TMUA practice · 20 questions · Free
TMUA Mock Paper 1: Applications of Mathematical Knowledge
Every question carries a worked solution behind a disclosure, so you commit to an answer before you see the key.
[01] Algebra and functions
Question 1
Evaluate $\left(\frac{27}{8}\right)^{-2/3} + \left(\frac{1}{16}\right)^{-3/4}$.
Reveal the answer and worked solution. Commit to an option first.
Answer: D
Key idea. A negative index inverts the base and a fractional index $\frac{p}{q}$ takes the $q$-th root then the $p$-th power. Invert first: the roots become small integers and nothing large is ever computed.
Fastest approach
Invert each base to clear the negative sign: $\left(\frac{27}{8}\right)^{-2/3} = \left(\frac{8}{27}\right)^{2/3} = \left(\frac{2}{3}\right)^{2} = \frac{4}{9}$
$\left(\frac{1}{16}\right)^{-3/4} = 16^{3/4} = 2^{3} = 8$
Add over a common denominator: $\frac{4}{9} + 8 = \frac{4 + 72}{9} = \frac{76}{9}$
Common mistake. Taking the power before the root, so $27^{2} = 729$ has to be cube-rooted by hand. The answer is the same; the arithmetic is not, and at 225 seconds a question that detour is what runs the clock down.
Why the other options are wrong
- A: Dropping the second term entirely after simplifying it to 8 and failing to add it. Incomplete Calculation
- B: Evaluating $16^{3/4}$ as 4, taking the fourth root and then squaring rather than cubing. Conceptual Misunderstanding
- C: Computing the second term as 8 but the first as $-\frac{4}{9}$, treating the negative index as a negative value. Sign Error
- E: Adding 8 as $\frac{24}{3}$ to $\frac{4}{9}$ without matching denominators. Arithmetic Slip
[02] Quadratic equations and inequalities
Question 2
The equation $kx^{2} + (k+3)x + 4 = 0$ has two distinct real roots. Which of the following describes the complete set of possible values of $k$?
Reveal the answer and worked solution. Commit to an option first.
Answer: A
Key idea. Two distinct real roots needs $b^{2} - 4ac > 0$, but 'the equation is a quadratic' is a second condition. At $k = 0$ the equation collapses to $3x + 4 = 0$, which has one root, not two.
Fastest approach
Discriminant: $(k+3)^{2} - 16k > 0$
$k^{2} + 6k + 9 - 16k > 0$
$k^{2} - 10k + 9 > 0$
$(k-1)(k-9) > 0 \implies k < 1 \text{ or } k > 9$
Now check the leading coefficient. If $k = 0$ the equation is $3x + 4 = 0$: linear, one root. $k = 0$ satisfies the discriminant inequality but not the question, so it must be excluded.
$k < 1$ or $k > 9$, with $k \neq 0$.
Common mistake. Stopping at the discriminant. $k = 0$ sits inside $k < 1$, so the discriminant-only interval is exactly the right answer to a question that was not asked.
Why the other options are wrong
- B: Using $\geqslant$, which admits the repeated-root cases $k = 1$ and $k = 9$ where the roots are not distinct. Boundary Error
- C: Taking only the upper branch of the quadratic inequality. Incomplete Calculation
- D: The discriminant condition alone, with $k = 0$ left in. This is the intended trap. Incomplete Calculation
- E: Solving $(k-1)(k-9) < 0$, the condition for no real roots. Sign Error
[03] Sequences and series
Question 3
An arithmetic progression has first term $5$ and common difference $d$. The sum of the first $20$ terms is equal to the sum of the first $30$ terms. Find $d$.
Reveal the answer and worked solution. Commit to an option first.
Answer: B
Key idea. $S_{20} = S_{30}$ says terms 21 to 30 sum to zero. Either use that directly, or set the two sum formulas equal.
