frac{3}{2}} - 20x \right) &= x - 16 - 5x^{\frac{3}{2}} + 20x \\ &= -5x^{\frac{3}{2}} + 21x - 16 \end{aligned}

Now integrate each term:

\begin{aligned} A &= \left[ -5 \left( \frac{x^{\frac{5}{2}}}{\frac{5}{2}} \right) + 21 \left( \frac{x^2}{2} \right) - 16x \right]_1^{16} \\ &= \left[ -2x^{\frac{5}{2}} + \frac{21}{2}x^2 - 16x \right]_1^{16} \end{aligned}

Evaluate at the upper limit $x = 16$ :

\begin{aligned} \text{At } x = 16: \quad & -2(16)^{\frac{5}{2}} + \frac{21}{2}(16)^2 - 16(16) \\ &= -2(1024) + \frac{21}{2}(256) - 256 \\ &= -2048 + 2688 - 256 \\ &= 384 \end{aligned}

Evaluate at the lower limit $x = 1$ :

\begin{aligned} \text{At } x = 1: \quad & -2(1)^{\frac{5}{2}} + \frac{21}{2}(1)^2 - 16(1) \\ &= -2 + 10.5 - 16 \\ &= -7.5 \end{aligned}

Calculate the area:

A = 384 - (-7.5) = 391.5

Alternatively, as a fraction:

A = 384 + \frac{15}{2} = \frac{768 + 15}{2} = \frac{783}{2}

Answer

391.5

Final answer

391.5

Detailed explanation

Walkthrough

The problem asks for the area of the shaded region bounded by the curve $y = 5x^{3/2} - 20x$ and the line $y = x - 16$ . The $x$ -coordinates of the intersection points are given as $x = 1$ and $x = 16$ .

Identify the upper and lower functions: Looking at the diagram , the straight line is above the curve in the shaded region. We can verify this by testing a point between $x=1$ and $x=16$ , for example $x=4$ . For the line, $y = 4 - 16 = -12$ . For the curve, $y = 5(4)^{3/2} - 20(4) = 5(8) - 80 = -40$ . Since $-12 > -40$ , the line is indeed the upper function $f(x) = x - 16$ and the curve is the lower function $g(x) = 5x^{3/2} - 20x$ .
Set up the definite integral: The area $A$ is given by $\int_1^{16} [f(x) - g(x)] \, dx$ . Substituting the expressions:
$A = \int_1^{16} \left[ (x - 16) - (5x^{3/2} - 20x) \right] \, dx$
Simplify the integrand: Combine like terms:
$x - 16 - 5x^{3/2} + 20x = -5x^{3/2} + 21x - 16$
Find the antiderivative: Use the power rule for integration $\int x^n \, dx = \frac{x^{n+1}}{n+1}$ :
- For $-5x^{3/2}$ : add 1 to the power to get $5/2$ , divide by $5/2$ (which is multiplying by $2/5$ ). So, $-5 \times \frac{2}{5} x^{5/2} = -2x^{5/2}$ .
- For $21x$ : becomes $\frac{21}{2}x^2$ .
- For $-16$ : becomes $-16x$ .
  The antiderivative is $-2x^{5/2} + \frac{21}{2}x^2 - 16x$ .
Apply the limits: Evaluate the antiderivative at $x = 16$ and $x = 1$ and subtract (Fundamental Theorem of Calculus):
- At $x = 16$ : $-2(16)^{5/2} + \frac{21}{2}(16)^2 - 16(16)$ . Note that $16^{5/2} = (\sqrt{16})^5 = 4^5 = 1024$ . So, $-2(1024) + 10.5(256) - 256 = -2048 + 2688 - 256 = 384$ .
- At $x = 1$ : $-2(1) + 10.5(1) - 16 = -2 + 10.5 - 16 = -7.5$ .
- Area $= 384 - (-7.5) = 391.5$ .

Key Takeaways

Area between curves: Always integrate (upper curve $-$ lower curve). The diagram is essential to determine which function is on top.
Fractional powers: Remember the power rule $\int x^n \, dx = \frac{x^{n+1}}{n+1}$ works for fractional $n$ as well. For $x^{3/2}$ , the next power is $5/2$ , and dividing by $5/2$ is multiplying by $2/5$ .
Evaluation: Be careful with arithmetic when evaluating large powers like $16^{5/2}$ . Break it down as $(16^{1/2})^5 = 4^5 = 1024$ .

Common Mistakes

Wrong order of subtraction: Integrating (curve $-$ line) instead of (line $-$ curve) will give a negative area ( $-391.5$ ). The area must be positive.
Integration errors with fractional powers: Forgetting to divide by the new power (e.g., integrating $5x^{3/2}$ as $5x^{5/2}$ instead of $2x^{5/2}$ ).
Arithmetic errors: Calculating $16^{5/2}$ incorrectly or making sign errors when subtracting the lower limit value (especially since the lower limit value is negative, $-7.5$ , so subtracting it adds $7.5$ ).

Things to Be Careful About

Domain: The function $x^{3/2}$ requires $x \geq 0$ . The limits $1$ and $16$ are both positive, so this is fine.
Mark scheme allowance: The mark scheme allows for calculating the area as $\int (\text{line}) \, dx - \int (\text{curve}) \, dx$ . Both methods yield the same result. Ensure signs are handled correctly when subtracting the second integral.
Final Answer: The answer $391.5$ or $\frac{783}{2}$ is required. CAO (Correct Answer Only) is noted in the mark scheme, so showing correct working is vital to reach this final value.

Techniques used

identify upper and lower curves from diagramset up definite integral for area between curvesintegrate terms with fractional powersapply limits of integration

Approach

Use the binomial theorem to expand $\left(1 - \frac{1}{2}x\right)^4$ , keeping only the terms up to $x^2$ .

Working

\left(1 - \frac{1}{2}x\right)^4 = 1^4 + \binom{4}{1}1^3\left(-\frac{1}{2}x\right) + \binom{4}{2}1^2\left(-\frac{1}{2}x\right)^2 + \cdots

= 1 - 4\left(\frac{1}{2}x\right) + 6\left(\frac{1}{4}x^2\right) + \cdots

= 1 - 2x + \frac{3}{2}x^2 + \cdots

Answer

The first three terms are $1 - 2x + \frac{3}{2}x^2$ .

Final answer

1 - 2x + 3/2 x^2

Detailed explanation

Walkthrough

Use the binomial expansion for $\left(1 - \frac{1}{2}x\right)^4$ . The binomial coefficients for $n = 4$ are $1$ , $4$ , and $6$ for the first three terms. Substitute $a = 1$ and $b = -\frac{1}{2}x$ , then simplify. The $x$ term is $4 \times \left(-\frac{1}{2}x\right) = -2x$ . The $x^2$ term is $6 \times \left(\frac{1}{4}x^2\right) = \frac{3}{2}x^2$ .

Key Takeaways

This question reinforces the binomial theorem with a fractional coefficient. It is important to square the fraction correctly when finding the $x^2$ term.

Common Mistakes

Squaring $-\frac{1}{2}x$ incorrectly; it should give $\frac{1}{4}x^2$ , not $-\frac{1}{4}x^2$ .
Forgetting the binomial coefficient $6$ for the $x^2$ term.
Writing the $x$ term as $+2x$ instead of $-2x$ .

Things to Be Careful About

The $x^2$ term is positive because a negative term is squared.
Keep fractions simplified: $6 \times \frac{1}{4} = \frac{3}{2}$ .
Only the first three terms are needed; ignore higher powers of $x$ .

