MCV4U Final Exam Review: Free Practice Exam

Free MCV4U Practice Exam

24 questions (27 parts), up to 3 from each of the 8 units, taken from the free tests on Martian Lab. Do them in one sitting, on paper, then open the answers.

1. Limits

   Determine $\displaystyle\lim_{x \to 2} 4$.

   Show answer

   $4$

2. Limits

   Determine $\displaystyle\lim_{x \to 2} 3^{-x}$.

   Show answer

   $\dfrac{1}{9}$

3. Limits

   Determine $\displaystyle\lim_{x \to 3} f(x)$, where $f(x) = \begin{cases} 2x-4, & x < 3 \\ 2, & x > 3 \end{cases}$.

   Show answer

   $2$

4. Derivatives

   Differentiate the following. Leave answers in simplified form with positive exponents only.

   1. $f(x) = \sqrt{3}x^4 + \dfrac{2}{\sqrt{x}}$
      Show answer

      $f(x) = \sqrt{3}x^4 + \dfrac{2}{\sqrt{x}} = \sqrt{3}\, x^4 + 2x^{-1/2}$ $f'(x) = 4\sqrt{3}\, x^3 - x^{-3/2}$ $= 4\sqrt{3}\, x^3 - \dfrac{1}{x^{5/2}}$
   2. $y = (7x^2 + 1)^3(2 - 5x)^7$
      Show solution

      $y' = 3(7x^2{+}1)^2 \cdot 14x \cdot (2{-}5x)^7 + (7x^2{+}1)^3 \cdot 7(2{-}5x)^6(-5)$ $= (7x^2{+}1)^2(2{-}5x)^6\bigl[42x(2{-}5x) - 35(7x^2{+}1)\bigr]$ $= (7x^2{+}1)^2(2{-}5x)^6(-455x^2{+}84x{-}35)$ $= -7\,(7x^2{+}1)^2(2{-}5x)^6(65x^2{-}12x{+}5)$

5. Derivatives

   Given the graph of $y=f(x)$, give a rough sketch of the graph of $y=f'(x)$.

   

   Show solution

   Rough sketch of $y=f'(x)$.

6. Derivatives

   If $f(x) = 2x^{10}$, what is $f^{(10)}(x)$?

   Show solution

   $f'(x) = 2 \times 10\, x^9$ $f''(x) = 2 \times 10 \times 9 \cdot x^8$ $f^3(x) = 2 \times 10 \times 9 \times 8 \cdot x^7$ $\therefore\; f^{(10)}(x) = 2 \times 10 \times 9 \times 8 \times 7 \cdots 1\, x^0$ $= 2 \times 10!$

7. Curve Sketching

   Given $f(x) = \dfrac{x-4}{x^2-x-2}$, complete the following over its entire domain. State the equation of the vertical asymptote(s) and use limits to analyze the behaviours of $f(x)$ near its vertical asymptote.

   Show solution

   $V.A. :\ x = 2\ ,\ x = -1$ $\displaystyle\lim_{x \to 2^+} f(x) = -\infty$ $\displaystyle\lim_{x \to 2^-} f(x) = +\infty$ $\displaystyle\lim_{x \to -1^+} f(x) = +\infty$ $\displaystyle\lim_{x \to -1^-} f(x) = -\infty$

8. Curve Sketching

   Given $f(x) = \dfrac{x-4}{x^2-x-2}$, complete the following over its entire domain. State the equation of the horizontal asymptote and use limits to analyze the end behaviours of $f(x)$

   Show solution

   $\displaystyle\lim_{x \to \infty} \dfrac{x-4}{x^2-x-2}$ $\displaystyle= \lim_{x \to \infty} \dfrac{x^2\!\left(\dfrac{1}{x} - \dfrac{4}{x^2}\right)}{x^2\!\left(1 - \dfrac{1}{x} - \dfrac{2}{x^2}\right)}$ $\displaystyle= \lim_{x \to \infty} \dfrac{0}{1}$ $= 0$ $\therefore\ H.A\ \text{is}\ y = 0$ $\lim_{x \to \infty} f(x) = 0^+$ $\therefore\ \text{as}\ x \to \infty,\ y \to 0\ \text{from above}$ $\lim_{x \to -\infty} f(x) = 0^-$ $\therefore\ \text{as}\ x \to -\infty,\ y \to 0\ \text{from below}$

9. Curve Sketching

   Given $f(x) = \dfrac{x-4}{x^2-x-2}$, complete the following over its entire domain. Locate all critical points. You may round to 1 decimal places.