Fastest approach
$S_n = \frac{n}{2}\left(2a + (n-1)d\right)$ with $a = 5$:
$10(10 + 19d) = 15(10 + 29d)$
$100 + 190d = 150 + 435d$
$-50 = 245d$
$d = -\frac{50}{245} = -\frac{10}{49}$
Common mistake. Writing $S_n = \frac{n}{2}(a + l)$ and then guessing at the last term $l$ instead of using $a + (n-1)d$.
Why the other options are wrong
- A: Doubling the numerator, from using $n$ rather than $\frac{n}{2}$ on one side only. Formula Misapplication
- C: Cancelling $\frac{50}{245}$ to $\frac{1}{7}$ by dividing numerator and denominator by different numbers. Arithmetic Slip
- D: Halving the numerator, from mis-cancelling $\frac{50}{245}$. Arithmetic Slip
- E: Correct magnitude, sign lost when dividing $-50$ by $245$. Sign Error
[04] Sequences and series
Question 4
A geometric series has second term $6$ and sum to infinity $32$. Given that the series converges, what are the possible values of the first term?
Reveal the answer and worked solution. Commit to an option first.
Answer: C
Key idea. Two facts, two unknowns. Use $S_\infty = \frac{a}{1-r}$ to write $a$ in terms of $r$, substitute into $ar = 6$, and the result is a quadratic in $r$ with two admissible roots.
Fastest approach
$\frac{a}{1-r} = 32 \implies a = 32(1-r)$
Substitute into $ar = 6$: $32(1-r)r = 6 \implies 32r - 32r^{2} = 6$
$16r^{2} - 16r + 3 = 0 \implies (4r-1)(4r-3) = 0$
$r = \frac{1}{4}$ or $r = \frac{3}{4}$, both with $|r| < 1$, so both are admissible.
$a = 32\left(1 - \frac{1}{4}\right) = 24$ or $a = 32\left(1 - \frac{3}{4}\right) = 8$
Common mistake. Solving the quadratic, finding two values of $r$, and then reporting only one value of $a$. Both roots satisfy $|r| < 1$, so the question has two answers.
Why the other options are wrong
- A: Solving $a + ar = 32$, treating the sum to infinity as the sum of the first two terms. Formula Misapplication
- B: Assuming the first term and second term differ by the common ratio rather than being multiplied by it. Conceptual Misunderstanding
- D: Taking only $r = \frac{3}{4}$ and stopping. Incomplete Calculation
- E: Taking only $r = \frac{1}{4}$ and stopping. Incomplete Calculation
[05] Binomial expansion
Question 5
Find the coefficient of $x^{3}$ in the expansion of $\left(2 - \frac{x}{2}\right)^{7}$.
Reveal the answer and worked solution. Commit to an option first.
Answer: B
Key idea. The $x^{3}$ term is $\binom{7}{3}(2)^{4}\left(-\frac{x}{2}\right)^{3}$. Both the binomial coefficient and both powers matter, and the minus sign is cubed, so it survives.
Fastest approach
$\binom{7}{3} \cdot 2^{4} \cdot \left(-\frac{1}{2}\right)^{3} = 35 \cdot 16 \cdot \left(-\frac{1}{8}\right)$
$16 \div 8 = 2$, so this is $35 \times (-2) = -70$
Common mistake. Losing the sign. $\left(-\frac{1}{2}\right)^{3}$ is negative because the power is odd; had the question asked for $x^{2}$ or $x^{4}$ it would not have been.
Why the other options are wrong
- A: Using $\left(-\frac{1}{2}\right)^{2}$ in place of the cube, halving the division. Arithmetic Slip
- C: Using $\binom{7}{3} = 35$ and $\left(-\frac{1}{2}\right)^{3}$ but forgetting the $2^{4}$ factor. Incomplete Calculation
- D: Correct magnitude with the sign dropped. Sign Error
- E: Computing $\binom{7}{3} \cdot 2^{4} \cdot \frac{1}{2}$, applying the fractional power only once. Formula Misapplication
[06] Coordinate geometry
Question 6
A circle has equation $x^{2} + y^{2} - 6x + 4y - 12 = 0$. Find the length of the tangent from the point $(8, 1)$ to this circle.
