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 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 $\frac{1}{4}$.

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 $\pi$.

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.

This range makes the discriminant of the derivative's numerator positive, which would result in the derivative changing signs and the function having turning points.

This results from incorrectly factoring the discriminant inequality $h^2 - 3h - 4 \le 0$ as $(h+4)(h-1) < 0$ instead of $(h-4)(h+1) \le 0$.

This results from incorrectly factoring the discriminant inequality $h^2 - 3h - 4 \le 0$ as $(h+4)(h-1) \le 0$.

This option is incorrect because the question explicitly asks for the form with "linear denominators", and $x^2+b$ is a quadratic denominator.

While mathematically similar, standard partial fraction decomposition absorbs the constant multiplier $a$ into the arbitrary numerator constants, making the $ax$ denominator unnecessary.

This option incorrectly places the constant $a$ inside the linear factors of $x^2+b$. The correct factors of $x^2+b$ are $x+\sqrt{|b|}$ and $x-\sqrt{|b|}$.

This option incorrectly assumes a repeated linear root. Since $b \neq 0$, $x^2+b$ has two distinct roots, so there should be no squared terms like $(x+\sqrt{|b|})^2$ in the denominators.

This value is incorrect. Based on the provided partial fractions, the numerator for the term with denominator $1+x$ is $-1$, not $-3$.

This value has the wrong sign. The numerator given in the partial fraction decomposition is $-1$.

This value is incorrect. It does not match the numerator $-1$ found in the term $\frac{-1}{1+x}$.