VCE Math Methods 2025 Exam 1

a. [1 mark]. Product rule with $u = x^2$ and $v = \cos(x)$:

b. [2 marks]. Write $f(x) = 6(x+1)^{\frac12} + 5$ and use the chain rule:

At $x = 8$: $f'(8) = \dfrac{3}{\sqrt 9} = 1$. The gradient of the tangent is 1.

From the report. Part a was well done (88% full marks), but many students left untidy double negatives or dropped brackets around terms. In part b some students mishandled the chain rule or the power $\tfrac12$, and some went on to find the equation of the tangent when only the gradient was asked for.

[2 marks]. Antidifferentiate. The inside function $2x + 3$ has derivative 2, so a factor of $\tfrac12$ is needed:

No absolute value is needed because $2x + 3 > 0$ on the given domain. Use $g(1) = 0$:

From the report. Only 40% scored full marks. The common slip was writing $\log_e(2x+3)$ without the factor $\tfrac12$; some students did not recognise that the antiderivative is a logarithm at all.

a. [1 mark]. $\cos(2x)$ takes every value in $[-1, 1]$ on this domain, so $2\cos(2x) + 1$ runs from $-1$ to $3$. The range is $[-1, 3]$.

b. [3 marks]. Rearrange: $\cos(2x) = -\tfrac12$. Since $x \in [0, 2\pi]$, the angle $2x \in [0, 4\pi]$, which is two full cycles. Cosine is negative in quadrants 2 and 3, with base angle $\tfrac{\pi}{3}$:

c. [2 marks]. The period is $\pi$, so the interval $\left[\tfrac{\pi}{2}, \tfrac{3\pi}{2}\right]$ is exactly one cycle.

• Endpoints: $f\!\left(\tfrac{\pi}{2}\right) = 2\cos(\pi) + 1 = -1$ and $f\!\left(\tfrac{3\pi}{2}\right) = 2\cos(3\pi) + 1 = -1$, so $\left(\tfrac{\pi}{2}, -1\right)$ and $\left(\tfrac{3\pi}{2}, -1\right)$.
• Maximum: $f(\pi) = 2\cos(2\pi) + 1 = 3$, at $(\pi, 3)$.
• $x$-intercepts from part b: $x = \tfrac{2\pi}{3}$ and $x = \tfrac{4\pi}{3}$.

From the report. In part b, some students chose the wrong quadrant or only gave two solutions, forgetting that $2x$ covers two cycles. In part c the common errors were mislabelled endpoints, drawing past the domain, missing the maximum at $(\pi, 3)$, and curves that looked parabolic or were not symmetric about $x = \pi$. Label points with round brackets.

a. [2 marks]. The probabilities must add to 1:

Multiply through by $25k$:

So $k = 10$ or $k = 15$, as required.

b. i. [1 mark]. With $k = 15$ the probabilities are $\tfrac{4}{15}, \tfrac{2}{5}, \tfrac{1}{5}, \tfrac{2}{15}$ (they sum to 1). Then

ii. [1 mark].

From the report. Part a is a "show that": the sum-to-1 equation, the quadratic and its solution all had to appear. Substituting 10 and 15 to check them did not earn the marks. Some students used the formula for $\text{E}(X)$ in part a instead of the sum of probabilities. In b.i the usual error was including $\Pr(X = 1)$, which gives the incorrect $\tfrac{11}{15}$; in b.ii the errors were fraction arithmetic.

a. [2 marks]. This is a quadratic in $e^x$. Factorise:

b. [2 marks]. For an inverse, $g$ must be one-to-one, so the domain must stop at the minimum.

$g$ is decreasing on $(-\infty, \log_e 4]$, so the largest value is $a = \log_e(4) = 2\log_e(2)$.

From the report. In part a some students wrongly discarded $x = 0$. In part b the classic trap was substituting $u = e^x$, finding $u = 4$, and then giving $a = 4$ instead of converting back to $x = \log_e(4)$.

a. [1 mark]. $\text{var}(X) = np(1-p) = 6 \times \tfrac14 \times \tfrac34 = \dfrac{9}{8}$.

b. [2 marks]. Add the two tail terms:

Since $4096 = 4^6 = 2^{12}$, the answer is $\dfrac{19}{2^{12}}$.