Reveal the answer and worked solution. Commit to an option first.
Answer: A
Key idea. The tangent, the radius to the point of contact, and the line from the external point to the centre form a right-angled triangle. So $L^{2} = d^{2} - r^{2}$.
Fastest approach
Complete the square: $(x-3)^{2} + (y+2)^{2} = 12 + 9 + 4 = 25$
Centre $(3, -2)$, radius $5$.
$d^{2} = (8-3)^{2} + (1+2)^{2} = 25 + 9 = 34$
$L = \sqrt{34 - 25} = \sqrt{9} = 3$
Common mistake. Getting $d^{2} = 34$ and stopping, or subtracting $r$ rather than $r^{2}$. The subtraction happens between squares, before the root.
Why the other options are wrong
- B: Reporting the radius. Conceptual Misunderstanding
- C: Reporting the distance to the centre instead of the tangent length. Conceptual Misunderstanding
- D: Adding the squared radius instead of subtracting: $34 + 25$. Sign Error
- E: Reporting $L^{2} = 9$ rather than $L$. Incomplete Calculation
[07] Trigonometry
Question 7
How many solutions does $3\sin^{2}x = \cos x + 1$ have in the interval $0 \leqslant x \leqslant 2\pi$?
Reveal the answer and worked solution. Commit to an option first.
Answer: C
Key idea. Replace $\sin^{2}x$ with $1 - \cos^{2}x$ to get a quadratic in $\cos x$. Then count how many $x$ each root of that quadratic supplies.
Fastest approach
$3(1 - \cos^{2}x) = \cos x + 1$
$3\cos^{2}x + \cos x - 2 = 0 \implies (3\cos x - 2)(\cos x + 1) = 0$
$\cos x = \frac{2}{3}$: two solutions in $[0, 2\pi]$, one in each of the first and fourth quadrants.
$\cos x = -1$: one solution, $x = \pi$.
Total: $3$.
Common mistake. Solving the quadratic correctly and then answering 2, on the assumption that each root of a trigonometric equation gives two angles. $\cos x = -1$ is a turning point of the cosine curve and gives only one.
Why the other options are wrong
- A: Discarding $\cos x = \frac{2}{3}$ as 'not a standard angle'. Conceptual Misunderstanding
- B: Counting two solutions for $\cos x = \frac{2}{3}$ and none for $\cos x = -1$. Incomplete Calculation
- D: Assuming both roots give two angles each. Conceptual Misunderstanding
- E: Counting the endpoints $0$ and $2\pi$ as additional solutions. Boundary Error
[08] Exponentials and logarithms
Question 8
Solve $\log_{2}x + \log_{4}x = 6$.
Reveal the answer and worked solution. Commit to an option first.
Answer: C
Key idea. $\log_{4}x = \frac{\log_{2}x}{\log_{2}4} = \frac{1}{2}\log_{2}x$. Once both terms share a base the equation is linear.
Fastest approach
$\log_{2}x + \frac{1}{2}\log_{2}x = 6$
$\frac{3}{2}\log_{2}x = 6 \implies \log_{2}x = 4$
$x = 2^{4} = 16$
Common mistake. Treating $\log_{2}x + \log_{4}x$ as $\log_{8}x^{2}$ by adding bases. Bases do not add; convert one to the other.
Why the other options are wrong
- A: Converting the wrong way: using $\log_{2}x = \frac{1}{2}\log_{4}x$. Formula Misapplication
- B: Solving $\log_{2}x = 3$, from dividing 6 by 2 rather than by $\frac{3}{2}$. Arithmetic Slip
- D: Solving $\log_{2}x = 5$ from an arithmetic slip in $6 \div \frac{3}{2}$. Arithmetic Slip
- E: Solving $\log_{2}x = 6$, ignoring the second term entirely. Incomplete Calculation
[09] Exponentials and logarithms
Question 9
Solve the inequality $2^{2x} - 5 \cdot 2^{x} + 4 < 0$.