   Show solution

   $f'(x) = \dfrac{1\cdot(x^2-x-2)-(x-4)(2x-1)}{(x^2-x-2)^2}$ $= \dfrac{x^2-x-2-2x^2+9x-4}{(x^2-x-2)^2}$ $= \dfrac{-x^2+8x-6}{(x^2-x-2)^2}$ $\therefore\ f'(x)=0$ $x = \dfrac{-8 \pm \sqrt{40}}{-2}$ $= 4 \pm \sqrt{10}$ $x \approx 7.16 \quad \text{or} \quad 0.84$

10. Optimization

    You are given a piece of sheet metal that is twice as long as it is wide and has an area of 800 m$^2$. Find the dimensions of the rectangular box that would contain a maximum volume if it were constructed from this piece of metal by cutting out squares of equal area at all four corners and folding up the sides. The box does not have a lid. Include a diagram with your setup/solution.

    Show solution

    Let $x$ be the width; $2x$ is the length. $2x \cdot x = 800$ $x^2 = 400$ $x = 20$ Let $y$ be the side length of the square that gets cut off. Maximize volume $V$. $V = (2x-2y)(x-2y)y$ $= (40-2y)(20-2y)y$ $= (800 - 120y + 4y^2)y$ $= 4y^3 - 120y^2 + 800y$, where $0 \le y \le 10$ $v' = 12y^2 - 240y + 800 = 0$ $3y^2 - 60y + 200 = 0$ $y \approx 4.2$ or $y = 15.7$; reject $y = 15.7$ since $0 \le y \le 10$. $V(0) = V(10) = 0$ $V(4.2) = 1540$ (maximum) $20 - 8.4 = 11.6$ $40 - 8.4 = 31.6$ $\therefore$ A box with dimensions $11.6\text{ m} \times 31.6\text{ m} \times 4.2\text{ m}$ would yield the maximum volume.

11. Optimization

    A ship is 25 kilometres north of a raft when it started sailing south at a rate of 4 km/h and the raft started to drift east at a speed of 5 km/h. Determine the shortest distance between the ship and the raft. Include a diagram in your setup/solution.

    Show solution

    Let $t$ be the time in hours after the ship and raft start moving. Minimize distance $d$. $d = \sqrt{x^2+y^2}$ $x = 5t$ $y = 25 - 4t$ $0 \le t \le \dfrac{25}{4}$ $d = \sqrt{(5t)^2 + (25-4t)^2}$ $= \sqrt{625 - 200t + 41t^2}$ $d'(t) = \dfrac{1}{2}\left(41t^2 - 200t + 625\right)^{-\frac{1}{2}} \cdot (82t - 200) = 0$ $t = \dfrac{200}{82} = \dfrac{100}{41}$ $d(0) = 25$ $d(6.25) = 31.25$ $d\!\left(\dfrac{100}{41}\right) \approx 19.5$ $\therefore$ The minimum distance between the ship and the raft is about 19.5 km.

12. Optimization

    A tall building is 27 m away from an 8 m wall that is parallel to it. A long ladder on the ground is to go over the wall and lean onto the building as shown. Determine the length of the shortest ladder that is required.

    

    Show solution

    Minimize the length of the ladder $L$. Divide the length $L$ into two parts as shown in the diagram. $L = x + y$ Use similar triangles: $\dfrac{y}{8} = \dfrac{x}{\sqrt{x^2 - 27^2}}$ $y = \dfrac{8x}{\sqrt{x^2-729}}$ $L = x + \dfrac{8x}{\sqrt{x^2-729}}$ $L'(x) = \dfrac{(x^2-729)^{3/2} - 5832}{(x^2-729)^{3/2}} = 0$ $(x^2 - 729)^{3/2} = 5832$ $x^2 - 729 = 324$ $x^2 = 1053$ $x \approx 32.4$ $L \approx 46.9$ $\therefore$ The minimum length of the ladder is about 46.9 m.