Techniques used

expand using the binomial theoremhandle fractional and negative termsextract the first three terms

(b)

Given that the coefficient of $x^2$ in the expansion of $(2 - px)^5 \left(1 - \frac{1}{2}x\right)^4$ is 93, find the possible values of the constant $p$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the first three terms from part (a) for each expansion. Multiply the two expansions and collect the coefficient of $x^2$ . Set this coefficient equal to 93 and solve the resulting quadratic for $p$ .

Working

The two expansions are:

(2 - px)^5 = 32 - 80px + 80p^2x^2 + \cdots

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

The coefficient of $x^2$ in the product comes from three contributions:

constant term from first $\times$ $x^2$ term from second: $32 \times \frac{3}{2} = 48$
$x$ term from first $\times$ $x$ term from second: $(-80p)(-2) = 160p$
$x^2$ term from first $\times$ constant term from second: $80p^2 \times 1 = 80p^2$

The coefficient of $x^2$ is

48 + 160p + 80p^2

Set it equal to 93:

48 + 160p + 80p^2 = 93

80p^2 + 160p - 45 = 0

Factorise:

5(4p + 9)(4p - 1) = 0

So

p = -\frac{9}{4} \quad \text{or} \quad p = \frac{1}{4}

Final answer

p = -9/4 and p = 1/4

Detailed explanation

Walkthrough

From part (a), we have the expansions up to $x^2$ . To find the coefficient of $x^2$ in the product, multiply the two expansions and collect all terms that produce $x^2$ : constant $\times$ $x^2$ , $x$ $\times$ $x$ , and $x^2$ $\times$ constant. Add these three contributions to get $48 + 160p + 80p^2$ . Set this equal to 93, simplify to $80p^2 + 160p - 45 = 0$ , and factorise to obtain $5(4p + 9)(4p - 1) = 0$ . Hence $p = -\frac{9}{4}$ or $p = \frac{1}{4}$ .

Key Takeaways

This question combines binomial expansion with solving a quadratic. It tests careful multiplication of expansions and the ability to collect coefficients accurately.

Common Mistakes

Omitting the cross term $(-80p)(-2)$ , which gives $160p$ .
Sign errors when multiplying negative terms.
Forgetting to subtract 93 before factorising.
Not showing the factorisation or quadratic formula, which is required for full marks.

Things to Be Careful About

The coefficient of $x^2$ has three contributions; all must be included.
After setting the coefficient equal to 93, rearrange to $80p^2 + 160p - 45 = 0$ before factorising.
The factorisation $5(4p + 9)(4p - 1) = 0$ gives two possible values of $p$ ; both are valid since the question asks for possible values.

Techniques used

multiply the two binomial expansionscollect the coefficient of x^2form a quadratic equationsolve the quadratic by factorisation

Q6MediumQuadratics

The equation of a curve is $2x^2 - kxy + 2 = 0$ and the equation of a line is $y = px + 3$ , where $k$ and $p$ are constants.

(a)

Given that $k = 2$ and $p = 11$ , find the coordinates of the points of intersection of the curve and the line.

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Substitute the given values of $k$ and $p$ into the equation of the curve, replacing $y$ by the line equation. This eliminates $y$ and gives a quadratic equation in $x$ , which we solve by factorisation. Each $x$ -value is then substituted back into the line equation to find the corresponding $y$ -coordinate.

Working

With $k=2$ and $p=11$ , the line is $y=11x+3$ . Substituting into the curve $2x^2 - kxy + 2=0$ :

2x^2 - 2x(11x+3)+2=0

Expand the bracket:

2x^2 - 22x^2 - 6x + 2 = 0

-20x^2 - 6x + 2 = 0

Multiplying by $-1$ gives

20x^2 + 6x - 2 = 0

or, dividing by $2$ ,

10x^2 + 3x - 1 = 0

Factorise:

(5x-1)(2x+1)=0

Hence

x = \frac{1}{5} \quad \text{or} \quad x = -\frac{1}{2}

Now substitute each value into $y = 11x+3$ .

For $x=\frac{1}{5}$ :

y = 11\left(\frac{1}{5}\right)+3 = \frac{11}{5}+\frac{15}{5} = \frac{26}{5}

For $x=-\frac{1}{2}$ :

y = 11\left(-\frac{1}{2}\right)+3 = -\frac{11}{2}+\frac{6}{2} = -\frac{5}{2}

Answer

The points of intersection are $\left(-\frac{1}{2}, -\frac{5}{2}\right)$ and $\left(\frac{1}{5}, \frac{26}{5}\right)$ .

Final answer

(-1/2, -5/2) and (1/5, 26/5)

Detailed explanation

Walkthrough

We are told the curve is $2x^2 - kxy + 2 = 0$ and the line is $y = px + 3$ . In part (a), the constants are fixed as $k = 2$ and $p = 11$ . To find where they meet, we substitute the line's expression for $y$ into the curve, because any point of intersection must satisfy both equations simultaneously. This turns the two equations into a single quadratic in $x$ .

After substituting, expanding and simplifying gives a quadratic equation. We factorise it to find the two $x$ -values, then use the line equation to get the associated $y$ -values. The coordinates are then the two pairs found.

Key Takeaways

The intersection of a line and a curve is found by eliminating one variable, usually by substituting the line into the curve.
Solving the resulting quadratic gives the $x$ -coordinates of the intersections.
Once $x$ is known, the line equation is the easiest way to find $y$ .
This is an example of a simultaneous equation with one linear and one quadratic equation.

Common Mistakes

Forgetting to multiply the $x(11x+3)$ term by the coefficient $k=2$ , leading to $2x^2 - 11x^2 - 3x + 2$ .
Sign errors when expanding $2x^2 - 2x(11x+3)$ .
Finding the $x$ -values but not substituting back to find $y$ , so losing the coordinate answer.
Not simplifying fractions such as $\frac{26}{5}$ (though here it is already simplified); the mark scheme requires simplified fractions.

Things to Be Careful About

Keep the algebra balanced: every transformation must preserve the equation.
Check that both $x$ -values correspond to points on the line; substitutions can be verified in the original curve.
The final coordinates should be written clearly: $\left(-\frac{1}{2}, -\frac{5}{2}\right)$ and $\left(\frac{1}{5}, \frac{26}{5}\right)$ .
The mark scheme awards A1 for either both $x$ -values or both coordinates of one point, but a complete solution should give all four coordinates.

Techniques used

substitute the line equation into the curveexpand and simplify a quadraticfactorise a quadratic expressioncompute corresponding y-values from the line

(b)

Given instead that $p = 4$ , find the set of values of $k$ for which the curve and the line do not intersect.

5M Similar

DifficultyMedium

Worked solution

Approach

Substitute $y = 4x+3$ into the curve, keeping $k$ as an unknown constant. This produces a quadratic equation in $x$ . The line and curve do not intersect exactly when this quadratic has no real roots, so its discriminant must be negative. Set up the discriminant inequality in $k$ , simplify, factorise and solve for the required range.

Working

The line is $y=4x+3$ . Substituting into $2x^2 - kxy + 2 = 0$ :

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

Expanding:

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

Group the $x^2$ -terms:

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

For no intersection, this quadratic in $x$ must have no real roots, so its discriminant must be negative. Here

a = 2-4k, \quad b = -3k, \quad c = 2

Thus

\begin{aligned} b^2 - 4ac &= (-3k)^2 - 4(2-4k)(2) \\ &= 9k^2 - 8(2-4k) \\ &= 9k^2 + 32k - 16 \end{aligned}

We need

9k^2 + 32k - 16 < 0

Factorise the quadratic:

(k+4)(9k-4) < 0

The critical values are $k = -4$ and $k = \frac{4}{9}$ . Since the coefficient of $k^2$ is positive, the quadratic expression is negative between its two roots:

-4 < k < \frac{4}{9}

Answer

For $p = 4$ , the curve and line do not intersect when $-4 < k < \frac{4}{9}$ .