Reveal the answer and worked solution. Commit to an option first.
Answer: A
Key idea. Put $u = 2^{x}$, noting $2^{2x} = (2^{x})^{2} = u^{2}$. Solve in $u$, then convert the $u$-interval back to $x$ using the fact that $2^{x}$ is increasing.
Fastest approach
Let $u = 2^{x}$: $u^{2} - 5u + 4 < 0 \implies (u-1)(u-4) < 0 \implies 1 < u < 4$
$1 < 2^{x} < 4$
Since $2^{x}$ is strictly increasing, take $\log_{2}$ throughout: $0 < x < 2$
Common mistake. Reporting $1 < u < 4$ as the answer. Those are values of $2^{x}$, not of $x$, and one of the options offered is exactly that error.
Why the other options are wrong
- B: Giving the interval for $u = 2^{x}$ rather than for $x$. This is the intended trap. Conceptual Misunderstanding
- C: Solving $(u-1)(u-4) > 0$, the wrong side of the inequality. Sign Error
- D: Using non-strict inequalities, which include the roots where the expression equals zero, not less than zero. Boundary Error
- E: Taking only the upper branch. Incomplete Calculation
[10] Differentiation
Question 10
The curve $y = x^{3} - 3x^{2} - 9x + 5$ has a local maximum. Find its $y$-coordinate.
Reveal the answer and worked solution. Commit to an option first.
Answer: D
Key idea. Differentiate, solve for the stationary points, then use the sign of $\frac{d^{2}y}{dx^{2}}$ to tell the maximum from the minimum. For a positive cubic the maximum is always the left-hand one.
Fastest approach
$\frac{dy}{dx} = 3x^{2} - 6x - 9 = 3(x-3)(x+1)$
Stationary at $x = -1$ and $x = 3$.
$\frac{d^{2}y}{dx^{2}} = 6x - 6$; at $x = -1$ this is $-12 < 0$, so $x = -1$ is the maximum.
$y(-1) = -1 - 3 + 9 + 5 = 10$
Common mistake. Finding both stationary points and substituting the wrong one. $x = 3$ gives $-22$, which is the local minimum and is offered as a distractor.
Why the other options are wrong
- A: The local minimum, at $x = 3$. Conceptual Misunderstanding
- B: Setting $y = 0$ instead of evaluating $y$ at the stationary point. Formula Misapplication
- C: The $y$-intercept, from substituting $x = 0$. Conceptual Misunderstanding
- E: A sign slip in $y(-1)$, taking $-(-1)^{3} - 3(-1)^{2}$ as $+1+3$. Sign Error
[11] Differentiation
Question 11
The tangent to the curve $y = x^{2}$ at the point where $x = a$ passes through $(0, -4)$. Find all possible values of $a$.
Reveal the answer and worked solution. Commit to an option first.
Answer: E
Key idea. Write the tangent at a general $x = a$, then substitute the point it must pass through. The $y$-intercept of the tangent to $y = x^{2}$ at $x = a$ is always $-a^{2}$, which makes the condition immediate.
Fastest approach
Gradient $\frac{dy}{dx} = 2x$, so at $x = a$ the gradient is $2a$ and the point of contact is $(a, a^{2})$.
Tangent: $y - a^{2} = 2a(x - a)$
At $x = 0$: $y = a^{2} - 2a^{2} = -a^{2}$
So $-a^{2} = -4 \implies a^{2} = 4 \implies a = \pm 2$
Common mistake. Taking only the positive root. Both tangents exist: the curve is symmetric about the $y$-axis, so a point on that axis below the curve has two tangents through it, not one.
Why the other options are wrong
- A: Solving $a^{2} = 4$ and reporting only the positive root. Incomplete Calculation
- B: Setting $a = 4$ directly from the $-4$ in the question without squaring. Conceptual Misunderstanding
- C: Solving $2a^{2} = 4$, from mis-substituting the gradient. Formula Misapplication
- D: Both errors above at once. Conceptual Misunderstanding
[12] Integration
Question 12
Find the area of the finite region enclosed between the curve $y = x^{2}$ and the line $y = 2x$.
