VCAA Specialist Mathematics Functions, relations and graphs

Substituting $A=-4$ and combining the fractions yields a numerator of $-4x + 3(x-2) = -x-6$, which does not match the required numerator $x-6$.

Substituting $A=2$ results in a combined numerator of $2x + 3(x-2) = 5x-6$, which is incorrect.

Substituting $A=4$ results in a combined numerator of $4x + 3(x-2) = 7x-6$, which is incorrect.

This is incorrect because if $c = 0$ or $c = -8$, the numerator shares a root with the denominator, resulting in a hole and leaving only one vertical asymptote.

This is incorrect because a rational function can have at most one horizontal asymptote. In this case, the horizontal asymptote is $y = 1$.

This is incorrect because if $c = 0$, the numerator becomes $x^2 + 2x = x(x+2)$, which cancels the $(x+2)$ in the denominator, leaving a vertical asymptote at $x = 2$, not $x = -2$.

This is incorrect because for most values of $c$, there are two vertical asymptotes. Additionally, if $c = -8$, the vertical asymptote is at $x = -2$, not $x = 2$.

This option incorrectly identifies the $x$-values where the absolute value expressions change sign ($x=3$ and $x=-3$), rather than the required values for the constant $a$.

This option incorrectly excludes $a=6$. When $a=6$, $f(x)=0$ for $x \in [-3,3]$, making $\frac{1}{f(x)}$ undefined on that interval, but it still has exactly two vertical asymptotes at $x=3$ and $x=-3$, plus the horizontal asymptote $y=0$.

For $a < 6$, the equation $|x-3|+|x+3|-a=0$ has no real solutions because the minimum value of $|x-3|+|x+3|$ is 6. This would result in zero vertical asymptotes.

This interval represents the range of $x$-values where the function $|x-3|+|x+3|$ reaches its minimum value of 6, not the set of values for the constant $a$.

If $t=\frac{s^2}{4}$, the discriminant of the denominator ($s^2-4t$) is zero. This means the denominator has exactly one real root, resulting in one vertical asymptote.

This condition does not correctly represent the discriminant of the quadratic denominator $x^2+sx+t$.

This inequality does not match the condition for the discriminant of the denominator to be negative.

If $t<\frac{s^2}{4}$, the discriminant ($s^2-4t$) is positive. This means the denominator has two distinct real roots, resulting in two vertical asymptotes.

This option gives the coordinates of the local minima. These occur when $\cos(ax) = 1$, which maximizes the denominator and results in a minimum value of $y = \frac{1}{(1+1)^2 + 3} = \frac{1}{7}$.

At $x = \frac{2\pi k}{a}$, $\cos(ax) = 1$. Substituting this into the equation gives $y = \frac{1}{7}$, not $\frac{1}{3}$.

These x-coordinates correspond to $\cos(ax) = 0$, which gives $y = \frac{1}{(0+1)^2 + 3} = \frac{1}{4}$. This point is neither a local maximum nor a local minimum.

While the x-coordinate correctly makes $\cos(ax) = -1$, substituting this into the equation yields a y-value of $\frac{1}{3}$, not $\frac{1}{4}$.

Incorrect. There are more than 2 integer values for $a$. This answer might result from only considering positive factors of 21.

Incorrect. There are exactly 4 integer values for $a$, corresponding to the 4 pairs of integer factors of -21.

Incorrect. There are only 4 pairs of integer factors of -21, so there can only be 4 possible values for $a$, not 5.

The factor $(x - 2)$ in the denominator cancels with the numerator, creating a removable singularity rather than a vertical asymptote.

As $x \to \pm\infty$, the function behaves like the line $y = x + 5$, meaning it approaches infinity rather than a horizontal asymptote.

Although the rational expression is undefined at $x = 2$, the piecewise definition explicitly fills this "hole" by defining $f(2) = 7$, making it continuous.

This incorrectly counts the solutions by only considering the interval $[0, 2\pi]$ for $3x$, missing the third solution in $[2\pi, 3\pi]$.

This might result from incorrectly assuming asymptotes occur where $\sec(3x)$ is undefined. However, as $\sec(3x) \to \pm\infty$, $f(x) \to 0$, creating removable discontinuities rather than vertical asymptotes.

This incorrectly counts the number of asymptotes, possibly by mistakenly including the points where $\sec(3x)$ is undefined as vertical asymptotes.

This overestimates the number of solutions, likely by incorrectly assuming there are two solutions to $\cos(3x) = -\frac{2}{3}$ in every interval of length $\pi$.

This is incorrect. Expanding and simplifying the numerator of the combined partial fractions yields $-7x + 2$, not $12x + 21$.

This is incorrect. Expanding and simplifying the numerator of the combined partial fractions yields $-7x + 2$, not $7x + 16$.

This is incorrect. Expanding and simplifying the numerator of the combined partial fractions yields $-7x + 2$, not $-5x + 4$.

This is incorrect. Expanding and simplifying the numerator of the combined partial fractions yields $-7x + 2$, not $3x + 6$.

Incorrect. This option incorrectly sets $b = 2$, which would result in an asymptote with a positive slope of $\frac{1}{2}$ instead of $-\frac{1}{2}$.

Incorrect. With $c = 1$ and $b = -2$, the $y$-intercept of the asymptote would be $-\frac{c}{b^2} = -\frac{1}{4}$, which contradicts the given value of $\frac{1}{4}$.

Incorrect. With $a = -2$ and $c = -1$, the $y$-intercept of the function would be $\frac{a}{c} = 2$, which contradicts the given $y$-intercept of $-2$.

Incorrect. If $a=-2$, the derivative at $x=0$ is $f'(0) = -4$, which means the y-intercept is not a stationary point.

Incorrect. This value might result from a sign error when solving the equation $5a - 6 = 0$ for $a$.

Incorrect. If $a=0$, the derivative at $x=0$ is $f'(0) = -1.5$, meaning the y-intercept is not a stationary point.

Incorrect. If $a=2$, the function simplifies to $f(x) = x+3$ (for $x \neq 2$), which has a constant derivative of $1$ and therefore no stationary points.

Setting $y=0$ gives $\frac{1}{1-x^2} = a$. For $0 \le a \le 1$, this equation has fewer than two solutions (e.g., no solutions for $a=0$ and one for $a=1$).

This range includes $0 \le a \le 1$ where the graph has fewer than two $x$-intercepts, and excludes $a < -1$ which correctly yield two intercepts.

This range includes $0 \le a \le 1$ which do not yield two $x$-intercepts, and misses valid values like $a = 2$ or $a = -2$.

While $a < 0$ yields two intercepts, $a = 1$ yields exactly one intercept at $x = 0$, so the inclusion of $a = 1$ makes this incorrect.

The function is undefined when the denominator $x^2 - 1 = 0$, which occurs at $x = 1$ and $x = -1$. Therefore, the domain is indeed $R \setminus \{-1, 1\}$.

As $x$ approaches $\pm\infty$, the denominator grows much faster than the numerator, meaning $f(x)$ approaches 0. Thus, the graph has a horizontal asymptote at $y = 0$.

As $x$ approaches $-1$, the numerator approaches 2 while the denominator approaches 0, which creates a vertical asymptote at $x = -1$.

The vertical axis intercept (y-intercept) is found by evaluating $f(0)$, which gives $\frac{|0 - 1|}{0^2 - 1} = -1$. The graph has a y-intercept at $(0, -1)$.