Instructions
- Please have your student card ready for inspection, turn off all cellular phones, and read all the instructions carefully.
- This test contains two parts: Part A (36 marks) contains nine short questions, and Part B (64 marks) contains four questions. All answers are to be given in this booklet.
- Please do not write anything on the QR codes that appear at the top of each page.
- Check that this test has 12 pages, including this cover page
- There is a formula sheet on page 11, and space for rough work. You can also use page 10 and 12 for rough work. You may tear off the formula sheet, but you must submit it together with the rest of the test.
Make sure to provide exact answers, using symbols such as \(\sqrt{ }, e\) and \(\pi\), if needed.
PART A (36 marks)
In this part, clearly indicate your final short answer in the appropriate box.
You must show your work (if any), even though only the final answer will be graded.
Simplify your answers as much as possible. Each question is worth 4 marks.
- Solve the inequality \(|3x-4|-5\geq0\). Provide your answer using the interval notation.
- Find the natural domain and the range of the function \(h(x)=\sqrt{x+1}-5\). Provide your answer using the interval notation.
- If \(f(x)=\cos x\) and \(g(x)=\frac{2x}{3}\), find \(f\circ g\left(\frac\pi4\right)\) and \(g\circ f\left(\frac{\pi}{4}\right)\) (note that \(\circ\) denotes composition). Simplify your answers, so that they do not include any trigonometric functions.
- Calculate the value of \(2\log_{8}(3)-\log_{8}(18)\). Your answer should be a single number, that does not include any logarithms.
- The function \(p(x)=\sin^{-1}(e^x-2)\) is one-to-one on its domain. Find the expression for its inverse \(p^{-1}(x)\).
- Calculate the value of \(\frac{9^{2/3}\cdot8^{5/6}}{3^{1/3}\cdot2^{1/2}}\). Your answer should be a single integer.
Evaluate the limit
\(\lim_{x\to-2}\frac{x^2+5x+6}{x^2+x-2}\)
Your answer should be a number, or DNE if the limit does not exist.
- Evaluate the limit \(\lim_{x\to0^-}\frac{\sin(8x)}{2\cdot|x|}\). Your answer should be a number, or DNE if the limit does not exist.
- Suppose that a function \(f\), with domain \(\mathbb{R}\), satisfies \(\lim_{x\to3^-}f(x)=f(3)+\frac12\) and \(\lim_{x\to3^+}f(x)=f(3)-\frac{1}{2}\). Is \(f\) continuous at \(x=3\)? Circle the correct answer.
- Yes. \(f\) is continuous at \(x=3\).
- No, \(f\) has a removable discontinuity at \(x=3\).
- No, \(f\) has a jump continuity at \(x=3\).
- No, \(f\) has an infinite discontinuity at \(x=3\).
- No, \(f\) has an oscillating discontinuity at \(x=3\).
PART B (64 marks)
In this part you are required to provide full solutions and to show all your work.
A correct answer obtained with false reasoning or with no reasoning will not receive any marks.
Each question is worth 16 marks.
- Here is a graph of a function \(f\). Answer the questions below. No explanation needed for parts (a) and (b).
(a) What are the domain and range of \(f\)?
(b) For which \(x\) value(s) is \(f(x)=2\)?
(c) Is \(f\) a one-to-one function? Explain.
(d) How many solutions does the equation \(f(x)=x\) have? Explain.
Consider the function \(g(x)=x\cdot\tan x\).
(a) Is the function \(g(x)\) even, odd or neither? Explain.
(b) We create a new function \(h(x)\) as follows: First, we stretch the graph of \(g(x)\) horizontally by a factor of 3 . Then, we reflect it about the \(x\)-axis. Finally, we shift the graph 2 units upwards. Find an explicit formula for \(h(x)\). Explain your answer.
- Suppose that a function \(f\) satisfies \(\cos x+1\leq f(x)\leq\frac{(\sqrt{x}+1)^2-1}{\sqrt{x}}\) for all \(x>0\). Use a theorem discussed in class to find the value of the limit \(\lim_{x\to0^+}f(x)\), or show that it does not exist.
- Use a theorem discussed in class to show that the equation \(3^x-x^3-x=0\) has at least two solutions in the interval \([1,4]\). Write a detailed solution and show all your work.
FORMULA SHEET
\[\sin(A+B)=\sin A\cos B+\cos A\sin B\]
\[\cos^2\theta=\frac{1+\cos(2\theta)}{2}\]
\[\tan\theta=\frac{\sin\theta}{\cos\theta}\]
\[\sin\left(\frac\pi4\right)=\cos\left(\frac\pi4\right)=\frac{\sqrt2}2\]
\[\sin\left({\frac{\pi}{6}}\right)=\cos\left({\frac{\pi}{3}}\right)={\frac{1}{2}}\]
\[\cos(A+B)=\cos A\cos B-\sin A\sin B\]
\[\sin^2\theta=\frac{1-\cos(2\theta)}2\]
\[\sin^2\theta+\cos^2\theta=1\]
\[\sin\left(\frac\pi3\right)=\cos\left(\frac\pi6\right)=\frac{\sqrt3}2\]
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