Reveal the answer and worked solution. Commit to an option first.
Answer: B
Key idea. Area between two curves is $\int (\text{upper} - \text{lower})$ between their intersections. Between $0$ and $2$ the line is above the parabola.
Fastest approach
Intersections: $x^{2} = 2x \implies x(x-2) = 0 \implies x = 0, 2$.
$\int_{0}^{2} (2x - x^{2})\,dx = \left[x^{2} - \frac{x^{3}}{3}\right]_{0}^{2}$
$= 4 - \frac{8}{3} = \frac{12 - 8}{3} = \frac{4}{3}$
Common mistake. Integrating $x^{2} - 2x$ and reporting $-\frac{4}{3}$, or dropping the sign and moving on. Subtract the lower curve from the upper one, and if the result is negative the order was wrong.
Why the other options are wrong
- A: Evaluating $\left[x^{2} - \frac{x^{3}}{3}\right]$ at $x = 1$ rather than $x = 2$. Boundary Error
- C: Integrating $2x$ alone over $[0,2]$ and forgetting to subtract. Incomplete Calculation
- D: Integrating $x^{2}$ alone over $[0,2]$. Incomplete Calculation
- E: Using limits $0$ to $4$, from misreading the intersection. Boundary Error
[13] Integration
Question 13
Evaluate $\displaystyle\int_{0}^{\pi/2} \sin^{2}x \,dx$.
Reveal the answer and worked solution. Commit to an option first.
Answer: B
Key idea. $\sin^{2}x$ cannot be integrated directly. Use $\sin^{2}x = \frac{1 - \cos 2x}{2}$, which turns it into two standard integrals.
Fastest approach
$\int_{0}^{\pi/2} \frac{1 - \cos 2x}{2}\,dx = \frac{1}{2}\left[x - \frac{\sin 2x}{2}\right]_{0}^{\pi/2}$
At $x = \frac{\pi}{2}$: $\sin \pi = 0$, so the bracket is $\frac{\pi}{2}$. At $x = 0$: the bracket is $0$.
$= \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$
Worth knowing as a fact: over a quarter period, $\sin^{2}$ averages $\frac{1}{2}$, so the integral is half the interval length.
Common mistake. Integrating $\sin^{2}x$ as $\frac{\sin^{3}x}{3}$, applying the reverse chain rule to a function that is not of the form $f'(x)[f(x)]^{n}$.
Why the other options are wrong
- A: Integrating $\sin^{2}x$ as $\frac{\sin^{3}x}{3}$ and evaluating. Formula Misapplication
- C: Integrating $\sin x$ instead of $\sin^{2}x$. Conceptual Misunderstanding
- D: Forgetting the factor of $\frac{1}{2}$ from the identity. Formula Misapplication
- E: Doubling rather than halving the interval contribution. Arithmetic Slip
[14] Graphs and transformations
Question 14
The graph of $y = f(x)$ has a local maximum at $(2, 5)$. Find the coordinates of the corresponding local maximum on the graph of $y = 3f(2x + 4) - 1$.
Reveal the answer and worked solution. Commit to an option first.
Answer: E
Key idea. Changes inside $f$ act on $x$ and run backwards; changes outside act on $y$ and run forwards. Solve $2x + 4 = 2$ for the new $x$, and apply $3(\cdot) - 1$ to the old $y$.
Fastest approach
The maximum occurs where the input to $f$ is $2$: $2x + 4 = 2 \implies x = -1$
The output is scaled then shifted: $y = 3(5) - 1 = 14$
So the point is $(-1, 14)$.
Common mistake. Applying the horizontal transformations forwards: halving then subtracting 4 gives $-3$, which is offered as a distractor. Inside the bracket, the operations undo in reverse order.
