10 free VCE Maths Methods questions, fully worked

1. Differentiate.\(f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)\)
2. Set $f'(x) = 0$.\(x = 1 \quad \text{or} \quad x = 3\)
3. Find the $y$-values.\(f(1) = 6, \qquad f(3) = 2\)
4. Use the second derivative to classify each one.\(f''(x) = 6x - 12\)
5. $f''(1) = -6 < 0$ and $f''(3) = 6 > 0$.\((1,6) \text{ is a local maximum}\)

\((1, 6) \text{ local maximum}; \;\; (3, 2) \text{ local minimum}\)

Where marks get lost: Finding the stationary points is half the marks. Classifying them is the other half — don't stop at f'(x) = 0.

1. Let $u = 2x^2+1$, so $y = u^5$.\(\frac{dy}{du} = 5u^4\)
2. Differentiate the inside.\(\frac{du}{dx} = 4x\)
3. Multiply them together.\(\frac{dy}{dx} = 5(2x^2+1)^4 \cdot 4x\)

\(\frac{dy}{dx} = 20x(2x^2+1)^4\)

Where marks get lost: The mark is lost by stopping at 5(2x²+1)⁴ and forgetting the 4x.

1. Read off $a = -3$.\(\text{dilation by factor } 3 \text{ from the } x\text{-axis}\)
2. The negative sign is a reflection.\(\text{reflection in the } x\text{-axis}\)
3. $(x+1)$ means $h = -1$.\(\text{translation 1 unit left}\)
4. $+4$ shifts it vertically.\(\text{translation 4 units up}\)

\(\text{dilate }3\text{ from }x\text{-axis, reflect in }x\text{-axis, then }1\text{ left and }4\text{ up}\)

Where marks get lost: (x + 1) moves the graph LEFT, not right. The sign inside the bracket is the opposite of the direction.

1. A logarithm needs a strictly positive argument.\(x - 3 > 0\)
2. Solve.\(x > 3\)
3. A log function takes every real value.\(\text{range} = R\)

\(\text{domain } (3, \infty), \quad \text{range } R\)

Where marks get lost: Strictly greater than 3 — log₂(0) is undefined, so 3 itself is not in the domain.

1. Antidifferentiate.\(\int (2x+1)dx = x^2 + x\)
2. Substitute the upper limit.\((3)^2 + 3 = 12\)
3. Substitute the lower limit and subtract.\(12 - \big((1)^2 + 1\big) = 12 - 2\)

\(= 10\)

Where marks get lost: No + c on a definite integral — it cancels when you subtract.

1. Amplitude is $|a|$.\(\text{amplitude} = 2\)
2. Period is $\frac{2\pi}{n}$ with $n = 3$.\(\text{period} = \frac{2\pi}{3}\)
3. The graph is centred on $y = -1$ and reaches $2$ either side.\(-1 \pm 2\)

\(\text{amplitude } 2, \;\; \text{period } \tfrac{2\pi}{3}, \;\; \text{range } [-3, 1]\)

Where marks get lost: Period divides by n. A bigger n gives a shorter, faster wave.

1. Write the definition.\(\Pr(A \mid B) = \frac{\Pr(A \cap B)}{\Pr(B)}\)
2. Substitute. Divide by what you were GIVEN.\(= \frac{0.2}{0.4}\)

\(\Pr(A \mid B) = 0.5\)

Where marks get lost: Divide by P(B), not by P(A) and not by 1. The condition sets the denominator.

1. Rearrange.\(\cos(x) = \frac{\sqrt{3}}{2}\)
2. First quadrant solution.\(x = \frac{\pi}{6}\)
3. Cosine is also positive in the fourth quadrant.\(x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}\)

\(x = \frac{\pi}{6} \;\text{ and }\; x = \frac{11\pi}{6}\)

Where marks get lost: Two solutions in the domain. Your calculator only hands you the first one.

1. This is binomial with $n=6$, $p=0.5$, $r=4$.\(\Pr(X=4) = \binom{6}{4}(0.5)^4(0.5)^2\)
2. $\binom{6}{4} = 15$ and $(0.5)^6 = \frac{1}{64}$.\(= 15 \times \frac{1}{64}\)

\(\Pr(X=4) = \frac{15}{64} \approx 0.234\)

Where marks get lost: The 15 counts the different orders the 4 heads could land in. Drop it and you're answering a different question.

1. No real solutions means $\Delta < 0$.\(\Delta = k^2 - 4(2)(8)\)
2. Simplify.\(\Delta = k^2 - 64\)
3. Solve the inequality.\(k^2 - 64 < 0 \Rightarrow k^2 < 64\)

\(-8 < k < 8\)

Where marks get lost: k² < 64 gives an interval, not just k < 8. Both ends matter.