MCR3U Functions Practice Exam — Free

Read the four features of a sinusoidal function off the two given points, then assemble the equation in the form $y = a\cos(k(x - d)) + c$.

Amplitude. Half the distance from the minimum up to the maximum: $a = \dfrac{11 - 3}{2} = 4$.

Axis. The midline is the average of the maximum and the minimum: $c = \dfrac{11 + 3}{2} = 7$.

Period. A maximum and the next minimum are half a cycle apart, so the half-period is $90^\circ - 30^\circ = 60^\circ$ and the full period is $120^\circ$. Then $k = \dfrac{360^\circ}{120^\circ} = 3$.

Phase shift. A basic cosine curve starts at its maximum, so a cosine model is shifted until its first maximum lands on $x = 30^\circ$: $d = 30^\circ$, giving $y = 4\cos(3(x - 30^\circ)) + 7$.

Check both points. At $x = 30^\circ$: $y = 4\cos 0^\circ + 7 = 11$. At $x = 90^\circ$: $y = 4\cos 180^\circ + 7 = 3$. Both match, so option C is correct. (A sine model describes the same graph: a sine curve reaches its maximum a quarter-period, here $30^\circ$, after it starts, so $y = 4\sin(3x) + 7$ is equivalent.)

Each distractor is one common slip. Option A, with $x + 30^\circ$, is the sign error on the phase shift: the form is $x - d$ with $d = 30^\circ$, so a shift to the right is $x - 30^\circ$; option A puts the maximum at $x = -30^\circ$ and gives $y = 3$ at $x = 30^\circ$. Option B uses $k = 6$, which comes from treating the $60^\circ$ gap between the maximum and the minimum as a whole period instead of half of one; its value at $x = 90^\circ$ is 11, not 3. Option D uses $a = 8$, the full max-to-min distance instead of half of it; that function swings between 15 and $-1$ rather than 11 and 3.

$y = 4\cos(3(x - 30^\circ)) + 7$ (option C).

Start with the function itself, because the inverse's domain and range come straight from it. The square root needs $x - 2 \ge 0$, so the domain of $f$ is $\{x \in \mathbb{R} \mid x \ge 2\}$. Since $\sqrt{x - 2} \ge 0$, the smallest output is $\sqrt{0} + 1 = 1$, so the range of $f$ is $\{y \in \mathbb{R} \mid y \ge 1\}$. (The graph is the basic square-root curve translated 2 units right and 1 unit up, with its endpoint at $(2, 1)$.)

To find the inverse, write $y = \sqrt{x - 2} + 1$, swap $x$ and $y$, then solve for $y$:

so $f^{-1}(x) = (x - 1)^2 + 2$.

An inverse swaps the roles of input and output, so the domain of $f^{-1}$ is the range of $f$, $\{x \in \mathbb{R} \mid x \ge 1\}$, and the range of $f^{-1}$ is the domain of $f$, $\{y \in \mathbb{R} \mid y \ge 2\}$. Geometrically, $f^{-1}$ is the right-hand half of a parabola with vertex $(1, 2)$: the reflection of the square-root graph in the line $y = x$.

The trap is to stop at $y = (x - 1)^2 + 2$ and leave the domain as all real numbers. That full parabola is not the inverse of $f$, because squaring in the step $x - 1 = \sqrt{y - 2}$ quietly discards the condition $x - 1 \ge 0$ that the square root forces. For example, $x = 0$ gives $(0 - 1)^2 + 2 = 3$, but $f(3) = 2$, not 0, so the left half of the parabola does not undo $f$. A quick check on the valid half: $f(6) = \sqrt{4} + 1 = 3$ and $f^{-1}(3) = (3 - 1)^2 + 2 = 6$.

$f^{-1}(x) = (x - 1)^2 + 2,\ x \ge 1$; domain $\{x \in \mathbb{R} \mid x \ge 1\}$, range $\{y \in \mathbb{R} \mid y \ge 2\}$

All three parts use the same interest rate per compounding period and the same number of periods. The rate is 4.8%/a compounded monthly, so $i = \dfrac{0.048}{12} = 0.004$ per month and $n = 4 \times 12 = 48$ months. The growth factor $(1.004)^{48} \approx 1.21120656$ appears in every part: compute it once and keep all the decimals in your calculator, rounding only at the end.

(a) Compound interest. A single deposit grows by the compound-interest formula $A = P(1 + i)^n$:

The deposit grows to about $3,028.02, so the interest earned is $A - P \approx 3028.02 - 2500 = 528.02$, about $528.02. Trap: using $i = 0.048$ with $n = 4$ gives $3,015.68, which is what annual compounding would pay; monthly compounding needs the monthly rate and the number of months.

(b) Future value of an annuity. The $100 deposits form an ordinary simple annuity: equal payments at the end of each month, with the payment interval matching the compounding period. Their total value at the end of the 4 years is

Add the lump sum from part (a): $A + FV \approx 3028.02 + 5280.16 = 8308.18$, so her savings total about $8,308.18. (She deposited $2,500 plus 48 deposits of $100, or $7,300 in all, so $1,008.18 of the total is interest.) The $2,500 is not part of the annuity: it was deposited once, today, so it is handled separately with the formula from part (a).

(c) Solving for the payment. The lump sum already grows to $3,028.02, so the monthly deposits must supply the rest, $12{,}000 - 3028.02 = 8971.98$. Set the annuity's future value equal to this amount and solve for $R$:

She needs to deposit about $169.92 at the end of every month. Check: $R \times 52.80164 \approx 169.92\times 52.80164 \approx 8972.05$, and adding $3,028.02 gives just over $12,000, so the rounded payment does reach the goal. The trap here is to set the whole $12,000 equal to the annuity's future value, which gives $227.27 per month, forgetting that the lump sum does part of the work.

(a) $3,028.02, of which $528.02 is interest; (b) $8,308.18; (c) about $169.92 per month.