VCAA Mathematical Methods Functions, relations and graphs

This option only includes the points where $g(0)$ reaches the global maximum of $f(x)$, which is $\frac{1}{4}$. It misses the intervals where $g(0)$ is a local maximum that remains the absolute maximum on the restricted interval $[0, 1]$.

This incorrectly assumes $a$ can be any value between the critical points. For $a \in \left(-\frac{1}{2}, \frac{1}{2}\right)$, the shifted interval $[-a, 1-a]$ will contain values of $x$ where $f(x) > f(-a)$, meaning $g(0)$ is not the maximum.

This option incorrectly includes $a \ge \frac{\sqrt{2}}{2}$. For these values, $-a \le -\frac{\sqrt{2}}{2}$, and the function $f(x)$ is strictly increasing immediately to the right of $-a$, so $g(0)$ cannot be the maximum on $[0, 1]$.

This incorrectly includes $a \in \left[-\frac{\sqrt{2}}{2}, -\frac{1}{2}\right]$. For these negative values of $a$, the function $f(x)$ increases as $x$ moves to the right of $-a$, meaning $g(x)$ will exceed $g(0)$ on the interval $[0, 1]$.

Incorrect. This is the standard period of $y = \cos(x)$. To find the period of this function, you must divide $2\pi$ by the coefficient of $x$, which is $2$.

Incorrect. This incorrectly divides the standard period $2\pi$ by the amplitude $3$. The period depends on the coefficient of $x$, not the amplitude.

Incorrect. This is the coefficient of $x$ (often denoted as $B$), which represents the angular frequency, not the period itself.

Incorrect. This is the amplitude of the function, which determines its maximum and minimum values, not its period.

The transformation scales $x$ by a constant factor, which would not result in a quadratic argument like $x(x + 2)$ for the sine function.

This option incorrectly multiplies by 2 instead of dividing by 2 when solving for the original $x$ coordinate, and has the wrong sign for the horizontal translation.

This option has the wrong sign for the horizontal translation. Solving $x' = 2x - 2$ for $x$ gives $x = \frac{1}{2}(x' + 2)$, not $\frac{1}{2}(x' - 2)$.

This option misses the reflection across the x-axis. The $-1$ in the transformation matrix means $y' = -y + 1$, which leads to a negative sine term.

This represents the union of the two domains. However, the domain of a sum function is the intersection of the individual domains, not the union.

This represents the intervals where only one of the functions is defined. For the sum function to be defined, both functions must be defined simultaneously.

This represents the union of the domains excluding the points $-1$ and $3$. The domain of $p + q$ requires finding the overlapping region (intersection) of both domains.

This incorrectly includes the endpoints $-1$ and $3$. The value $-1$ is not in the domain of $q$, and $3$ is not in the domain of $p$, so they cannot be in the domain of $p + q$.

This is the value of the function $N(3) = 15 \ln(22) \approx 46$. This represents the number of koalas (or the population increase) at year 3, rather than the rate at which the population is changing.

This value appears to be the result of calculating $N(3) - 20 \approx 26$. This subtracts the initial population from the model's value at $t=3$, which does not represent the instantaneous rate of change.

This is an incorrect value. It does not correspond to the derivative at $t=3$ or the function value, likely resulting from a calculation error.

While $n=1$ results in a linear function with exactly one $x$-intercept, it is not the highest value among the options.

When $n=2$, the function is a parabola which may have zero, one, or two $x$-intercepts depending on the specific value of $p$.

While $n=3$ guarantees exactly one $x$-intercept because the function is strictly increasing, it is not the highest valid value provided.

While $n=5$ guarantees exactly one $x$-intercept, there is a higher valid value in the options.

The asymptotes of $y = \tan(ax)$ occur at $x = \frac{(2k+1)\pi}{2a}$. If $a=4$, $x = \frac{(2k+1)\pi}{8}$, which cannot equal $\frac{\pi}{6}$ for any integer $k$.

If $a=6$, the asymptotes are at $x = \frac{(2k+1)\pi}{12}$. Setting this to $\frac{\pi}{6}$ requires $2k+1 = 2$, which has no integer solution for $k$.

If $a=10$, the asymptotes are at $x = \frac{(2k+1)\pi}{20}$, which cannot equal $\frac{\pi}{6}$ for any integer $k$.

If $a=12$, the asymptotes are at $x = \frac{(2k+1)\pi}{24}$. Setting this to $\frac{\pi}{6}$ requires $2k+1 = 4$, which has no integer solution for $k$.

This incorrectly assumes the horizontal transformation is $2x + 3 = 1$, which would give $x = -1$. However, the argument is $2x - 3$.

This option uses an incorrect $x$-value by solving $2x + 3 = 1$ and an incorrect $y$-value calculation.

While this correctly identifies $x = 2$, it incorrectly calculates the $y$-value as $y = -1 + 3 = 2$ instead of $y = 1 - 3 = -2$.

This option is incorrect because while $f'(-1)=0$, the second derivative $f''(-1) = e^{-1}(-1+2) = e^{-1}$ is positive, not negative.

This option is incorrect because the first derivative evaluates to zero at $x=-1$, not a negative value, and the second derivative is positive.

This option is incorrect because the first derivative $f'(-1)$ equals $0$, not a value less than $0$.

This incorrect value is half of the correct answer, which may result from a calculation error during the multiplication of fractions or evaluating the trigonometric ratio.

This answer results from evaluating $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ instead of $\cos\left(\frac{\pi}{3}\right)$ in the derivative, or incorrectly assuming the derivative of sine is sine.

This option is incorrect and likely results from misapplying the chain rule or arithmetic errors when combining the constants.

This is the y-coordinate of the vertex (the minimum value of the function), not the x-coordinate required for the domain.

This is one of the roots of the quadratic equation ($x=-5$), not the x-coordinate of the vertex that determines where the function changes direction.

The interval $(-\infty, 0]$ includes the vertex at $x = -1.5$, meaning the function decreases and then increases within this domain, so it is not one-to-one.

This is the other root of the quadratic equation ($x=2$). The interval $(-\infty, 2]$ includes the vertex, so the function is not one-to-one.

Incorrect. The function has a vertical asymptote at $x = -3$, which is outside the interval $[0, 5]$, meaning it is continuous on the given interval.

Incorrect. The square root function is continuous for all values where its argument is non-negative ($x \ge -3$), which fully includes the interval $[0, 5]$.

Incorrect. The cube root function is defined and continuous for all real numbers, so it is continuous on the interval $[0, 5]$.

Incorrect. The sine function is continuous for all real numbers, so its square is also continuous everywhere, including the interval $[0, 5]$.

This transformation results in $x' = \frac{1}{2}x - 4$. Substituting $x = 2x' + 8$ into $y = \cos(x)$ would map the graph to $y = \cos(2x + 8)$, not $y = \cos(2x + 4)$.

This transformation results in $x' = \frac{1}{2}(x - 2) = \frac{1}{2}x - 1$. This would map the original graph to $y = \cos(2x + 2)$, which is incorrect.

This transformation results in $x' = 2(x + 2) = 2x + 4$. This represents a horizontal stretch rather than a compression, mapping the graph to $y = \cos(\frac{1}{2}x - 2)$.

This transformation results in $x' = 2x + 2$. This applies a horizontal stretch instead of a compression, mapping the graph to $y = \cos(\frac{1}{2}x - 1)$.