QCE Maths Methods EA strategy: 2026 guide

Step 1. Identify the structure

This is a product of two functions: $f(x) = x^2$ and $g(x) = \sin(3x)$. Product rule applies.

Step 2. Differentiate each factor

$f'(x) = 2x$. For $g(x) = \sin(3x)$ the chain rule gives $g'(x) = 3 \cos(3x)$, since the inner derivative is $3$.

Step 3. Apply the product rule

Substitute into $f'g + f g'$:

Step 4. Tidy. Group the $3$ with $x^2$:

Answer. $2x \sin(3x) + 3x^2 \cos(3x)$. Both terms required for full marks under the QCAA "mathematical reasoning" criterion.

Step 1. Recognise the form

The integrand is $\cos(\text{linear in } x)$, which has standard antiderivative $\sin$ times a reciprocal-coefficient factor.

Step 2. Write the candidate antiderivative

Try $\sin(5x + 1)$. Differentiating, $\frac{d}{dx}[\sin(5x + 1)] = 5 \cos(5x + 1)$, which is $5$ times the integrand.

Step 3. Correct for the chain factor

Divide by $5$ to undo the excess factor:

Step 4. Include $+C$. The indefinite integral always carries the arbitrary constant.

Answer. $\frac{1}{5} \sin(5x + 1) + C$.

Step 1. Isolate the exponential

Divide both sides by $3$: $e^{2x} = 4$.

Step 2. Apply the natural logarithm

Since $\ln$ and $\exp$ are inverses: $2x = \ln 4$.

Step 3. Solve for $x$

Divide by $2$: $x = \frac{\ln 4}{2}$.

Step 4. Simplify to canonical exact form

Use $\ln 4 = 2 \ln 2$, so $x = \frac{2 \ln 2}{2} = \ln 2$.

Answer

$x = \ln 2$. Exact form expected on Paper 1 (technology-free).

Step 1. Read the scenario

Population $P(t) = \frac{L}{1 + A e^{-kt}}$ with carrying capacity $L = 1000$, initial population $P(0) = 100$, and growth constant $k = 0.2$ per year. Find $t$ when $P(t) = 500$.

Step 2. Use the initial condition to find $A$

Substituting $t = 0$: $100 = \frac{1000}{1 + A}$, so $1 + A = 10$ and $A = 9$.

Step 3. Write the equation to solve

With $A = 9$:

Step 4. Rearrange algebraically

Cross-multiply: $500 (1 + 9 e^{-0.2 t}) = 1000$, so $1 + 9 e^{-0.2 t} = 2$ and $9 e^{-0.2 t} = 1$. Hence $e^{-0.2 t} = 1/9$.

Step 5. Take logs

$-0.2 t = \ln(1/9) = -\ln 9$, so $t = \frac{\ln 9}{0.2} = 5 \ln 9$.

Step 6. Evaluate

$t = 5 \ln 9 \approx 10.99$ years.

Step 7. State with units and reasonableness

The population reaches $500$ (half of capacity, the inflection point) at $t \approx 11.0$ years, which sits between the initial small population and the saturation phase. Reasonable.

Answer

$t \approx 11.0$ years.

Step 1. Confirm the model

$X \sim \text{Bin}(20, 0.3)$: $n = 20$ independent trials, success probability $p = 0.3$, $X$ counts successes.

Step 2. State what is required

Cumulative probability $P(X \le 5)$.

Step 3. Set up the calculator command

On most approved graphics calculators this is a binomcdf function, for example binomcdf(20, 0.3, 5) (some models ask for a lower and upper bound instead, so check yours). Always report the setup as well as the output.

Step 4. Execute

Output: $0.4164$ (to 4 decimal places).

Step 5. Report supporting statistics

Mean $= np = 20 \times 0.3 = 6$. Variance $= np(1 - p) = 20 \times 0.3 \times 0.7 = 4.2$. Standard deviation $\approx 2.05$.