13. Derivatives of Trig, Exp and Log

    Is the derivative of a sinusoidal function always periodic? Explain why or why not with an example.

    Show solution

    The derivative of $\sin(x)$ is $\cos(x)$ and the derivative of $\cos(x)$ is $-\sin(x)$. Both derivative functions are periodic. Since all sinusoidal functions can be expressed as $\sin(x)$ or $\cos(x)$, their derivative would also be periodic. $y = \tan x$, $y' = \sec^2 x$ is also periodic.

14. Derivatives of Trig, Exp and Log

    If $f(x) = \dfrac{1}{3}e^{6x} + 4\sin(x)$, find $f''\!\left(\dfrac{\pi}{3}\right)$.

    Show solution

    $f'(x) = \dfrac{1}{3}e^{6x} \cdot 6 + 4\cos(x) = 2 \cdot e^{6x} + 4\cos(x)$ $f''(x) = 2e^{6x} \cdot 6 - 4\sin(x) = 12e^{6x} - 4\sin(x)$ $f''\!\left(\dfrac{\pi}{3}\right) = 12 \cdot e^{2\pi} - 4 \cdot \sin\!\left(\dfrac{\pi}{3}\right) = 12e^{2\pi} - 2\sqrt{3}$.

15. Derivatives of Trig, Exp and Log

    Find the equation of the tangent line to the graph of $f(x) = e^{-\frac{x}{2}}$ that is perpendicular to the line $x - 2y - 6 = 0$. Express your answer in EXACT value.

    Show solution

    $2y = x - 6$, $y = \dfrac{1}{2}x - 3$ $\therefore\; m_\perp = -2$ $f'(x) = e^{-\frac{x}{2}} \cdot \dfrac{1}{2} = -2$ $e^{-\frac{x}{2}} = 4$ $\dfrac{-x}{2} = \ln 4$ $x = -2\ln 4$ $y = e^{-\frac{1}{2}(-2\ln 4)} = 4$ $y = mx + b$ $4 = (-2)(-2\ln 4) + b$ $b = 4 - 4\ln 4$ $\therefore$ The equation of tangent is $y = -2x + (4 - 4\ln 4)$.

16. Geometric Vectors

    Describe how the vectors $3\vec{v}$ and $-3\vec{v}$ are related $\vec{v}$. Include diagrams in your description.

    Show solution

    They are collinear or parallel. They have different magnitude. $\vec{v}$ and $3\vec{v}$ have the same direction. $\vec{v}$ and $-3\vec{v}$ have opposite directions.

17. Geometric Vectors

    Given $\vec{v} = 12\,\text{m/s}$ at a bearing of $020^{\circ}$. What are the horizontal and vertical components, $\vec{v}_h$ and $\vec{v}_v$ respectively, of the vector?

    Show solution

    $|\vec{v}_y| = 12 \times \sin 70^{\circ}$ $\approx 11.28\,\text{m/s}$ $|\vec{v}_x| = 12 \cos 70^{\circ}$ $\approx 4.10\,\text{m/s}$

18. Geometric Vectors

    Given $|\vec{a}| = 8$, $|\vec{b}| = 12$, and $|\vec{a} - \vec{b}| = 10$, determine the angle between the vectors $\vec{a}$ and $\vec{b}$ to the nearest degree. Include a diagram in your answer.

    Show solution

    $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2 \times |\vec{a}|\,|\vec{b}|\cos\theta$ $100 = 64 + 144 - 2 \times 8 \times 12 \times \cos\theta$ $\cos\theta = \dfrac{108}{192},\; \theta \approx 55.8^{\circ}$ $\therefore$ Angle between the two vectors is about $55.8^{\circ}$.