From the report. In part a some students gave the standard deviation instead of the variance. In part b many found only $\Pr(X = 5)$, or stopped at $\tfrac{19}{4096}$ without writing $4096$ as $2^{12}$, which the required form demanded.

a. [1 mark]. $f(5) = 125 - 25 - 80 - 20 = 0$, so $x = 5$ is a solution.

b. [2 marks]. Divide $f(x)$ by $(x - 5)$ (long division, synthetic division or equating coefficients):

So $d = 2$. Check: $(x-5)(x^2 + 4x + 4) = x^3 + 4x^2 + 4x - 5x^2 - 20x - 20 = x^3 - x^2 - 16x - 20$.

c. [1 mark]

From the factorised form: $(-2, 0)$ (a turning point on the axis), $(5, 0)$, and $(0, -20)$.

d. i. [1 mark]

$f(x)g(x) = (x + 2)^3(x - 5)$. A cubed (odd, repeated) factor gives a stationary point of inflection on the axis, at $(-2, 0)$.

ii. [1 mark]

Sign of $(x+2)^3(x-5)$: positive for $x < -2$ (negative times negative), negative for $-2 < x < 5$, positive for $x > 5$, and zero at both roots. So

From the report. Parts a to c went well, though some dropped the negative signs on the intercepts. Part d.i (44% correct) was often answered with the equation of the product rather than a coordinate pair, and students who expanded the product instead of using $(x+2)^3(x-5)$ made it much harder. Part d.ii was poorly answered (27% correct), often because of sign slips such as dropping the negative sign in front of the 2.

a. [3 marks]. Integrate from $k$ to the top of the domain:

This equals $\tfrac{1}{16}(4 - 3k)^2$ (the area of the triangle under the line from $x = k$ to $x = \tfrac43$). Set it equal to $\tfrac{9}{16}$:

Only $k = \tfrac13$ lies in $\left[0, \tfrac43\right]$, so $k = \dfrac13$.

b. [2 marks]. Split the integral. The pdf integrates to 1 over its domain:

From the report. In part a, students integrating $(4 - 3x)$ as a bracket raised the power to 2 but then divided by 2 instead of 6 (forgetting the chain-rule factor of $-3$), and many used the quadratic formula where factorising was quicker. Most correctly rejected $k = \tfrac73$. In part b a frequent error was taking $m$ out of the whole integral, including the $n$ term; students who used "total probability is 1" were generally successful.

a. [3 marks]. With $w = -3$: $\dfrac{9}{(x-1)^2} = (x + 3)^2$, so $\big[(x-1)(x+3)\big]^2 = 9$. Rather than expanding to a quartic, take square roots:

• $x^2 + 2x - 3 = 3 \implies x^2 + 2x - 6 = 0 \implies x = -1 \pm \sqrt7$.
• $x^2 + 2x - 3 = -3 \implies x(x + 2) = 0 \implies x = 0$ or $x = -2$.

None equals 1, so all four are valid: $x = -2,\ 0,\ -1 - \sqrt7,\ -1 + \sqrt7$.

b. i. [2 marks]. The parabola $y = (x-1)(x-w)$ has roots 1 and $w$, so by symmetry its minimum is halfway between them, at $x = \tfrac{w+1}{2}$. Then

The minimum is at $\left(\dfrac{w+1}{2},\ -\dfrac{(w-1)^2}{4}\right)$.

b. ii. [2 marks]. $f(x) = g(x)$ becomes $\big[(x-1)(x-w)\big]^2 = w^2$, so

• The first equation simplifies to $x^2 - (w+1)x = 0$, giving $x = 0$ or $x = w + 1$: always two solutions for $w > 0$.
• So we need the second equation to have exactly one solution: the horizontal line $y = -w$ must just touch the minimum from part i.

Both values are positive, and $x = 1$ never satisfies either equation (it would need $0 = \pm w$), so $w = 3 - 2\sqrt2$ or $w = 3 + 2\sqrt2$.

From the report. This was the hardest question on the paper: 94% of students scored 0 on b.ii. In part a many students expanded to a quartic and could not finish, or ignored the $\pm$ when taking square roots. In b.i the $y$-coordinate was often lost to errors combining fractions. The key insight for b.ii is to split into the $+w$ and $-w$ cases and ask how many solutions each can give.