Concordia MATH 203 Midterm Practice Mock

9 multi-part questions • 100 illustrative marks • Suggested practice time: 90 minutes

This practice paper covers selected topics from weeks 1–6 of Concordia MATH 203 in Fall 2026.

• Work without the solutions or lesson links on your first attempt. Use separate paper and show your reasoning. Give exact answers unless the question says otherwise.
• For a timed attempt, allow about 90 minutes: 82 minutes to work and 8 minutes to check your answers. You can also try the questions without a time limit first.
• All variables and coordinates are real and dimensionless except the position/time quantities in question 7. Angles are in radians. A square root is the nonnegative real root; $\ln$ is the natural logarithm; $e$ is its exponential base.
• A prime denotes differentiation with respect to the function’s input. Ordinary derivatives in this paper are two-sided unless stated otherwise. $\mathbb R$ denotes the real numbers; a square interval bracket includes its endpoint and a round bracket excludes it.
• The questions can be answered exactly without a calculator. For the actual assessment, follow the course’s current calculator and other exam rules.

The official outline lists a 90-minute common midterm on November 1, 2026, starting at 17:30, with weeks 1–6 coverage. Its calculated end time is 19:00. Check Moodle and your instructor’s latest announcements for current arrangements. Source: Fall 2026 outline, page 2.

Question paper

MATH 203 Mock: Marking Guide and Explained Solutions

Illustrative total: 100 marks. Use this suggested marking guide to check your work. Accept mathematically equivalent methods unless a question requires a specific method. Give points for correct reasoning, and count a carried arithmetic error only once when later reasoning is valid. Use the mistakes you find to choose what to practise next.

Compare the earliest incorrect or unjustified step in each answer before reading its next-step link. An answer reconstructed from the solution is not yet an unaided result.

1. Transform and compose

(a) The square-root input must satisfy $3-x\ge0$. Thus the domain is $( -\infty,3]$. The square root can take every nonnegative value, so $1-2\sqrt{3-x}$ takes every value at most 1. The range is $( -\infty,1]$; the maximum 1 occurs at $x=3$.

(b) If $(u,v)$ lies on the base graph, set $3-x=u$. Then $x=3-u$ and the new height is $1-2v$. The coordinate map is

The three images are $(3,1)$, $(2,-1)$ and $(-1,-3)$. Substituting their first coordinates into $T$ verifies their heights.

(c) Substitute $x^2$ into the entire input slot:

The requirement $3-x^2\ge0$ is equivalent to $-\sqrt3\le x\le\sqrt3$. Both endpoints are included because the square root of zero exists.

2. Invert and solve

(a) Since $e^{-x}$ ranges over $(0,\infty)$, $F$ ranges over $( -\infty,3)$. The upper bound 3 is never attained. If $x_1<x_2$, then $e^{-x_1}>e^{-x_2}$; multiplying by $-2$ reverses the inequality. Hence $F(x_1)<F(x_2)$, which proves that $F$ is strictly increasing and one-to-one without a derivative.

(b) Solve $y=3-2e^{-x}$ for $x$:

After exchanging the input/output names,

Its domain is $( -\infty,3)$, the range of $F$. Its range is $\mathbb R$, the domain of $F$. The logarithm argument is positive precisely when $x<3$.

(c) Both logarithm arguments must be positive: $3-u>0$ and $1+u>0$, so $-1<u<3$. On this domain the product rule for logarithms gives

The logarithm is one-to-one on positive inputs, so $6-2u=1+u$. Thus $u=5/3$, which satisfies the restrictions. At that input both sides equal $\ln(8/3)$.

3. Read trigonometric structure

(a) Write the argument as $2(x-\pi/6)$. The amplitude is 2, the period is $2\pi/2=\pi$, and the horizontal shift is $\pi/6$ to the right.

(b) The equation is $\cos(2x-\pi/3)=1/2$. Set $\theta=2x-\pi/3$. The required input interval becomes

All angles in this interval with cosine $1/2$ are $-\pi/3$, $\pi/3$ and $5\pi/3$. Solving $x=(\theta+\pi/3)/2$ gives

Both interval endpoints are solutions and must be retained. The general-angle form $\theta=2k\pi\pm\pi/3$, with integer $k$, followed by interval filtering is equally valid.

4. Read a graph

(a) The filled point on the vertical line $x=0$ is $(0,1)$, so $g(0)=1$. The branch on the left approaches height 2, while the branch on the right approaches height $-1$:

Because the one-sided limits differ, the two-sided limit does not exist. The isolated value 1 changes neither limit.

(b) Both adjacent segments approach $(2,1)$, so the two-sided limit is 1. The actual filled point is $(2,3)$, hence $g(2)=3$. Since the limit and value differ, $g$ is not continuous at 2.

(c) Differentiability would require continuity, so the only possible replacement value is 1. With that value the new function is continuous at 2, but its left and right derivative limits are the slopes of its straight segments:

The slopes disagree, giving a corner. Thus no choice of the single value at 2 makes the function differentiable there: every other choice already fails continuity.

5. Limits and asymptotes

(a) The expression requires $x\ge-1/3$ and $x\ne0$. For nonzero inputs close to 0, multiply by the conjugate:

The cancellation uses $x\ne0$. The simplified expression tends to $3/2$. The initial form $0/0$ is indeterminate; it is not the limit.