Final answer

-4 < k < 4/9

Detailed explanation

Walkthrough

We now keep $k$ unknown and set $p = 4$ . Substituting $y = 4x+3$ into the curve gives a quadratic in $x$ whose coefficients depend on $k$ . A quadratic equation $ax^2+bx+c=0$ has no real roots when $b^2-4ac < 0$ . Intuitively, if there is no real $x$ satisfying the substituted equation, the line never meets the curve.

We identify $a = 2-4k$ , $b=-3k$ and $c=2$ , compute the discriminant, and require it to be negative. This gives a quadratic inequality in $k$ . After simplifying and factorising, we find the roots $k=-4$ and $k=\frac{4}{9}$ . Because the graph of $9k^2+32k-16$ is a U-shaped parabola, it is negative between the roots. Hence the answer is $-4<k<\frac{4}{9}$ .

Key Takeaways

Solving simultaneous equations by substitution can lead to a quadratic with parameters.
The discriminant $b^2-4ac$ determines the number of intersections: negative discriminant means no real intersections.
A quadratic inequality is solved by factorising and using the sign of the parabola.
The parameter values make the question harder, but the principle is the same as a numerical case.

Common Mistakes

Omitting the $(2-4k)$ coefficient when computing $4ac$ ; the mark scheme specifically requires the $a$ term to have two components.
Forgetting the sign of the discriminant: for no intersection it is $<0$ , not $=0$ or $>0$ .
Solving $9k^2+32k-16=0$ incorrectly or failing to factorise.
Giving $k \le -4$ or $k \ge \frac{4}{9}$ by choosing the wrong region of the quadratic inequality.
The mark scheme has a special case (SC B1) if no method is shown for solving the quadratic; always show the factorisation.

Things to Be Careful About

If $2-4k=0$ , the quadratic would become linear, but this value $k=\frac{1}{2}$ is not in the final interval, so it does not cause an issue here.
The mark scheme says A0 if the correct answer follows an incorrect quadratic, so make sure the quadratic is correct before using the discriminant.
Check the inequality direction: the leading coefficient of the quadratic in $k$ is positive, so the expression is negative between the roots.
Use interval notation or inequality notation consistently; the mark scheme accepts $k$ and allows other correct notation.

Techniques used

substitute the line equation into the curveform the discriminant of a quadraticrequire the discriminant to be negativefactorise and solve a quadratic inequality

Q7MediumDifferentiation

The equation of a curve is $y = 4x^2 + \frac{9}{x^2} - 8$ .

(a)

A point $P$ is moving along the curve in such a way that its $y$ -coordinate is decreasing at 5 units per second.

Find the rate at which the $x$ -coordinate of point $P$ is changing when $x = 2$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Differentiate $y$ with respect to $x$ , evaluate at $x = 2$ , then use the chain rule to connect the given rate $\frac{dy}{dt}$ to the required rate $\frac{dx}{dt}$ .

Working

\begin{aligned} y &= 4x^2 + \frac{9}{x^2} - 8 \\ &= 4x^2 + 9x^{-2} - 8 \end{aligned}

Differentiate:

\frac{dy}{dx} = 8x - 18x^{-3} = 8x - \frac{18}{x^3}

At $x = 2$ :

\frac{dy}{dx} = 16 - \frac{18}{8} = 16 - \frac{9}{4} = \frac{55}{4}

The $y$ -coordinate is decreasing at 5 units per second, so $\frac{dy}{dt} = -5$ . Using the chain rule:

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

-5 = \frac{55}{4} \cdot \frac{dx}{dt}

\frac{dx}{dt} = -5 \cdot \frac{4}{55} = -\frac{20}{55} = -\frac{4}{11}

Answer

The $x$ -coordinate is changing at $-\frac{4}{11}$ units per second, so it is decreasing at $\frac{4}{11}$ units per second.

Final answer

dx/dt = -4/11 units per second (decreasing at 4/11 units per second)

Detailed explanation

Walkthrough

We are told the $y$ -coordinate is decreasing at 5 units per second. Decreasing means the rate is negative, so $\frac{dy}{dt} = -5$ . The curve gives $y$ in terms of $x$ , so to connect $\frac{dy}{dt}$ and $\frac{dx}{dt}$ we differentiate $y$ with respect to $x$ first.

Rewrite $\frac{9}{x^2}$ as $9x^{-2}$ so the power rule applies cleanly: $4x^2 \to 8x$ and $9x^{-2} \to -18x^{-3}$ . Therefore $\frac{dy}{dx} = 8x - \frac{18}{x^3}$ .

At $x = 2$ , substitute: $8(2) - \frac{18}{8} = 16 - \frac{9}{4} = \frac{55}{4}$ . This is the slope of the curve at that point.

The chain rule says $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$ . Substitute the known values: $-5 = \frac{55}{4} \cdot \frac{dx}{dt}$ . Dividing by $\frac{55}{4}$ gives $\frac{dx}{dt} = -5 \cdot \frac{4}{55} = -\frac{4}{11}$ . The negative sign means the $x$ -coordinate is also decreasing, at $\frac{4}{11}$ units per second.

Key Takeaways

This problem combines differentiation with the chain rule for related rates. The sign of a rate matters: decreasing means negative. It also shows the importance of rewriting negative powers before differentiating.

Common Mistakes

Using $\frac{dy}{dt} = +5$ instead of $-5$ .
Forgetting to take the reciprocal when solving $\frac{dx}{dt} = \frac{dy}{dt} \div \frac{dy}{dx}$ .
Differentiating $\frac{9}{x^2}$ incorrectly, e.g. writing $+\frac{18}{x^3}$ instead of $-\frac{18}{x^3}$ .
Giving $\frac{dy}{dx}$ as the final answer instead of $\frac{dx}{dt}$ .

Things to Be Careful About

The mark scheme awards the derivative B1 even if unsimplified. Evaluating at $x = 2$ gains M1. The chain rule step with $\pm 5$ and their $\frac{dy}{dx}$ gains M1. The final answer must be $\frac{dx}{dt} = -\frac{4}{11}$ or "decreasing at $\frac{4}{11}$ units per second"; $-0.364$ is also accepted. Always include the negative sign or state decreasing.

Techniques used

differentiate y with respect to xevaluate the derivative at x = 2apply the chain rule for related rates

(b)

Find the coordinates of the stationary points of the curve and determine their nature.

5M Similar

DifficultyMedium

Worked solution

Approach

Set $\frac{dy}{dx} = 0$ to find the $x$ -coordinates of stationary points, substitute to find $y$ , then use the second derivative to determine their nature.

Working

From part (a),

\frac{dy}{dx} = 8x - \frac{18}{x^3}

Set it equal to zero:

8x - \frac{18}{x^3} = 0

8x = \frac{18}{x^3}

8x^4 = 18

x^4 = \frac{9}{4}

x = \pm \sqrt{\frac{3}{2}} = \pm \frac{\sqrt{6}}{2}

For both values, $x^2 = \frac{3}{2}$ , so:

\begin{aligned} y &= 4\left(\frac{3}{2}\right) + \frac{9}{\frac{3}{2}} - 8 \\ &= 6 + 6 - 8 \\ &= 4 \end{aligned}

Now find the second derivative:

\frac{d^2y}{dx^2} = 8 + \frac{54}{x^4}

At the stationary points, $x^4 = \frac{9}{4}$ , so:

\frac{d^2y}{dx^2} = 8 + \frac{54}{\frac{9}{4}} = 8 + 24 = 32 > 0

Since the second derivative is positive at both stationary points, both are minima.

