MHF4U Advanced Functions Practice Exam — Free

The plan is to rewrite $\log_2\left(\dfrac{45}{8}\right)$ using only logarithms of $3$, $5$ and $2$, because those are the values we know: $\log_2 3 = a$, $\log_2 5 = b$ and $\log_2 2 = 1$.

Start with the quotient law, $\log_c\left(\dfrac{M}{N}\right) = \log_c M - \log_c N$:

Now factor each number. Since $3^2 \times 5 = 45$, the product law and then the power law give

Since $2^3 = 8$, we have $\log_2 8 = \log_2 2^3 = 3\log_2 2 = 3$; in words, "2 to what power gives 8?" Putting the pieces together:

A quick numerical check: $a = \log_2 3 \approx 1.585$ and $b = \log_2 5 \approx 2.322$, so $2a + b - 3 \approx 2.492$, and indeed $2^{2.492} \approx 5.625 = \dfrac{45}{8}$.

Each distractor is a specific error. Option A, $\dfrac{2a + b}{3}$, comes from treating the logarithm of a quotient as a quotient of logarithms, $\log_2 45 \div \log_2 8$; there is no such law. Option B, $2a + b - 8$, subtracts $8$ itself instead of $\log_2 8 = 3$: the quotient law subtracts the logarithm of the denominator, not the denominator. Option D, $a^2 + b - 3$, writes $\log_2 9$ as $(\log_2 3)^2$; the power law moves the exponent in front as a multiplier, $\log_2 3^2 = 2\log_2 3$, and never squares the logarithm.

$\log_2\left(\dfrac{45}{8}\right) = 2a + b - 3$ (option C).

The equation mixes the angle $2x$ with the angle $x$, so the first job is to get everything in terms of $x$. Use the double-angle identity $\sin 2x = 2\sin x\cos x$:

Both terms contain $\cos x$, so factor it out rather than dividing by it:

A product is zero when one of its factors is zero, which gives two separate equations.

Case 1: $\cos x = 0$. On $0\le x<2\pi$ this happens at $x = \dfrac{\pi}{2}$ and $x = \dfrac{3\pi}{2}$.

Case 2: $2\sin x + 1 = 0$, so $\sin x = -\dfrac12$. The related acute angle is $\dfrac{\pi}{6}$, because $\sin\dfrac{\pi}{6} = \dfrac12$. Sine is negative in quadrants III and IV, so $x = \pi + \dfrac{\pi}{6} = \dfrac{7\pi}{6}$ and $x = 2\pi - \dfrac{\pi}{6} = \dfrac{11\pi}{6}$.

Check each candidate in the original equation. At $x = \dfrac{7\pi}{6}$: $\sin\dfrac{7\pi}{3} = \sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$ and $\cos\dfrac{7\pi}{6} = -\dfrac{\sqrt{3}}{2}$, so the sum is $0$. At $x = \dfrac{11\pi}{6}$: $\sin\dfrac{11\pi}{3} = \sin\dfrac{5\pi}{3} = -\dfrac{\sqrt{3}}{2}$ and $\cos\dfrac{11\pi}{6} = \dfrac{\sqrt{3}}{2}$, again $0$. At $x = \dfrac{\pi}{2}$ and $x = \dfrac{3\pi}{2}$, both $\sin 2x$ and $\cos x$ equal $0$.

The trap is to divide both sides of $2\sin x\cos x = -\cos x$ by $\cos x$. That leaves $\sin x = -\dfrac12$ and only the two quadrant III and IV answers; the solutions $\dfrac{\pi}{2}$ and $\dfrac{3\pi}{2}$, where $\cos x = 0$, are thrown away. Dividing by an expression that can equal zero always risks losing solutions, so factor instead. A second common slip is to stop after $\dfrac{7\pi}{6}$ and forget the quadrant IV angle $\dfrac{11\pi}{6}$; sine is negative in two quadrants, so there are two angles with $\sin x = -\dfrac12$.

$x = \dfrac{\pi}{2},\ \dfrac{7\pi}{6},\ \dfrac{3\pi}{2},\ \dfrac{11\pi}{6}$

(a) The factor theorem says $x - k$ is a factor of $f(x)$ exactly when $f(k) = 0$. Any rational zero has the form $\dfrac{p}{q}$ with $p$ a factor of the constant term $6$ and $q$ a factor of the leading coefficient $2$, so the candidates are $\pm 1, \pm 2, \pm 3, \pm 6, \pm\dfrac12, \pm\dfrac32$. Testing: $f(1) = 2 - 3 - 11 + 6 = -6$, so $x - 1$ is not a factor, but $f(3) = 54 - 27 - 33 + 6 = 0$, so $x - 3$ is. Synthetic division by $x - 3$ on the coefficients $2$, $-3$, $-11$, $6$ gives $2$, $3$, $-2$ with remainder $0$, so

The quadratic factors by decomposition: $2x^2 + 3x - 2 = 2x^2 + 4x - x - 2 = (2x - 1)(x + 2)$. Therefore

with zeros $-2$, $\dfrac12$ and $3$. Expanding the product back out is the check.

(b) A polynomial can only change sign at a zero, so the three zeros split the number line into four intervals. Test one point in each: $f(-3) = -42 < 0$, $f(0) = 6 > 0$, $f(1) = -6 < 0$ and $f(4) = 42 > 0$. This matches a cubic with a positive leading coefficient: negative on the far left, positive on the far right, crossing at each simple zero. Because the inequality is $\ge 0$, the zeros are included:

The common slip is to answer $x \ge -2$ after the first zero, forgetting that the cubic dips below the axis again between $\dfrac12$ and $3$.

(c) $h(x) = \dfrac{2x - 1}{x + 2}$ is undefined where the denominator is $0$, so the domain is $\{x \in \mathbb{R} \mid x \ne -2\}$ and the vertical asymptote is $x = -2$ (the numerator is $-5$ there, not $0$, so it is an asymptote, not a hole). Numerator and denominator have the same degree, so the horizontal asymptote is the ratio of leading coefficients, $y = 2$.

To solve $h(x) > 2$, do not multiply both sides by $x + 2$: its sign is unknown, so you cannot tell whether the inequality flips. A student who multiplies anyway gets $2x - 1 > 2x + 4$, that is $-1 > 4$, and wrongly concludes there is no solution. Instead, bring everything to one side and combine:

The numerator is a negative constant, so the fraction is positive exactly when the denominator is negative: $x + 2 < 0$, giving $x < -2$. Check: $h(-3) = \dfrac{-7}{-1} = 7 > 2$, while $h(0) = -\dfrac12$ is not greater than $2$. The graph agrees: writing $h(x) = 2 - \dfrac{5}{x + 2}$ shows the curve sits above its horizontal asymptote $y = 2$ only on the left branch, where $x + 2 < 0$.

(a) $f(x) = (x + 2)(2x - 1)(x - 3)$; (b) $x \in [-2, \tfrac12] \cup [3, \infty)$; (c) domain $x \ne -2$, vertical asymptote $x = -2$, horizontal asymptote $y = 2$, and $h(x) > 2$ exactly when $x < -2$, that is $x \in (-\infty, -2)$.