MCV4U Calculus and Vectors Practice Exam — Free

The function is a product of two pieces, $u = e^{2x}$ and $v = \cos(3x)$, so start with the product rule, $f'(x) = u'v + uv'$. Each piece is itself a composition, so each of $u'$ and $v'$ needs the chain rule.

For $u = e^{2x}$: the derivative of $e^{g(x)}$ is $e^{g(x)}\cdot g'(x)$, so $u' = e^{2x}\cdot 2 = 2e^{2x}$. The exponential survives unchanged; only the inner derivative $2$ comes out in front.

For $v = \cos(3x)$: the derivative of $\cos(g(x))$ is $-\sin(g(x))\cdot g'(x)$, so $v' = -\sin(3x)\cdot 3 = -3\sin(3x)$. Two things must be tracked here: the negative sign that always comes with differentiating cosine, and the inner derivative $3$.

Assemble with the product rule:

Factoring out $e^{2x}$ is optional, but it makes the result easy to compare with the choices.

Each wrong option is a specific slip. Option A keeps both chain-rule factors but drops the minus sign from the derivative of cosine, the single most common trig-derivative error. Option B applies the product rule correctly but forgets the chain rule entirely, as if the derivatives of $e^{2x}$ and $\cos(3x)$ were simply $e^{2x}$ and $-\sin(3x)$. Option D multiplies the two derivatives together, $2e^{2x}\cdot\left(-3\sin(3x)\right)$, instead of using the product rule; the derivative of a product is never just the product of the derivatives.

A quick sanity check: at $x = 0$ the formula gives $f'(0) = 2\cos 0 - 3\sin 0 = 2$, and estimating the slope of $f$ between $x = -0.001$ and $x = 0.001$ also gives $2.00$. Option B gives $1$ at $x = 0$ and option D gives $0$, so both fail immediately. Option A happens to agree at $x = 0$ because the sine term vanishes there, so to separate it from option C you would test any point where $\sin(3x) \ne 0$.

$f'(x) = e^{2x}\left(2\cos(3x) - 3\sin(3x)\right)$ (option C).

Let the printed region be $w$ cm wide and $h$ cm tall, so $wh = 600$ and $h = \dfrac{600}{w}$. Adding the side margins ($3$ cm on each side) makes the page $w + 6$ cm wide, and adding the top and bottom margins ($2$ cm each) makes it $h + 4$ cm tall. The quantity to minimize is the page area:

Differentiate and find the critical number:

rejecting $w = -30$ because a width must be positive. Then $h = \dfrac{600}{30} = 20$.

Confirm that this is a minimum, which a marker expects to see. The second derivative is $P''(w) = \dfrac{7200}{w^3}$, which is positive for every $w > 0$, so the graph of $P$ is concave up on the whole domain and $w = 30$ gives the absolute minimum. (A first-derivative sign chart works too: $P'(29) < 0$ and $P'(31) > 0$, so $P$ decreases and then increases.)

Finally, answer the question that was asked. The page is $w + 6 = 36$ cm wide and $h + 4 = 24$ cm tall, with area $P = (36)(24) = 864$ square centimetres. Check: $P(30) = 624 + 120 + 120 = 864$, and the printed region $wh = (30)(20) = 600$ fits as required.

The common trap is to stop at $30$ cm by $20$ cm. Those are the dimensions of the printed text, not of the page; the question asks for the page, so the margins must be added back. A second slip is attaching the margins to the wrong dimensions, using a page of $(w + 4)$ by $(h + 6)$. That gives $w = 20$, $h = 30$ and a page $24$ cm wide by $36$ cm tall. The area is still $864\ \text{cm}^2$, which makes the slip easy to miss, but the orientation is wrong: the $3$ cm margins are on the sides, so the page must be the wider one, $36$ cm by $24$ cm.

The page should be $36$ cm wide by $24$ cm tall; the minimum page area is $A = 864\ \text{cm}^2$.

(a) A scalar equation $Ax + By + Cz + D = 0$ needs a normal vector $\vec{n} = (A, B, C)$ and one point on the plane. Two direction vectors in the plane are

Their cross product is perpendicular to both, so it is a normal to the plane:

So $5x + 3y + 4z + D = 0$. Substituting $A(1, 0, 2)$ gives $5(1) + 3(0) + 4(2) + D = 0$, so $D = -13$ and the plane is $5x + 3y + 4z - 13 = 0$. Always check with the points you did not use: $B$ gives $5(2) + 3(1) + 4(0) - 13 = 0$ and $C$ gives $5(0) + 3(3) + 4(1) - 13 = 0$.

(b) The direction vector of $L$ is $\vec{d} = (2, -1, 1)$. Since $\vec{n} \cdot \vec{d} = 10 - 3 + 4 = 11 \ne 0$, the direction of $L$ is not perpendicular to the normal, so $L$ is not parallel to $\pi$ and must cross it at exactly one point. Write $L$ in parametric form, $x = 3 + 2t$, $y = -t$, $z = 5 + t$, and substitute into the plane:

Substituting back gives $Q(3 - 4,\ 2,\ 5 - 2) = Q(-1, 2, 3)$. Check: $5(-1) + 3(2) + 4(3) - 13 = 0$. The trap here is stopping at $t = -2$; the parameter is not the answer, the point is.

(c) The distance from $P(x_0, y_0, z_0)$ to the plane $Ax + By + Cz + D = 0$ is

Now $\overrightarrow{PQ} = (-4, 2, -2)$ has length $\sqrt{16 + 4 + 4} = \sqrt{24} = 2\sqrt{6} \approx 4.90$. That is longer because $PQ$ runs along the line $L$, which meets the plane at an angle rather than at right angles. The distance from a point to a plane is measured along the perpendicular, in the direction of $\vec{n}$, and the perpendicular is the shortest segment from $P$ to any point of the plane. Reporting $|\overrightarrow{PQ}|$ as the distance is the common error here. (As a check, the foot of the perpendicular is $F(0.8, -1.32, 3.24)$, which satisfies the plane equation, and $|\overrightarrow{PF}| \approx 3.11$.)

(a) $5x + 3y + 4z - 13 = 0$; (b) $\vec{n}\cdot\vec{d} = 11 \ne 0$, and $L$ meets $\pi$ at $Q(-1, 2, 3)$ (when $t = -2$); (c) $\dfrac{11\sqrt{2}}{5} \approx 3.11$ units, which is less than $|\overrightarrow{PQ}| = 2\sqrt{6} \approx 4.90$ because $PQ$ is not perpendicular to the plane.