Answer

The stationary points are $\left(\sqrt{\frac{3}{2}}, 4\right)$ and $\left(-\sqrt{\frac{3}{2}}, 4\right)$ , and both are minima.

Final answer

(sqrt(3/2), 4) and (-sqrt(3/2), 4); both minima

Detailed explanation

Walkthrough

Stationary points occur where the gradient is zero, so start from $\frac{dy}{dx} = 8x - \frac{18}{x^3}$ and set it equal to zero.

Solve $8x - \frac{18}{x^3} = 0$ . Multiply through by $x^3$ : $8x^4 - 18 = 0$ , so $x^4 = \frac{9}{4}$ . Taking the fourth root gives two real values, $x = \pm \sqrt{\frac{3}{2}}$ , because the fourth root of a positive number has both a positive and a negative value. These are also $\pm \frac{\sqrt{6}}{2}$ .

To find $y$ , notice $x^2 = \frac{3}{2}$ for both values. Substitute: $4\left(\frac{3}{2}\right) + \frac{9}{\frac{3}{2}} - 8 = 6 + 6 - 8 = 4$ . So both stationary points have the same $y$ -coordinate, $4$ .

To determine nature, differentiate $\frac{dy}{dx}$ : $\frac{d^2y}{dx^2} = 8 + \frac{54}{x^4}$ . At the stationary points, $x^4 = \frac{9}{4}$ , so $\frac{d^2y}{dx^2} = 8 + 54 \cdot \frac{4}{9} = 8 + 24 = 32$ , which is positive. A positive second derivative means the curve is locally concave up, so both points are minima.

Key Takeaways

Stationary points are found by solving $\frac{dy}{dx} = 0$ . Solving an equation of the form $x^4 = k$ gives both positive and negative roots. The second derivative test uses the sign of $\frac{d^2y}{dx^2}$ : positive means minimum, negative means maximum.

Common Mistakes

Forgetting the $\pm$ when taking the fourth root, giving only one stationary point.
Substituting $x$ instead of $x^2$ when finding $y$ , or making an arithmetic slip with $\frac{9}{x^2}$ .
Differentiating $\frac{dy}{dx}$ incorrectly, especially the sign of the $\frac{54}{x^4}$ term.
Concluding maximum instead of minimum because of a sign error in the second derivative.

Things to Be Careful About

$x = 0$ is not in the domain of the curve, so multiplying by $x^3$ does not introduce an extra solution. Since $x^4 = \frac{9}{4}$ is positive, the second derivative is positive at both stationary points, so both are minima. The mark scheme accepts $x = \pm \sqrt{\frac{3}{2}}$ or $\pm \frac{\sqrt{6}}{2}$ , and $y = 4$ for both. For the second derivative, at least one correct term is needed for M1; the final nature mark is awarded for both minima.

Techniques used

set the derivative equal to zerosolve the resulting x^4 equationsubstitute to find the y-coordinatesuse the second derivative test to determine nature

Q88MMedium-HardCoordinate Geometry

The circle with equation $x^2 + y^2 - 6x + 10y - 27 = 0$ intersects the line $x = -2$ at the points $P$ and $Q$ .

Find the area of the triangle formed by the tangents to the circle at $P$ and $Q$ , and the line $x = -2$ .

DifficultyMedium-Hard

Worked solution

Approach

Complete the square to find the centre of the circle. Substitute the line $x=-2$ to find the points $P$ and $Q$ . Since each tangent is perpendicular to its radius, use the gradients of the radii to write the tangent equations. Find where the two tangents meet; then take $PQ$ as the base and the horizontal distance from the third vertex to $x=-2$ as the height.

Working

Centre of the circle. Rewrite the circle equation:

x^2 - 6x + y^2 + 10y - 27 = 0

Complete the square:

(x-3)^2 - 9 + (y+5)^2 - 25 - 27 = 0

(x-3)^2 + (y+5)^2 = 61

So the centre is $C(3, -5)$ .

Intersection points with $x=-2$ . Substitute $x=-2$ :

4 + y^2 + 12 + 10y - 27 = 0

y^2 + 10y - 11 = 0

(y - 1)(y + 11) = 0

Hence $y = 1$ and $y = -11$ . Taking $P(-2,1)$ and $Q(-2,-11)$ :

PQ = 1 - (-11) = 12

Tangent at $P$ . The gradient of radius $PC$ is

\frac{-5 - 1}{3 - (-2)} = -\frac{6}{5}

so the tangent gradient is the negative reciprocal, $\frac{5}{6}$ :

y - 1 = \frac{5}{6}(x + 2)

y = \frac{5}{6}x + \frac{8}{3}

Tangent at $Q$ . The gradient of radius $QC$ is

\frac{-5 - (-11)}{3 - (-2)} = \frac{6}{5}

so the tangent gradient is $-\frac{5}{6}$ :

y + 11 = -\frac{5}{6}(x + 2)

y = -\frac{5}{6}x - \frac{38}{3}

Intersection of the two tangents. Equate the two expressions for $y$ :

\frac{5}{6}x + \frac{8}{3} = -\frac{5}{6}x - \frac{38}{3}

\frac{10}{6}x = -\frac{46}{3}

\frac{5}{3}x = -\frac{46}{3}

x = -\frac{46}{5}

The horizontal distance from this point to the line $x=-2$ is

-2 - \left(-\frac{46}{5}\right) = \frac{36}{5}

Area of the triangle.

\text{Area} = \frac{1}{2} \times 12 \times \frac{36}{5} = \frac{216}{5}

Answer

\frac{216}{5} \quad \text{or} \quad 43.2

Final answer

216/5 or 43.2

Detailed explanation

Walkthrough

Start by putting the circle equation into centre-radius form. Completing the square on $x^2 - 6x$ and $y^2 + 10y$ shows the centre is $(3,-5)$ and the radius is $\sqrt{61}$ . This centre is essential because every tangent is perpendicular to the radius drawn to its point of contact.

Next, the line $x = -2$ cuts the circle where its $y$ -coordinates satisfy the quadratic obtained by substituting $x=-2$ into the circle equation. The quadratic factors cleanly, giving $y=1$ and $y=-11$ , so the two points are $P(-2,1)$ and $Q(-2,-11)$ . Their vertical separation is the base of the required triangle, $12$ .

For the tangent at $P$ , compute the gradient of the radius joining the centre $C(3,-5)$ to $P(-2,1)$ . It is $-6/5$ . A tangent is perpendicular to the radius, so its gradient is the negative reciprocal, $5/6$ . Using the point-slope form gives the tangent equation. The same process at $Q$ gives a radius gradient of $6/5$ and therefore a tangent gradient of $-5/6$ .

The two tangent lines form the two slanted sides of the triangle. Their intersection is the third vertex. Solving the two tangent equations together gives $x=-46/5$ , and substituting back gives $y=-5$ , so the third vertex lies on the horizontal line through the centre. The height of the triangle is the perpendicular distance from this vertex to the base line $x=-2$ , which is $36/5$ . Finally, use $\text{Area} = \frac12 \times \text{base} \times \text{height}$ : $\frac12(12)\left(\frac{36}{5}\right) = \frac{216}{5}$ .

Key Takeaways

This question brings together several coordinate geometry ideas: completing the square to read the centre of a circle, intersecting an algebraic curve with a line, using the perpendicular relationship between a radius and a tangent, writing straight-line equations in point-slope form, and computing the area of a triangle from its base and perpendicular height.

When a line is vertical, the perpendicular distance from a point to the line is simply the horizontal difference between the $x$ -coordinates. Recognising that $PQ$ lies on the vertical line $x=-2$ makes the base and height easy to identify.