(b) Factor $x^2$ inside the root, keeping $\sqrt{x^2}=|x|$. For negative $x$, $|x|=-x$, so

As $x\to-\infty$, the numerator tends to $-2$ and the denominator to 1. The limit is $-2$. The sign is consistent with a positive numerator and a negative denominator for large negative inputs.

(c) Factoring gives

The original domain is $\mathbb R\setminus\{-1,2\}$. On that domain,

Near 2 the numerator of the reduced fraction is positive. The denominator approaches zero from the negative side on the left and the positive side on the right:

Thus $x=2$ is the only vertical asymptote. At the other excluded input,

There is a removable discontinuity (a hole) at $(-1,2/3)$, not a vertical asymptote. Finally, $1/(x-2)\to0$ at both positive and negative infinity, so $y=1$ is the horizontal asymptote in both directions. No other denominator zeros or infinite-end limits can add further vertical or horizontal asymptotes.

6. Two claims about zeros

(a) Define $H(x)=x^3-2x-2$. As a polynomial, $H$ is continuous on the whole closed interval $[1,2]$. Its endpoint values are

Zero lies strictly between those values. By the intermediate value theorem, there is at least one $c\in(1,2)$ for which $H(c)=0$. This argument establishes existence; it does not by itself establish uniqueness.

(b) The endpoint values are indeed $r(0)=-1$ and $r(2)=1$. However, the intermediate value theorem requires continuity on the entire closed interval $[0,2]$. The function is undefined at 1, so that condition fails.

A failed theorem condition alone would not prove that no zero exists. Here we can settle the question directly: $1/(x-1)=0$ is impossible for every permitted input, since its numerator is the nonzero constant 1. Therefore the claim is false and $r$ has no zero in $(0,2)$, or anywhere else in its domain.

7. Position and rate

(a) The endpoint positions are $s(0)=0$ m and $s(2)=3$ m. The average velocity is

This compares a change in position with a change in time; it is not an average of the two positions.

(b) Use nonzero time increments $h$ close to 0. Both positive and negative increments are allowed near $t=2$, while keeping $2+h\ge0$. Since $s(2)=3$, the numerical difference quotient is

Combining the numerator over its common denominator gives

Taking the limit gives

The cancellation precedes the limit; setting $h=0$ in the original quotient would produce the indeterminate form $0/0$.

(c) The point on the graph is $(2,3)$, with time in seconds and position in metres. The tangent line is

Its slope is the instantaneous velocity, $3/4$ metre per second. Its positive sign means the particle is moving in the chosen positive direction at that instant. The slope is a rate, not the position 3 m. It need not equal the average velocity over the earlier two seconds.

8. Differentiate three functions

(a) The two factors are $v(x)=e^{-x^2}$ and $w(x)=\sin(3x)$. The product rule gives $A'=v'w+vw'$. Apply the chain rule within each factor:

The base-e exponential keeps its form when differentiated, then gains the inner derivative $-2x$. The sine gains the inner derivative 3. Substituting both contributions gives one derivative:

Equivalently,

The function and its derivative are defined for every real input, so both domains are $\mathbb R$.

(b) For the numerator $v=x^3+2$ and denominator $w=x^2+1$, the quotient rule gives

Here $v'=3x^2$ and $w'=2x$, so

An equivalent simplified numerator is $x^4+3x^2-4x$. Because $x^2+1>0$ for every real $x$, both domains are $\mathbb R$.

(c) The original function requires $5-x^2\ge0$, giving the closed domain $[-\sqrt5,\sqrt5]$. At interior inputs the chain rule gives

This formula is defined for $-\sqrt5<x<\sqrt5$. The endpoints belong to the original function’s domain, but not to the domain of its ordinary two-sided derivative. Thus the derivative domain is $(-\sqrt5,\sqrt5)$.

9. An implicit curve

(a) At $(1,2)$, $1^2(2)+2^3=2+8=10$, so the point belongs. Write $y'=dy/dx$ and differentiate before substituting the point:

The product $x^2y$ contributes both $2xy$ and $x^2y'$. The chain rule supplies $y'$ when differentiating $y^3$. Collecting gives

For real coordinates, $x^2+3y^2$ is zero only at $(0,0)$. That point does not satisfy the curve equation. Thus the denominator is strictly positive at every point on this curve, so this division loses no curve points.

(b) At $P$, the slope is

The tangent line is

It passes through $P$; its slope also satisfies the differentiated equation $4+13y'=0$ at that point.

(c) A horizontal tangent requires $y'=0$. The denominator is nonzero, so $xy=0$. Consider both possibilities. If $y=0$, the original equation becomes $0=10$, which is impossible. If $x=0$, the original equation gives $y^3=10$, hence $y=\sqrt[3]{10}$.

Therefore the only point is $(0,\sqrt[3]{10})$. Its derivative denominator is $3(\sqrt[3]{10})^2>0$, and its numerator is zero, confirming a horizontal tangent. Both cases of $xy=0$ have been checked.

Topics covered

The paper samples functions and transformations, elementary trigonometric/exponential/inverse/logarithmic functions, rates and tangents, limits including one-sided and infinite-end behaviour, continuity, derivative definitions, and differentiation through the product, quotient, chain and implicit rules. A single paper cannot assess every possible skill in these topics.

The official sequence places derivatives of inverse/logarithmic functions and inverse trigonometric functions in week 7, and related rates and linearization in week 8. Those later topics are not part of this paper. Base-e exponential differentiation is included with the product and chain rules. Confirm any additional rules in your instructor’s assigned work.

Course outline: Fall 2026 weekly topics, page 3, and later topics, page 4.