19. Cartesian Vectors

    Consider position vector $\vec{u} = [-2, 3, -4]$,

    1. Draw $\vec{u}$ by showing it with a prism.
       
       Show solution

       
    2. Find the exact value of the magnitude of $\vec{u}$.
       Show answer

       $|\vec{u}| = \sqrt{(-2)^2 + (3)^2 + (-4)^2}$ $= \sqrt{4 + 9 + 16}$ $= \sqrt{29}$
    3. Find the unit vector that is in the direction of $\vec{u}$.
       Show answer

       $\dfrac{1}{|\vec{u}|}\vec{u}$ $= \dfrac{1}{\sqrt{29}}[-2, 3, -4]$ $= \left[-\dfrac{2}{\sqrt{29}}, \dfrac{3}{\sqrt{29}}, -\dfrac{4}{\sqrt{29}}\right]$

20. Cartesian Vectors

    Given that $\vec{u} = [-4, 1, -8]$ and $\vec{v} = [8, -2, 1]$, determine the angle between $\vec{u}$ and $\vec{v}$.

    Show solution

    $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta$ $\cos\theta = \dfrac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|}$ $= \dfrac{[-4, 1, -8][8, -2, 1]}{9\sqrt{69}}$ $= \dfrac{-42}{9\sqrt{69}}$ $\theta = \cos^{-1}\!\left(\dfrac{-42}{9\sqrt{69}}\right)$ $\doteq 124.18^\circ$ $|\vec{u}| = \sqrt{(-4)^2 + (1)^2 + (-8)^2}$ $= 9$ $|\vec{v}| = \sqrt{(8)^2 + (-2)^2 + (1)^2}$ $= \sqrt{69}$ $\therefore\ \theta = 124.18^\circ$

21. Cartesian Vectors

    Given $\vec{u} = [2, -3, 3]$ and $\vec{v} = [3, 1, 4]$, determine the vector projection of $\vec{v}$ onto $\vec{u}$.

    Show solution

    $\operatorname{proj}_{\vec{u}}\vec{v} = \left(\dfrac{\vec{v} \cdot \vec{u}}{\vec{u} \cdot \vec{u}}\right)\vec{u}$ $= \dfrac{[3, 1, 4][2, -3, 3]}{[2, -3, 3][2, -3, 3]}\,[2, -3, 3]$ $= \dfrac{15}{22}\,[2, -3, 3]$ $= \left[\dfrac{15}{11},\; -\dfrac{45}{22},\; \dfrac{45}{22}\right]$ $\therefore\ \operatorname{proj}_{\vec{u}}\vec{v} = \left[\dfrac{15}{11},\; -\dfrac{45}{22},\; \dfrac{45}{22}\right]$

22. Lines and Planes

    Which vector equation represents the line perpendicular to the line shown?

    

    1. $\vec{r} = \left(0,-1\right)+s\!\left(1,2\right), s \in \mathbf{R}$
    2. $\vec{r} = \left(0,-1\right)+s\!\left(-2,1\right), s \in \mathbf{R}$
    3. $\vec{r} = \left(0,-1\right)+s\!\left(2,1\right), s \in \mathbf{R}$
    4. none of these

    Show answer

    b) $\vec{r} = \left(0,-1\right)+s\!\left(-2,1\right), s \in \mathbf{R}$

23. Lines and Planes

    Which of the following is a direction vector for the line with scalar equation $3x +2y - 1 = 0$?

    1. $\vec{m} = [3,2]$
    2. $\vec{m} = [-2,3]$
    3. $\vec{m} = [3,-2]$
    4. none of these

    Show answer

    b) $\vec{m} = [-2,3]$

24. Lines and Planes

    Which of the following vectors is perpendicular to the line passing through the points $P\!\left(2,3\right)$ and $Q\!\left(7,10\right)$?

    1. $\vec{v} = [5,7]$
    2. $\vec{v} = [7,-5]$
    3. $\vec{v} = [-3,2]$
    4. $\vec{v} = [-10,7]$

    Show answer

    b) $\vec{v} = [7,-5]$

Unit-by-Unit Checklist

Tick what you can do without notes. Anything you cannot tick, open the unit page and do its full free test.

MCV4U Exam Review FAQ

Start Practicing

No credit card required to start.