Common Mistakes

The mark scheme requires the centre to be seen or implied; forgetting to complete the square correctly gives the wrong centre and wrong tangent gradients. It also requires the quadratic from substituting $x=-2$ to be formed; leaving it unsimplified can hide the factorisation and lose the method mark. Students often use the gradient of the tangent itself as the radius gradient, so be careful to take the negative reciprocal. A common error is to use the intersection distance to the wrong vertical line or to treat the $46/5$ horizontal distance as the height. The mark scheme condones using $46/5$ only in the area formula with an accompanying error, but the correct perpendicular height is $36/5$ .

Things to Be Careful About

When substituting $x=-2$ , every term must be evaluated carefully: $(-2)^2=4$ and $-6x=12$ . The centre comes from completing the square with constants $-9$ and $-25$ ; the radius is $\sqrt{61}$ , although it is not needed for the area. The tangent at $P$ and the tangent at $Q$ have opposite gradients, so their equations are symmetric in structure. Check that the intersection point is to the left of $x=-2$ : $x=-46/5$ is less than $-2$ , so the height is $\frac{36}{5}$ not $\frac{46}{5}$ . Finally, give the area either as a fraction or a decimal; the mark scheme accepts $\frac{216}{5}$ or $43.2$ .

Techniques used

complete the square to identify the centre of a circlesubstitute a vertical line into a circle equation to find intersection pointsuse the perpendicular gradient of a radius to write tangent equationssolve two tangent equations simultaneouslyfind triangle area from base and perpendicular height

Approach

Use the formula for the area of a sector, $\frac{1}{2}r^2\alpha$ , to find $r$ . Then use the formula for the area of a triangle, $\frac{1}{2}r^2\sin\alpha$ , to find $\alpha$ . The shaded segment area is the sector area minus the triangle area.

Working

Step 1: Find $r$ using the sector area.

The area of sector $ABC$ is given as $8\alpha$ cm $^2$ . Using the sector area formula:

\frac{1}{2}r^2\alpha = 8\alpha

Since $\alpha > 0$ , divide both sides by $\alpha$ :

\frac{1}{2}r^2 = 8

r^2 = 16

r = 4 \text{ cm}

Step 2: Find $\alpha$ using the triangle area.

The area of triangle $ABC$ is given as $4$ cm $^2$ . Using the triangle area formula with two sides and the included angle:

\frac{1}{2}r^2\sin\alpha = 4

Substitute $r = 4$ :

\frac{1}{2}(4)^2\sin\alpha = 4

8\sin\alpha = 4

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

Since $0 < \alpha < \frac{\pi}{2}$ :

\alpha = \frac{\pi}{6}

Step 3: Calculate the area of the shaded segment.

The shaded segment is the region between the chord $BC$ and the arc $BC$ . Its area equals the sector area minus the triangle area:

\text{Area of segment} = \text{Area of sector} - \text{Area of triangle}

= \frac{1}{2} \times 4^2 \times \frac{\pi}{6} - 4

= \frac{1}{2} \times 16 \times \frac{\pi}{6} - 4

= \frac{8\pi}{6} - 4

= \frac{4\pi}{3} - 4

Answer

\frac{4\pi}{3} - 4 \text{ cm}^2

Final answer

(4π/3 - 4) cm²

Detailed explanation

Walkthrough

Step 1: The sector area formula is $\frac{1}{2}r^2\alpha$ . We are told this equals $8\alpha$ . Since $\alpha$ is positive (given $0 < \alpha < \frac{\pi}{2}$ ), we can safely divide both sides by $\alpha$ , leaving $\frac{1}{2}r^2 = 8$ , which gives $r = 4$ .

Step 2: The triangle area formula with two sides and the included angle is $\frac{1}{2}ab\sin C$ . Here both sides are $r = 4$ and the included angle is $\alpha$ , so the area is $\frac{1}{2}r^2\sin\alpha = 8\sin\alpha$ . Setting this equal to $4$ gives $\sin\alpha = \frac{1}{2}$ . The only solution in $(0, \frac{\pi}{2})$ is $\alpha = \frac{\pi}{6}$ .

Step 3: The shaded segment is the part of the sector that lies outside the triangle. So we subtract: sector area $8\alpha = 8 \times \frac{\pi}{6} = \frac{4\pi}{3}$ , minus triangle area $4$ , giving $\frac{4\pi}{3} - 4$ .

Key Takeaways

The sector area formula $\frac{1}{2}r^2\alpha$ can be used to find the radius when the area is given in terms of $\alpha$ .
The triangle area formula $\frac{1}{2}r^2\sin\alpha$ links the radius, angle, and area of a triangle formed by two radii and a chord.
The area of a circular segment equals the sector area minus the triangle area.

Common Mistakes

Forgetting to divide by $\alpha$ and trying to solve $\frac{1}{2}r^2\alpha = 8\alpha$ as a quadratic in $r$ with $\alpha$ still present.
Using the wrong triangle area formula (e.g., $\frac{1}{2} \times \text{base} \times \text{height}$ without finding the height first).
Not simplifying the fraction $\frac{8\pi}{6}$ to $\frac{4\pi}{3}$ .
Subtracting in the wrong order (triangle minus sector instead of sector minus triangle), which would give a negative answer.

Things to Be Careful About

The constraint $0 < \alpha < \frac{\pi}{2}$ ensures $\sin\alpha = \frac{1}{2}$ has only one solution: $\alpha = \frac{\pi}{6}$ . Without this constraint, $\alpha = \frac{5\pi}{6}$ would also be valid.
The answer must be exact, so leave $\pi$ in the answer rather than computing a decimal.
The fraction $\frac{4\pi}{3} - 4$ can also be written as $\frac{4\pi - 12}{3}$ , which is an equivalent form.

Techniques used

apply sector area formulaapply triangle area formula with sinesolve trigonometric equation for anglesubtract triangle area from sector area

(b)

It is given instead that the length of the chord $BC$ is $\frac{1}{\sqrt{2}}r\text{ cm}$ but the area of the triangle $ABC$ is still $4\text{ cm}^2$ .

Find the area of the shaded segment. Give your answer correct to 3 significant figures.

4M Similar

DifficultyMedium

Worked solution

Approach

Use the cosine rule in triangle $ABC$ to find $\cos\alpha$ from the given chord length. Then find $\alpha$ and use the triangle area formula to find $r$ . Finally, compute the segment area as sector area minus triangle area.

Working

Step 1: Find $\cos\alpha$ using the cosine rule.

In triangle $ABC$ , the sides are $AB = r$ , $AC = r$ , and $BC = \frac{1}{\sqrt{2}}r$ . Applying the cosine rule:

BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos\alpha

\left(\frac{1}{\sqrt{2}}r\right)^2 = r^2 + r^2 - 2r^2\cos\alpha

\frac{r^2}{2} = 2r^2 - 2r^2\cos\alpha

Divide through by $r^2$ (since $r > 0$ ):

\frac{1}{2} = 2 - 2\cos\alpha

2\cos\alpha = 2 - \frac{1}{2} = \frac{3}{2}

\cos\alpha = \frac{3}{4}

Step 2: Find $\alpha$ .

\alpha = \cos^{-1}\left(\frac{3}{4}\right) \approx 0.7227 \text{ rad}

(This is approximately $41.4°$ .)

Step 3: Find $r$ using the triangle area.

The area of triangle $ABC$ is $4$ cm $^2$ . Using the formula:

\frac{1}{2}r^2\sin\alpha = 4

First find $\sin\alpha$ . Since $\cos\alpha = \frac{3}{4}$ and $0 < \alpha < \frac{\pi}{2}$ :

\sin\alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}

Substitute:

\frac{1}{2}r^2 \cdot \frac{\sqrt{7}}{4} = 4

\frac{r^2\sqrt{7}}{8} = 4

r^2 = \frac{32}{\sqrt{7}} \approx 12.096

r \approx 3.478 \text{ cm}

Step 4: Calculate the area of the shaded segment.