Step 6. Interpret in context

Probability of at most $5$ successes is about $0.416$, slightly below half because the mean is $6$.

Answer

$P(X \le 5) \approx 0.4164$.

Step 1. Identify the parameter

$p$ = unknown population proportion. Sample size $n = 400$, observed successes $144$.

Step 2. Compute the sample proportion

$\hat{p} = 144 / 400 = 0.36$.

Step 3. Choose the critical value

For a 95 percent CI, $z^* = 1.96$.

Step 4. Compute the standard error

$SE = \sqrt{\hat{p}(1 - \hat{p}) / n} = \sqrt{0.36 \times 0.64 / 400}$.

Step 5. Compute the margin of error

$E = z^* \cdot SE = 1.96 \times 0.024 \approx 0.047$.

Step 6. Form the interval

$\hat{p} \pm E = 0.36 \pm 0.047$, giving $(0.313, 0.407)$.

Step 7. Communicate in context

We are 95 percent confident that the true population proportion lies between $0.313$ and $0.407$. Note the QCAA EA expects this interpretation as part of the "communicate" descriptor.

Answer

95 percent CI: $(0.313, 0.407)$.

These eight questions cover the chain rule for logs, reverse-chain antidifferentiation, solving log equations, definite integrals, calculator normal and binomial calculations, confidence intervals, and exponential decay.

Step 1: Q1. Differentiate $y = \ln(x^2 + 1)$ using the chain rule

The derivative of $\ln(u)$ is $\frac{1}{u} \cdot u'$. Here $u = x^2 + 1$, so $u' = 2x$.

Step 2: Q2. Antidifferentiate $(3x - 2)^4$ by reverse chain

Try raising the power by 1 and dividing by the new power: $\frac{(3x - 2)^5}{5}$. Differentiating this gives $(3x - 2)^4 \cdot 3$, which is $3$ times too large, so divide by $3$ to compensate.

Step 3: Q3. Solve $2\ln x = \ln(x + 6)$ exactly

Use the log power rule to write $2\ln x = \ln x^2$. When two logs with the same base are equal, their arguments must be equal. Then solve the resulting quadratic and check the domain.

Candidates: $x = 3$ or $x = -2$. Since $\ln x$ requires $x > 0$, reject $x = -2$.

The solution is $x = 3$.

Step 4: Q4. Evaluate $\displaystyle\int_0^{\pi/2} \sin(2x)\, dx$

The antiderivative of $\sin(2x)$ is $-\frac{1}{2}\cos(2x)$ (the factor of $\frac{1}{2}$ compensates for the inner derivative of $2x$). Apply the fundamental theorem.

Step 5: Q5. Calculator: probability for $X \sim N(170, 10^2)$ between 180 and 195

On your graphics calculator use the normal cumulative distribution function with lower bound 180, upper bound 195, mean 170, and standard deviation 10.

About $15.24\%$ of values fall in this interval.

Step 6: Q6. Calculator: $P(X \ge 2)$ for $X \sim \text{Binomial}(50, 0.04)$

Use the complement rule: $P(X \ge 2) = 1 - P(X \le 1)$. Evaluate the cumulative binomial on the calculator (binomcdf).

Step 7: Q7. 90% confidence interval for a proportion

Compute the sample proportion, the standard error, and the margin of error using $z^* = 1.645$ for 90%.

90% confidence interval: $(0.52 - 0.052,\ 0.52 + 0.052) = (0.468, 0.572)$.

Step 8: Q8. Exponential decay: find $k$ and then the time to reach $10\%$

First find the decay constant $k$ from the half-life condition $A_0/2 = A_0 e^{-5.27k}$. Then use $k$ to find the time for the amount to fall to $10\%$ of the original.

For $10\%$ remaining: $0.1 = e^{-kt}$, so

Final answer: the substance falls to $10\%$ of its original amount after approximately $17.5$ years.