VCE Specialist Mathematics 2025 Exam 1

[4 marks]. Differentiate implicitly with respect to $x$ (product rule on both terms):

At $(4, -2)$: $e^{-2y} = e^4$ and $e^x = e^4$, so

The tangent is $y + 2 = \tfrac{5}{12}(x - 4)$, that is

From the report. Implicit differentiation was usually done well, but arithmetic slips cost the gradient for some, and a few found the gradient and then stopped without writing the tangent equation.

[3 marks]. Use a different parameter for each line:

Equate components:

• $x$: $2 + \lambda = 1 - \mu$, so $\lambda + \mu = -1$.
• $y$: $3 + 2\lambda = 3 - \mu$, so $2\lambda + \mu = 0$.

Subtracting gives $\lambda = 1$, then $\mu = -2$. Check $z$: $1 - 1 = 0$ and $2 + (-2) = 0$. The lines meet at $(3, 5, 0)$.

From the report. The common error was using the same parameter for both lines, which gives equations with no consistent solution and was not eligible for full marks.

a. [1 mark]. $x(t) = \displaystyle\int\frac{t}{\sqrt{t^2 + k}}\,dt = \sqrt{t^2 + k} + c$. Since $x(0) = 0$, $\sqrt k + c = 0$, so $c = -\sqrt k$ and $x(t) = \sqrt{t^2 + k} - \sqrt k$.

b. [2 marks]. By the quotient rule,

At $t = 0$: $a(0) = \dfrac{k}{k^{3/2}} = \dfrac{1}{\sqrt k}$ m s$^{-2}$.

c. [2 marks]. $s(3) - x(3) = 1$:

Squaring: $9 + k = 4 + 4\sqrt k + k$, so $\sqrt k = \tfrac54$ and $k = \dfrac{25}{16}$.

From the report. In part a some did not use the initial condition to find the constant. In part b the quotient rule was not always applied correctly; a correct value from an incorrect derivative did not earn full marks. Part c was weak (28% full marks), with many unproductive attempts at rearranging and squaring.

a. [3 marks]. $\text{E}(T) = \dfrac{3}{2\log_e 2}\displaystyle\int_0^1\frac{t}{(t + 1)(2 - t)}\,dt$. Use partial fractions:

$t = -1$ gives $A = -\tfrac13$; $t = 2$ gives $B = \tfrac23$. Then

So $\text{E}(T) = \dfrac{3}{2\log_e 2} \times \dfrac{\log_e 2}{3} = \dfrac12$.

b. [2 marks]. $\bar T$ is approximately normal with mean $\tfrac12$ and standard deviation $\tfrac{0.3}{\sqrt{25}} = 0.06$. Then 0.44 is exactly one standard deviation below the mean:

From the report. In part a some missed that partial fractions were needed, left the $t$ out of the numerator, or got the coefficients wrong; the working towards the given answer had to be clear. In part b most found the mean and standard deviation of the sample mean, and a sketch of the normal curve helped.

a. [1 mark]. The $\mathbf{j}$ components agree at $t = 1$: $-1 = 2 + c + 1$, so $c = -4$.

b. [2 marks]. The $\mathbf{i}$ components agree at $t = 1$: $1 + a = b + 2$, so $b = a - 1$. Velocities at $t = 1$:

Perpendicular means the dot product is zero: $(3 + 2a)(a + 1) = 0$, so $a = -\tfrac32$ or $a = -1$.

c. [1 mark]. $\ddot{\mathbf{r}}_P = (6t + 2a)\mathbf{i}$ and $\ddot{\mathbf{r}}_Q = 4\mathbf{j}$. At $t = 1$, $|6 + 2a| = 4$, so $a = -1$ or $a = -5$. Only $a = -1$ also satisfies part b, so $a = -1$ and $b = a - 1 = -2$.

From the report. Part b needed the dot product of the velocities to be zero together with the collision condition $b = a - 1$. Part c was poorly done (25% correct): answers that also listed $a = -5$, $b = -6$ did not get the mark.

[4 marks].

Substitute $u = \arctan(x)$, $\tfrac{du}{dx} = \tfrac{1}{1 + x^2}$; the terminals become $u = \tfrac{\pi}{4}$ and $u = \tfrac{\pi}{3}$:

So $a = 7$, $b = 3$, $c = 288$.

From the report. Most recognised that a substitution (or integration by parts) was needed. Common errors were dropping the factor $\pi$, not changing the terminals after substituting (or changing them wrongly), and arithmetic slips.

[4 marks]

Let $P(n)$ be the statement $\displaystyle\sum_{i=1}^{n}(i + 1)^2 = \tfrac16n\left(2n^2 + 9n + 13\right)$.

Base case, $n = 1$

LHS $= 2^2 = 4$. RHS $= \tfrac16(1)(2 + 9 + 13) = 4$. So $P(1)$ is true.

Inductive step

Assume $P(k)$ is true for some $k \in N$: $\displaystyle\sum_{i=1}^{k}(i + 1)^2 = \tfrac16k\left(2k^2 + 9k + 13\right)$. Then

The RHS of $P(k + 1)$ is

These are equal, so $P(k)$ true implies $P(k + 1)$ true. Since $P(1)$ is true, by mathematical induction $P(n)$ is true for all $n \in N$.

From the report. Common errors were not verifying the base case properly, misstating the assumption (it must be for a particular $k$, not "for all $n$"), and starting the inductive step by assuming the equality to be proved.

a. [1 mark]. $z_1$ is at $(1, 2)$ and $\overline{z_1} = 1 - 2i$ at $(1, -2)$.

b. [2 marks]. $f$ has real coefficients, so the conjugate $1 - 2i$ is also a solution:

c. [2 marks]. Equating coefficients (or noticing $f(z) = (z^2 + 5)^2 - 4z^2$):

$z^2 + 2z + 5 = (z + 1)^2 + 4 = 0$ gives $z = -1 \pm 2i$. The remaining solutions are $1 - 2i$, $-1 + 2i$ and $-1 - 2i$.

From the report. Most used the conjugate root to build the quadratic factor. In part c, equating coefficients was more reliable than long or synthetic division, and completing the square was more reliable than the quadratic formula.

a. [2 marks]. Factorise and cancel the common factor $(x - 1)$ (allowed since $x \ne 1$):

b. [1 mark]. $k = \displaystyle\lim_{x \to 1}f(x) = -1 - 1 + \tfrac12 = -\tfrac32$.

c. [3 marks]. Asymptotes $x = -1$ and $y = -x - 1$. Intercepts: $f(x) = -\tfrac{x(x + 2)}{x + 1} = 0$ at $x = 0$ and $x = -2$, so $(0, 0)$ and $(-2, 0)$. There is no asymptote at $x = 1$: the graph has a hole there, shown by an open circle at $\left(1, -\tfrac32\right)$.

From the report. In part a the working had to genuinely produce the given form. Part b needed a number (46% correct). In part c the open circle at $\left(1, -\tfrac32\right)$ was often missing or misplaced, some drew a false vertical asymptote at $x = 1$, and some curves missed the intercepts or bent away from the asymptotes. Use a ruler for asymptotes.