You’ve studied hard all semester and you are nearly at the end. Let us help you do your best when it counts the most: the final exam. There’s nothing better than doing a practice math exam to prepare yourself. You need the grade 11 math exam questions and their answers. Doing this exercise with a private tutor will help your brain “chunk” the information. The terms and strategies will come easy during the exam getting you more results and less stress.
What You'll See On The Final Exam
What Type Of Math Is In Grade 11?
The math curriculum in Grade 11 is also called algebra II. The the U.S. students follow a standard class sequence: algebra I in 8th grade, and geometry in 9th grade. Different districts have different math curriculums. In any case, you have to know polynomial functions, radicals, and trigonometric functions by heart.
What You Can Expect On The MCR3U Grade 11 Math Final Exam
The final exam is 3 hours long, has 25 questions, and is worth 25% of the student’s total mark. There are no retakes of a final exam without retaking the course.
MCR3U is the university-track grade 11 math final exam offered in Canada. Students can expect 6 unit tests, 2 assignments and a final exam. See an example gradebook here. Everything in the course syllabus can appear on your final exam. In general, you can expect to see:
Grade 11 Math Final Exam Sample Questions And Their Full Solutions
Are you looking to study material for your Grade 11 math exam? We’ve got you covered. Below you will find questions and answers to actual previous math exams.
Evaluating Functions, Composing Functions and Taking Differences Of A Functions
Evaluating functions is easy when you know that they are like vending machines. You plug in a value to the function and get and output, like you press a button and get a drink.
Question: If f(x) = -x+5 and g(x) = sqrt(2x+1)
then find the following, expanding fully if necessary.
i). f(5) =
ii). g(-a) =
iii). g(g(x)) =
iv). 2*f(3)-g(4)=
Answer:
- f(5) = 0,
- g(-a)=Sqrt(2(-a)+1),
- g(g(x))= sqrt(2*(sqrt(2x+1))+1),
- 2*f(3)-g(4) = -4
Simplify Polynomial Functions And State Their Domain Restrictions
One of the main topics of Grade 11 Math is the analysis of polynomial functions. A part of that analysis is the simplification of math equations involving polynomials.
Question: Simplify the function
F(x) = (x+5)/(x+1) + (x+2)/(x-2)
and state its restrictions.
To simplify polynomial functions, you need to know three things:
📍 That you can only add polynomials if they have a common denominator.
📍The trick to make two terms have a common denominator (Multiply by 1)
📍 How to find the restrictions by seeing what part of the domain is undefined (i.e. When there is a division by 0!).
The answer is 2(x-1)(x+4)/(x+1)(x-2), x cannot be -1, or 2.
Coterminal Angles, What They Are And How To Find Them
By Grade 11, students are expected to know the unit circle by heart and how to use it. The concept of coterminal is about different ways to express the same angle.
Question: State A Coterminal Angle To π/5.
To answer the question “State A Coterminal Angle To π/5”, you need to know three things:
📍 What a coterminal angle means?
📍What the unit circle is (and the ability to recall it).
📍 That π/5 and 11π/5 are the same angle.
The answer is 11π/5, but there’s like infinite answers to this question. Do you see why?
Simplify Rational Expressions/Rational Functions, Leaving Only Positive Exponents
In Grade 11 you will have studied rational functions and should be able to simplify complicated expressions. You will have to use tricks that you have learned throughout your math education, like rewriting perfect squares and cancelling out factors.
Question: Simplify leaving only positive exponents.
To simplify an equation like this question asks, you need to know three things:
📍What perfect squares are and to recall them by heart,
📍What the exponent laws are and how to use them
📍How cancellations work.
The answer to simplifying (x^4/49) * ((49^(1/2)/x^(3/2))^3 is 7/sqrt(x).
Watch this tutorial video or find the written solution below:
It can seem complicated. That’s why we have to break it down into smaller problems.