\text{Area of segment} = \frac{1}{2}r^2\alpha - 4

= \frac{1}{2} \times \frac{32}{\sqrt{7}} \times 0.7227 - 4

= \frac{16 \times 0.7227}{\sqrt{7}} - 4

\approx \frac{11.563}{2.6458} - 4

\approx 4.371 - 4

\approx 0.371 \text{ cm}^2

Answer

0.371 \text{ cm}^2

Final answer

0.371 cm²

Detailed explanation

Walkthrough

Step 1: The cosine rule states $c^2 = a^2 + b^2 - 2ab\cos C$ . Here, the side opposite angle $\alpha$ is the chord $BC = \frac{r}{\sqrt{2}}$ , and the two adjacent sides are both $r$ . Substituting gives $\frac{r^2}{2} = 2r^2 - 2r^2\cos\alpha$ . Dividing by $r^2$ and rearranging yields $\cos\alpha = \frac{3}{4}$ .

Step 2: Since $0 < \alpha < \frac{\pi}{2}$ , we can find $\alpha = \cos^{-1}(0.75) \approx 0.7227$ rad. We keep extra digits during intermediate calculations to avoid rounding errors.

Step 3: The triangle area is $\frac{1}{2}r^2\sin\alpha = 4$ . We compute $\sin\alpha = \sqrt{1 - (3/4)^2} = \sqrt{7}/4$ . Then $\frac{1}{2}r^2 \cdot \frac{\sqrt{7}}{4} = 4$ gives $r^2 = 32/\sqrt{7} \approx 12.096$ , so $r \approx 3.478$ .

Step 4: The segment area is sector area minus triangle area: $\frac{1}{2}r^2\alpha - 4 \approx \frac{1}{2}(12.096)(0.7227) - 4 \approx 4.371 - 4 = 0.371$ cm $^2$ .

Key Takeaways

The cosine rule can be applied to a triangle formed by two radii and a chord to relate the chord length to the central angle.
When $\cos\alpha$ is known, $\sin\alpha$ can be found using $\sin^2\alpha + \cos^2\alpha = 1$ .
The segment area is always sector area minus triangle area, regardless of whether values are exact or approximate.
Carrying extra digits in intermediate steps is essential when the final answer must be given to a specific number of significant figures.

Common Mistakes

Using $\alpha = \frac{\pi}{6}$ or $r = 4$ from part (a) — these values do not apply here since the conditions are different.
Forgetting to square $\frac{1}{\sqrt{2}}r$ correctly: $\left(\frac{r}{\sqrt{2}}\right)^2 = \frac{r^2}{2}$ , not $r^2$ or $\frac{r}{2}$ .
Rounding $\alpha$ or $r$ too early, which can lead to an incorrect final answer.
Using degrees instead of radians in the sector area formula $\frac{1}{2}r^2\alpha$ — the angle must be in radians.

Things to Be Careful About

The sector area formula $\frac{1}{2}r^2\alpha$ requires $\alpha$ in radians. If you compute $\alpha$ in degrees ( $41.4°$ ), you must convert: $\frac{1}{2}r^2 \times \frac{\pi}{180} \times 41.4$ .
The answer is required to 3 significant figures, so $0.371$ is correct. Do not write $0.37$ or $0.3707$ .
The triangle area is still $4$ cm $^2$ as stated in the question, so you do not need to recalculate it — it is given directly.

Techniques used

apply cosine rule to find anglesolve trigonometric equation for angleuse triangle area formula to find radiussubtract triangle area from sector area

Q10MediumFunctions

The functions $f$ and $g$ are defined by

\begin{aligned} f(x) &= \sqrt{x} && \text{for } x \ge 0, \\ g(x) &= 3\sqrt{x+2} - 5 && \text{for } x \ge -2. \end{aligned}

(a)

Describe fully a sequence of transformations which transforms the graph of $y = f(x)$ to the graph of $y = g(x)$ . You should make clear the order in which the transformations are applied.

5M Similar

DifficultyMedium

Worked solution

Approach

We compare the definitions of $f(x)$ and $g(x)$ to identify the transformations. Starting from $f(x) = \sqrt{x}$ , we build up to $g(x) = 3\sqrt{x+2} - 5$ step by step.

Working

The function $g(x)$ can be written in terms of $f$ as:

\begin{aligned} g(x) &= 3\sqrt{x+2} - 5 \\ &= 3f(x+2) - 5 \end{aligned}

This expression tells us the sequence of transformations:

Vertical stretch: The factor of $3$ multiplying the function indicates a stretch by scale factor $3$ parallel to the $y$ -axis (or in the $y$ -direction). Applying this to $f(x)$ gives $y = 3f(x)$ .
Translation: The terms $x+2$ inside the function and $-5$ outside indicate a translation. The $x+2$ shifts the graph $2$ units to the left, and the $-5$ shifts it $5$ units down. This corresponds to a translation vector of $\begin{pmatrix} -2 \\ -5 \end{pmatrix}$ .

Applying the translation after the stretch:

\begin{aligned} y &= 3f(x) \\ \xrightarrow{\text{translate } \begin{pmatrix} -2 \\ -5 \end{pmatrix}} y &= 3f(x+2) - 5 = 3\sqrt{x+2} - 5 = g(x) \end{aligned}

Alternative sequence:

Translation by $\begin{pmatrix} -2 \\ -\frac{5}{3} \end{pmatrix}$ : $y = f(x+2) - \frac{5}{3}$
Vertical stretch by scale factor $3$ : $y = 3\left(f(x+2) - \frac{5}{3}\right) = 3f(x+2) - 5$

Both sequences are valid as long as the order is consistent with the algebra.

Answer

A stretch by scale factor $3$ parallel to the $y$ -axis, followed by a translation by $\begin{pmatrix} -2 \\ -5 \end{pmatrix}$ .

Final answer

Stretch by scale factor 3 parallel to the y-axis, then translation by (-2, -5)

Detailed explanation

Walkthrough

To find the transformations, we express $g(x)$ in terms of $f(x)$ . Since $f(x) = \sqrt{x}$ , we see that $g(x) = 3\sqrt{x+2} - 5 = 3f(x+2) - 5$ .

The algebraic form $3f(x+2) - 5$ reveals two operations:

The multiplication by $3$ outside the function is a vertical stretch with scale factor $3$ .
The $+2$ inside the argument is a horizontal shift left by $2$ , and the $-5$ outside is a vertical shift down by $5$ . Together, this is a translation by the vector $\begin{pmatrix} -2 \\ -5 \end{pmatrix}$ .

Order matters: if we translate first by $\begin{pmatrix} -2 \\ -5 \end{pmatrix}$ , we get $f(x+2) - 5$ . Stretching this vertically by $3$ gives $3(f(x+2) - 5) = 3f(x+2) - 15$ , which is incorrect. Therefore, the stretch must come first, or the translation must be adjusted to $\begin{pmatrix} -2 \\ -5/3 \end{pmatrix}$ if applied before the stretch.

Key Takeaways

Transformations applied outside the function (multiplication/addition) affect the $y$ -coordinates (vertical stretch/shift).
Transformations applied inside the function (addition/subtraction to $x$ ) affect the $x$ -coordinates (horizontal shift).
The order of transformations matters; a vertical stretch after a vertical translation will scale the translation distance as well.

Common Mistakes

Applying the translation before the stretch without adjusting the vertical component of the translation vector.
Confusing the direction of horizontal shifts (e.g., thinking $x+2$ means a shift right by $2$ ).

