SCSA Mathematics Specialist Functions and sketching graphs

This option is incorrect because it does not properly account for the domain of the inverse cosine function, which requires the argument to be between $-1$ and $1$.

This would be the domain if $b=1$ and the inequality was incorrectly set to $0 \le \log_e(x) \le 1$ instead of $-1 \le \log_e(x) \le 1$.

This incorrectly assumes the domain of $\cos^{-1}(u)$ is $[0, 1]$ instead of $[-1, 1]$, which would lead to solving $0 \le \log_e(bx) \le 1$.

This incorrectly uses the range of $\cos^{-1}(u)$, which is $[0, \pi]$, instead of its domain $[-1, 1]$ to set up the inequality $0 \le \log_e(bx) \le \pi$.

This option incorrectly calculates the composite function $g(f(x))$ instead of $f(g(x))$ and provides an incorrect range.

This option incorrectly calculates the composite function $g(f(x))$ instead of $f(g(x))$ and provides an incorrect range.

While the rule $\sin(x)\cos(x)$ is equivalent to $f(g(x))$, the range is incorrect because for $0 < x < \frac{\pi}{2}$, the function only takes positive values up to $0.5$.

The rule is correct, but the range incorrectly excludes the maximum value of $\frac{1}{2}$, which is achieved when $x = \frac{\pi}{4}$.

This expression represents the function when $x \le \frac{1}{2}$, where both $2x - 1$ and $x - 3$ are negative, resulting in $y = -(2x - 1) - -(x - 3) = -x - 2$.

This expression represents the function when $x \ge 3$, where both $2x - 1$ and $x - 3$ are positive, resulting in $y = (2x - 1) - (x - 3) = x + 2$.

This is incorrect and likely results from an arithmetic error, such as incorrectly adding the constants when combining the terms.

This is incorrect and likely results from incorrectly evaluating the signs of the absolute value expressions or making an arithmetic error.

The absolute value function ensures all outputs are non-negative, so the range cannot include negative values like $-a$.

This assumes the minimum value is $|-a|=a$. However, since the inner function's range $[-a, b\pi-a]$ includes $0$, the minimum of the absolute value is $0$.

The maximum value of the inner function is $b(\pi)-a = b\pi-a$. The value $b\pi+a$ would only be possible if the inner function was $b\cos^{-1}(x)+a$.

The absolute value function cannot produce negative values, and these bounds do not correctly reflect the transformation of the arccosine range.

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 $\frac{1}{4}$.

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 $-2$.

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. 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 $(h+4)(h-1) \le 0$.