We have two terms, one on the left and one on the right.
Let’s look at the first term. We should rewrite 49 as 7^2 like this:
Putting it back together, we have X to the four over seven squared
The second term can be rewritten as:
Can we rewrite root X to the nine in terms of x? Yes. Why? Because. X to the nine is equal to root X times root x, nine times. We have four pairs of root X that make four X’s whole and four x times x times x is just X to the four.
We can cancel out common terms like this:
How To Transform A Function
Grade 11 students will be expected to know how functions can transform, and will be asked to do it on their final exam.
Question: Given the point at x=4, what is the transformation of the point -f(2x)-3?
There are three types of transformations that you need to know to answer this question:
- Reflection (associated with variable a)
- Stretching/Compression (associated with variable b)
- Translation (associated with variable k)
Solution: By reading the function you can see what are the values for A, B, and K.
This minus in front means A is equal to minus one. That means the y-values will turn into their opposites and the function will be reflected on the x-axis.
Within the brackets, the 2x means B is equal to 2. That is compressing the graph by a factor of 2.
Now, look at -3. That means K is -3, and that translates the function down by 3 points. Every point will be 3 points lower on the y-axis.
We can use this formula to determine the transformation of the point at X equals four.
After drawing the final result after three transformations, you can read the final resting point of the original point for x=4 is (2,-3). You can also get it by the formula:
Simplify Equations That Have Perfect Squares As Factors.
You can expect to be asked to simplify equations that have square roots in them. For example, a recent Grade 11 math exam asked to simplify √48+5*√3.
Solution: Recognize that √48 can be rewritten as √16*3. Now you separate them into two factors, so √16*3 = 4*√3. Now you can directly evaluate the question which is now:
4*√3 – 5*√3 = -√3
Many students do not know that they can do the last step, but it makes perfect sense if you think about it visually.
Create Functions That Calculate Simple Or Compounding Interest
Grade 11 Math covers financial modeling of simple and compounding interest. Below is an exam question that tests a student’s ability to apply the simple interest formula.
Question: Find the total amount in an account that initially had $1000, after 3 years of simple interest at 4%.
To answer the question, you need to know three things:
📍The difference between simple interest and compound interest,
📍The Simple Interest Formula: Total = Principal * (1 + (Interest Rate * Years))
📍The correct order of operations (BEDMAS: Brackets, Exponents, Division, Multiplication, Addition, Subtraction)
Solution: Recognize that √48 can be rewritten as √16*3. Now you separate them into two factors, so √16*3 = 4*√3. Now you can directly evaluate the question which is now:
4*√3 – 5*√3 = -√3
Many students do not know that they can do the last step, but it makes perfect sense if you think about it visually.
How To Analyze Rational Polynomial Functions And Find Their Characteristics
Every Grade 11 Math Student can expect to be asked to analyze a rational function and give its characteristics. Some characteristics that will be asked are:
- Horizontal Asymptote,
- Vertical Asymptote,
- Y-intercepts,
- X-intercepts,
- Domain of the function,
- Range of the function
Question: For the function f(x) = (5*(2x-1))/(x+5(2x-1)), state the following if they exist:
- Horizontal Asymptote(s)
- Y-intercepts(s)
- Domain
- Range
To answer the question, you need to know:
📍What are the terms Horizontal Asymptote, Y-intercepts, Domain and Range.
📍The way to find these characteristics.
📍How to simplify a rational function by canceling like factors.
Answer:
First, simplify the expression by canceling out (2x-1) in the numerator and denominator.
Now recall the algorithm to find the Horizontal Asymptote:
- If the degree of the numerator (N) is less than the degree of the denominator (D) then the horizontal asymptote is y = 0.
- If N = D then the horizontal asymptote is y = ratio of leading coefficients
- If N > D then there is no horizontal asymptote.
In this case N = 0 and D = 1. Therefore the horizontal asymptote is y = 0, since N < D.