Why the other options are wrong
- A: Transforming $x$ correctly but leaving $y$ untouched. Incomplete Calculation
- B: Applying the inside transformations in the forward direction rather than inverting them. Conceptual Misunderstanding
- C: Solving $2x - 4 = 2$, misreading the sign inside the bracket. Sign Error
- D: Computing $3(5) - 1$ as $15 - 0$, dropping the vertical shift. Arithmetic Slip
[15] Quadratic equations and inequalities
Question 15
Solve $|2x - 3| < x + 1$.
Reveal the answer and worked solution. Commit to an option first.
Answer: C
Key idea. $|A| < B$ is equivalent to $-B < A < B$, which quietly requires $B > 0$. Both branches must hold at once, so the answer is an intersection, not a union.
Fastest approach
$-(x+1) < 2x - 3 < x + 1$
Left: $-x - 1 < 2x - 3 \implies 2 < 3x \implies x > \frac{2}{3}$
Right: $2x - 3 < x + 1 \implies x < 4$
Both must hold: $\frac{2}{3} < x < 4$.
The condition $x + 1 > 0$ is implied by $x > \frac{2}{3}$, so it adds nothing here.
Common mistake. Solving only $2x - 3 < x + 1$ and answering $x < 4$. The modulus makes this two inequalities, and the lower bound is the half that gets dropped.
Why the other options are wrong
- A: Solving only the negative branch. Incomplete Calculation
- B: Using non-strict inequalities where the original is strict. Boundary Error
- D: Solving only the positive branch. Incomplete Calculation
- E: Sign slip on the left branch: solving $-x-1 < 2x-3$ as $x > -\frac{2}{3}$. Sign Error
[16] Counting and probability
Question 16
How many distinct arrangements are there of the letters of the word STATISTICS?
Reveal the answer and worked solution. Commit to an option first.
Answer: B
Key idea. With $n$ letters of which one repeats $p$ times, another $q$ times and so on, the count is $\frac{n!}{p!\,q!\cdots}$. Each repeated group is indistinguishable, so its internal orderings must be divided out.
Fastest approach
STATISTICS has 10 letters: S appears 3 times, T appears 3 times, I appears 2 times, and A and C once each.
$\frac{10!}{3!\,3!\,2!} = \frac{3\,628\,800}{6 \times 6 \times 2} = \frac{3\,628\,800}{72} = 50\,400$
Common mistake. Miscounting the letters. There are three S and three T, not two of each; the word is ten letters long and it is worth writing them out before dividing.
Why the other options are wrong
- A: Dividing by an extra $2!$ for the A and C, which appear once each and need no division. Formula Misapplication
- C: Dividing by $3!\,3!$ and forgetting the two I. Incomplete Calculation
- D: Dividing by $3!\,2!\,2!$, from counting only two T. Miscount
- E: Reporting $10!$ with no division at all. Conceptual Misunderstanding
[17] Counting and probability
Question 17
Two fair six-sided dice are rolled. What is the probability that the sum of the two scores is a prime number?
Reveal the answer and worked solution. Commit to an option first.
Answer: D
Key idea. Work through the achievable sums $2$ to $12$, keep the primes, and count the ways to make each. The sample space is 36 equally likely ordered pairs.
Fastest approach
Primes among the possible sums $2$ to $12$: $2, 3, 5, 7, 11$.
Ways to make each: $2 \to 1$, $3 \to 2$, $5 \to 4$, $7 \to 6$, $11 \to 2$
Total favourable: $1 + 2 + 4 + 6 + 2 = 15$
$P = \frac{15}{36} = \frac{5}{12}$
Common mistake. Including $9$ as a prime, or excluding $2$ on the grounds that it is even. $2$ is prime and $9 = 3 \times 3$ is not.
Why the other options are wrong
- A: Counting 11, from treating the dice as indistinguishable and undercounting the ordered pairs. Sample Space Error
- B: Counting 12 favourable outcomes, from omitting the sum of 2 and one way of making 11. Miscount
- C: Counting 14, from omitting the sum of 2. Miscount
- E: Counting 18, from including 9 as prime. Conceptual Misunderstanding
[18] Algebra and functions
Question 18
The polynomial $f(x) = 2x^{3} + ax^{2} + bx - 6$ is exactly divisible by $(x-1)$ and leaves a remainder of $12$ when divided by $(x+2)$. Find $a - b$.