Things to Be Careful About

Always check the order: if a vertical stretch is applied after a vertical translation, the translation amount is also stretched. The mark scheme accepts either order as long as the translation vector is adjusted accordingly.
Specify the direction of the stretch clearly (e.g., "parallel to the $y$ -axis" or "vertically").

Techniques used

identify vertical stretchidentify translation vectordetermine correct order of transformations

(b)

Fig. 10.1

The diagram shows the graph of $y = g(x)$ .

On the diagram sketch the graph of $y = g^{-1}(x)$ together with any relevant mirror line.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The graph of $y = g^{-1}(x)$ is the reflection of the graph of $y = g(x)$ in the line $y = x$ . We identify key points on $g(x)$ and reflect them to sketch $g^{-1}(x)$ .

Working

The graph of $y = g(x) = 3\sqrt{x+2} - 5$ has its starting point (endpoint) at $x = -2$ :

g(-2) = 3\sqrt{-2+2} - 5 = -5

So the graph starts at $(-2, -5)$ . It passes through other points such as $(2, 1)$ and $(7, 4)$ .

To sketch $y = g^{-1}(x)$ :

Draw the mirror line $y = x$ , which is a straight line through the origin at $45^\circ$ .
Reflect the starting point $(-2, -5)$ across $y = x$ to get $(-5, -2)$ . This is the starting point of $g^{-1}(x)$ .
Reflect $(2, 1)$ to get $(1, 2)$ , and $(7, 4)$ to get $(4, 7)$ .
Draw a smooth curve through these reflected points. The curve should be increasing and concave down (like a sideways square root), starting at $(-5, -2)$ and passing through the first quadrant.

Answer

Sketch showing $y = g^{-1}(x)$ as the reflection of $y = g(x)$ in the line $y = x$ , starting at $(-5, -2)$ and not passing through the origin.

Final answer

Sketch of y = g^{-1}(x) reflected in y = x, starting at (-5, -2)

Detailed explanation

Walkthrough

The fundamental property of inverse functions is that their graphs are mirror images of each other across the line $y = x$ . This is because if $(a, b)$ is on the graph of $y = g(x)$ , then $(b, a)$ must be on the graph of $y = g^{-1}(x)$ .

We identify the endpoint of $g(x)$ : since the domain is $x \ge -2$ , the minimum $x$ -value is $-2$ , giving $g(-2) = -5$ . The point $(-2, -5)$ reflects to $(-5, -2)$ , which is the starting point of $g^{-1}(x)$ .

We also note that $g(x)$ passes through $(2, 1)$ because $g(2) = 3\sqrt{4} - 5 = 1$ . This reflects to $(1, 2)$ on $g^{-1}(x)$ .

The curve $g^{-1}(x)$ is a reflection of $g(x)$ , so it will have the same general shape but oriented differently. Since $g(x)$ is a square root curve opening to the right, $g^{-1}(x)$ will be a square root curve opening upwards.

Key Takeaways

The graph of $f^{-1}$ is always the reflection of $f$ in the line $y = x$ .
The domain of $f$ becomes the range of $f^{-1}$ , and the range of $f$ becomes the domain of $f^{-1}$ .
When sketching, reflect key points (endpoints, intercepts) across $y = x$ to guide the curve.

Common Mistakes

Drawing the inverse curve passing through the origin $(0, 0)$ , which is incorrect since $g(0) = 3\sqrt{2} - 5 \neq 0$ .
Forgetting to draw the mirror line $y = x$ .
Drawing the inverse curve with the wrong shape (e.g., making it concave up instead of concave down).

Things to Be Careful About

The sketch must be in the correct quadrants. Since $g(x)$ starts in the third quadrant and goes into the first, $g^{-1}(x)$ starts in the second quadrant (at $(-5, -2)$ ) and goes into the first.
The curve must not come back on itself; it must be a function (pass the vertical line test).
No label is needed on the sketch, but the line $y = x$ should be clearly drawn.

Techniques used

identify mirror line y = xreflect curve across mirror linecheck key points and quadrants

(c)

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

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

To find $g^{-1}(x)$ , we set $y = g(x)$ , swap $x$ and $y$ , and solve for $y$ .

Working

Let $y = 3\sqrt{x+2} - 5$ .

Isolate the square root term:

\begin{aligned} y + 5 &= 3\sqrt{x+2} \\ \frac{y + 5}{3} &= \sqrt{x+2} \end{aligned}

Square both sides to eliminate the square root:

\left(\frac{y + 5}{3}\right)^2 = x + 2

Solve for $x$ :

x = \left(\frac{y + 5}{3}\right)^2 - 2

Now swap $x$ and $y$ to express the inverse function:

y = \left(\frac{x + 5}{3}\right)^2 - 2

Therefore:

g^{-1}(x) = \left(\frac{x + 5}{3}\right)^2 - 2

Answer

g^{-1}(x) = \left(\frac{x + 5}{3}\right)^2 - 2

Final answer

g^{-1}(x) = ((x + 5)/3)^2 - 2

Detailed explanation

Walkthrough

Finding the inverse function involves algebraic manipulation to express $x$ in terms of $y$ , then swapping the variables.

Start with $y = 3\sqrt{x+2} - 5$ . The goal is to isolate $x$ .

Add $5$ to both sides: $y + 5 = 3\sqrt{x+2}$ .
Divide by $3$ : $\frac{y+5}{3} = \sqrt{x+2}$ .
Square both sides: $\left(\frac{y+5}{3}\right)^2 = x + 2$ . Note that squaring is valid here because the original function $g(x)$ has range $y \ge -5$ , so $\frac{y+5}{3} \ge 0$ , meaning the square root is non-negative and squaring does not introduce extraneous solutions in this context.
Subtract $2$ : $x = \left(\frac{y+5}{3}\right)^2 - 2$ .

Finally, replace $y$ with $x$ and $x$ with $y$ to write the inverse function in standard form: $g^{-1}(x) = \left(\frac{x+5}{3}\right)^2 - 2$ .

Key Takeaways

To find an inverse, set $y = f(x)$ , solve for $x$ in terms of $y$ , then swap $x$ and $y$ .
Always check that the original function is one-one (or restrict the domain) before finding an inverse.
When squaring both sides, be aware of domain restrictions that may eliminate extraneous solutions.

Common Mistakes

Forgetting to swap $x$ and $y$ at the end, leaving the answer as $x = \left(\frac{y+5}{3}\right)^2 - 2$ .
Writing the answer as $y = \left(\frac{x+5}{3}\right)^2 - 2$ without explicitly stating $g^{-1}(x) = \dots$ .
Algebraic errors when isolating the square root or squaring both sides.

Things to Be Careful About

The final expression must be in terms of $x$ , not $y$ .
The mark scheme requires the answer in the form $g^{-1}(x) = \dots$ , not $x = \dots$ .
Although not required here, the domain of $g^{-1}(x)$ is $x \ge -5$ (the range of $g(x)$ ), and the range is $y \ge -2$ (the domain of $g(x)$ ).

Techniques used

substitute y for g(x)isolate the square root termsquare both sides to solve for xswap x and y to write inverse function

(d)

State the range of $g^{-1}$ .

1M Similar

DifficultyEasy

Worked solution

Approach

The range of $g^{-1}(x)$ is equal to the domain of $g(x)$ .

Working

From the definition of $g(x)$ :

g(x) = 3\sqrt{x+2} - 5 \quad \text{for } x \ge -2

The domain of $g(x)$ is $x \ge -2$ .

Therefore, the range of $g^{-1}(x)$ is:

y \ge -2

This can also be written as $[-2, \infty)$ .