Now, to find the y-intercept, we plug x = 0 into the function to get 5/5 = 1. Thus, the y-intercept is y = 1.
To find the domain, we have to consider all possible x values, and which ones are undefined in the function. We can see that if x = -5 then f(-5) = 5/0 which is an undefined output! That means the domain is all real numbers except -5.
To find the Range, we have to consider what are all the possible y values. Since all possible y-values can be created with this function, the range is all real numbers.
How to Find the Ratios Of Trigonometric Functions Using The Pythagorean Theorem
Students should expect to be able to solve trigonometry questions on their grade 11 math final exam.
Question: If csc(x) = 17/15 is in quadrant 2, state the ratio for cos(x).
To answer this trigonometry question you need to know four things:
📍What do quadrants mean on a cartesian plane
📍What are the 6 trigonometric functions (Sin, Cos, Tan, Arcsin, Csc, Arctan)
📍How they can be applied to a triangle using the “SOHCAHTOA” mnemonic
📍How to use the Pythagorean theorem to find the length of a side of a right-angled triangle.
Solution: Draw a cartesian plane and locate the 4th quadrant in the bottom right. Draw a right-angle triangle.
Recognize that Csc(x) = 1/Sin(x), so if Csc(x) = 17/15, then Sin(x) =15/17 because the reciprocal of 17/15 is 15/17.
Using SOHCAHTOA, we can see that Sin = Opposite/Hypotenuse.
In quadrant 4, draw a right-angled triangle with the opposite side with a length of 15, and the hypotenuse with a length of 17.
We need to determine the adjacent side because cos(x) = Adjacent/Hypotenuse.
Use the Pythagorean theorem to find the adjacent side is 8.
Therefore the answer is cos(x) = 8/17.
Complex Functions Word Problems
If you have not seen complex numbers, these questions may not appear on you Grade 11 Math Exam.
Complex Function Word Problems For Advanced Math Students
Question: Patricia is studying a polynomial function f(x). three given roots of f(x) are negative 11 minus startroot 2 endroot i, 3 + 4i, and 10. patricia concludes that f(x) must be a polynomial with degree 4. which statement is true?
Answer: Without plotting the function, you can say that is not true that f(x) has degree 4. We can say this with confidence because there are 5 roots. Three roots are given, but two are complex. The complex conjugate root theorem tells us that complex roots are always found in pairs. The complex roots are always in the form a+/-bi. So, both 3-4i and -11+2i must also be roots.
- The root x+11sqrt(2)i has the conjugate pair x-11sqrt(2)i
- The root
both 3-4i and -11+2i must also be roots. Every root contains an “x”, so if we expanded it out using FOIL (First, Outer, Inner Last) we will get x^3 as the highest degree. That means the function is a polynomial with degree 5 not 4, because
Simplifying Multivariable Polynomial Functions And Determining The Degree
Question: which statement is true about the polynomial –3x4y3 + 8xy5 – 3 + 18x3y4 – Question: 3xy5 after it has been fully simplified? it has 3 terms and a degree of 5. it has 3 terms and a degree of 7. it has 4 terms and a degree of 5. it has 4 terms and a degree of 7.
Answer: Simplify the polynomial function by factoring and cancelling factors. Then you will find the result when you cannot simplify further has 4 terms, and a degree of 7. Therefore the statement “it has 4 terms and a degree of 7” is true.
Applying Algebra To Functions Of Probability
Question: Ximena answered 72 questions correctly on her multiple choice history final and earned a grade of 36%. how many total questions were on the final exam?
To find the total number of questions on the final exam, you can use the following formula:
Total questions = (Number of correct answers / Percentage) * 100
In this case, you know that Ximena answered 72 questions correctly and earned a grade of 36%, so you can plug those values into the formula:
Total questions = (72 / 36%) * 100 = 200
Answer: there were 200 questions on the final exam.
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