Reveal the answer and worked solution. Commit to an option first.
Answer: E
Key idea. Exactly divisible by $(x-1)$ means $f(1) = 0$; remainder 12 on division by $(x+2)$ means $f(-2) = 12$. Two equations, two unknowns.
Fastest approach
$f(1) = 2 + a + b - 6 = 0 \implies a + b = 4$
$f(-2) = -16 + 4a - 2b - 6 = 12 \implies 4a - 2b = 34 \implies 2a - b = 17$
Adding the first to the second: $3a = 21 \implies a = 7$, so $b = -3$.
$a - b = 7 - (-3) = 10$
Common mistake. Using $f(2) = 12$ for division by $(x+2)$. The remainder theorem evaluates at the root of the divisor, which is $x = -2$.
Why the other options are wrong
- A: Correct magnitude with the subtraction reversed: $b - a$. Sign Error
- B: Reporting $-b$ alone. Incomplete Calculation
- C: Reporting $a + b$, which is 4, instead of $a - b$. Misread Question
- D: Reporting $a$ alone. Incomplete Calculation
[19] Trigonometry
Question 19
Find the maximum value of $\dfrac{1}{3\sin x + 4\cos x + 7}$.
Reveal the answer and worked solution. Commit to an option first.
Answer: D
Key idea. $3\sin x + 4\cos x = 5\sin(x + \alpha)$, so it ranges over $[-5, 5]$. The whole denominator therefore ranges over $[2, 12]$, and a fraction is largest when its denominator is smallest.
Fastest approach
$3\sin x + 4\cos x = R\sin(x+\alpha)$ with $R = \sqrt{3^{2}+4^{2}} = 5$.
Denominator range: $[7-5,\ 7+5] = [2, 12]$.
The denominator is never zero, so the expression is defined everywhere, and the maximum of the fraction occurs at the minimum of the denominator: $\frac{1}{2}$
Common mistake. Maximising the denominator instead of minimising it, giving $\frac{1}{12}$. For a positive denominator the fraction moves the opposite way.
Why the other options are wrong
- A: Using the maximum of the denominator, which gives the minimum of the fraction. Conceptual Misunderstanding
- B: Ignoring the trigonometric part entirely and using the constant 7. Incomplete Calculation
- C: Taking $R = 4$ rather than $5$, from using only the larger coefficient. Formula Misapplication
- E: Reporting the minimum denominator rather than the value of the fraction. Misread Question
[20] Algebra and functions
Question 20
Real numbers $x$ and $y$ satisfy $x + y = 5$ and $x^{3} + y^{3} = 35$. Find $xy$.
Reveal the answer and worked solution. Commit to an option first.
Answer: D
Key idea. Never solve for $x$ and $y$ individually here. The identity $x^{3} + y^{3} = (x+y)^{3} - 3xy(x+y)$ turns the problem into one linear equation in $xy$.
Fastest approach
$x^{3} + y^{3} = (x+y)^{3} - 3xy(x+y)$
$35 = 125 - 15xy$
$15xy = 90 \implies xy = 6$
Common mistake. Using $x^{3}+y^{3} = (x+y)(x^{2}-xy+y^{2})$ and then getting stuck converting $x^{2}+y^{2}$, which needs $(x+y)^{2} - 2xy$ as a second step. Both routes work; the cube identity is one step shorter.
Why the other options are wrong
- A: Sign error in $125 - 15xy = 35$, solving $15xy = -90$. Sign Error
- B: Arithmetic slip: $125 - 35 = 90$ mis-divided by 15 as 4. Arithmetic Slip
- C: Assuming $xy$ equals $x + y$ by symmetry. Conceptual Misunderstanding
- E: Solving $(x+y)^{3} - 3xy = 35$, dropping the factor of $(x+y)$ from the identity. Formula Misapplication