Answer

y \ge -2

Final answer

y >= -2

Detailed explanation

Walkthrough

A fundamental property of inverse functions is that the domain and range are swapped. If $f: A \to B$ , then $f^{-1}: B \to A$ . This means:

Domain of $f^{-1}$ = Range of $f$
Range of $f^{-1}$ = Domain of $f$

Here, $g(x)$ is defined for $x \ge -2$ , so the domain of $g$ is $[-2, \infty)$ . Therefore, the range of $g^{-1}$ is $[-2, \infty)$ , or $y \ge -2$ .

We can verify this using the expression for $g^{-1}(x)$ found in part (c):

g^{-1}(x) = \left(\frac{x+5}{3}\right)^2 - 2

Since $\left(\frac{x+5}{3}\right)^2 \ge 0$ for all real $x$ , the minimum value of $g^{-1}(x)$ is $0 - 2 = -2$ . Thus, $g^{-1}(x) \ge -2$ , confirming the range is $y \ge -2$ .

Key Takeaways

The range of an inverse function is always the domain of the original function.
This is a quick way to find the range of an inverse without analyzing the inverse expression.

Common Mistakes

Confusing domain and range (e.g., stating the range is $x \ge -2$ instead of $y \ge -2$ ).
Using strict inequality $y > -2$ instead of $y \ge -2$ .

Things to Be Careful About

Use $y \ge -2$ or interval notation $[-2, \infty)$ , not $x \ge -2$ .
The mark scheme accepts various notations including $[-2, \infty)$ , but must include $-2$ with a non-strict inequality.

Techniques used

recognize range of inverse equals domain of original function

(e)

The function $h$ is defined by

h(x) = x - 2 \quad \text{for } x \ge 0.

Find the value of $g^{-1}h(4)$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

To find $g^{-1}h(4)$ , we evaluate the inner function $h(4)$ first, then substitute the result into $g^{-1}$ .

Working

The function $h$ is defined by:

h(x) = x - 2 \quad \text{for } x \ge 0

Evaluate $h(4)$ :

h(4) = 4 - 2 = 2

Now find $g^{-1}(2)$ using the expression from part (c):

g^{-1}(x) = \left(\frac{x + 5}{3}\right)^2 - 2

Substitute $x = 2$ :

\begin{aligned} g^{-1}(2) &= \left(\frac{2 + 5}{3}\right)^2 - 2 \\ &= \left(\frac{7}{3}\right)^2 - 2 \\ &= \frac{49}{9} - 2 \\ &= \frac{49}{9} - \frac{18}{9} \\ &= \frac{31}{9} \end{aligned}

Answer

g^{-1}h(4) = \frac{31}{9}

Final answer

31/9

Detailed explanation

Walkthrough

Composite functions are evaluated from the inside out. For $g^{-1}h(4)$ , this means we first compute $h(4)$ , then use that result as the input to $g^{-1}$ .

Step 1: Evaluate $h(4)$ .
Since $h(x) = x - 2$ for $x \ge 0$ , and $4 \ge 0$ is in the domain:
$h(4) = 4 - 2 = 2$ .

Step 2: Evaluate $g^{-1}(2)$ .
Using the inverse function formula from part (c): $g^{-1}(x) = \left(\frac{x+5}{3}\right)^2 - 2$ .
Substitute $x = 2$ : $g^{-1}(2) = \left(\frac{2+5}{3}\right)^2 - 2 = \left(\frac{7}{3}\right)^2 - 2 = \frac{49}{9} - 2 = \frac{49 - 18}{9} = \frac{31}{9}$ .

We can also verify that $x = 2$ is in the domain of $g^{-1}$ . The domain of $g^{-1}$ is the range of $g$ , which is $y \ge -5$ . Since $2 \ge -5$ , this is valid.

Key Takeaways

Composite function notation $f g(x)$ means $f(g(x))$ , evaluated from right to left (inside out).
Always check that the output of the inner function is within the domain of the outer function.

Common Mistakes

Evaluating $g^{-1}(4)$ first instead of $h(4)$ .
Algebraic errors when substituting into the inverse function formula, especially with fractions and squares.
Forgetting to subtract $2$ at the end.

Things to Be Careful About

The notation $g^{-1}h(4)$ means $g^{-1}(h(4))$ , not $g^{-1}(4) \cdot h(4)$ .
Ensure the value $h(4) = 2$ is within the domain of $g^{-1}$ (which is $x \ge -5$ ). It is, so the composition is valid.
The final answer can be given as a fraction $\frac{31}{9}$ or as a decimal $3.44$ (to 3 significant figures), but exact form is preferred.

Techniques used

evaluate inner function firstsubstitute result into outer functionapply inverse function formula

(f)

Explain why the composite function $hg^{-1}$ cannot be formed.

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

For the composite function $hg^{-1}$ to be formed, the range of $g^{-1}$ must be a subset of the domain of $h$ .

Working

The function $h$ is defined by:

h(x) = x - 2 \quad \text{for } x \ge 0

The domain of $h$ is $x \ge 0$ .

From part (d), the range of $g^{-1}(x)$ is $y \ge -2$ .

For $hg^{-1}$ to be defined, every value in the range of $g^{-1}$ must be a valid input for $h$ . However, the range of $g^{-1}$ includes values such as $-2, -1, 0$ , which are not in the domain of $h$ (since $h$ requires $x \ge 0$ ).

Specifically, values in $[-2, 0)$ are in the range of $g^{-1}$ but not in the domain of $h$ .

Therefore, the composite function $hg^{-1}$ cannot be formed because the range of $g^{-1}$ is not entirely within the domain of $h$ .

Answer

The range of $g^{-1}$ is $y \ge -2$ , which is not within the domain of $h$ ( $x \ge 0$ ). Therefore, $hg^{-1}$ cannot be formed.

Final answer

The range of g^{-1} (y >= -2) is not within the domain of h (x >= 0)

Detailed explanation

Walkthrough

A composite function $f(g(x))$ can only be formed if the range of $g$ is a subset of the domain of $f$ . In other words, every output produced by $g$ must be a valid input for $f$ .

Here, we are considering $h(g^{-1}(x))$ :

The inner function is $g^{-1}$ , with range $y \ge -2$ (from part d).
The outer function is $h$ , with domain $x \ge 0$ .

For $hg^{-1}$ to be valid, we need: range of $g^{-1}$ $\subseteq$ domain of $h$ .

Is $y \ge -2$ a subset of $x \ge 0$ ? No, because values like $-2, -1, -0.5$ are in the range of $g^{-1}$ but are not valid inputs for $h$ (since $h$ only accepts $x \ge 0$ ).

Since there are values in the range of $g^{-1}$ that fall outside the domain of $h$ , the composite function $hg^{-1}$ cannot be formed for all $x$ in the domain of $g^{-1}$ . Therefore, it is undefined as a complete function.

Key Takeaways

For $f(g(x))$ to be formed, the range of $g$ must be within the domain of $f$ .
Always check domain and range conditions when dealing with composite functions involving inverses.

Common Mistakes

Stating that the domain of $g^{-1}$ is not within the domain of $h$ (confusing domain with range).
Saying "the domains don't match" without specifying the correct relationship (range of inner vs. domain of outer).

Things to Be Careful About

The key phrase is: "the range of $g^{-1}$ is not within the domain of $h$ ".
Be precise: it's not that they are completely disjoint (they overlap for $x \ge 0$ ), but that the range of $g^{-1}$ extends outside the domain of $h$ (for $-2 \le x < 0$ ).
The mark scheme accepts "the range of $g^{-1}$ is not within the domain of $h$ " as a minimum acceptable explanation.

Techniques used

compare range of inner function with domain of outer function

Mathematics 9709/11 — May/June 2025

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