Math 3A Exam II Practice Answers

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These are the answers to the Exam II Practice.

P-level

1. 1. \(2\)
   2. \(0\)
   3. Diverges to \(\infty\)
   4. If \(a_n\) is a positive null sequence, then the limit diverges to \(\infty\). If \(a_n\) is a negative null sequence, then the limit diverges to \(-\infty\). Otherwise, inconclusive.
   5. Diverges to \(\infty\)
   6. \(1\)
   7. \(0\)
   8. \(1\)
   9. \(0\)
   10. \(1\)
   11. If \(a_n\) is a positive null sequence, then the limit diverges to \(\infty\). If \(a_n\) is a negative null sequence, then the limit diverges to \(-\infty\). Otherwise, inconclusive.
   12. Diverges to \(\infty\)
   13. \(1\)
   14. \(0\)
   15. \(0\)
   16. \(1\)
   17. \(-1\)
   18. \(0\)
2. 1. \(0\)
   2. \(\tfrac{3}{5}\)
   3. \(\tfrac{2}{5+\mathrm{e}^2}\)
   4. \(\arctan(5)+\ln(3)\)

3. \(\mathcal{A}(\Delta p_1p_2p_3)=\tfrac{7}{2}\) and the triangle is positively oriented.

4. \(\mathcal{A}(\Delta p_1p_2p_3)=\tfrac{7}{2}\)

5. The path \(c\) defined on \([0,2\pi]\) is given by \[c(t)=\left(8\cos(-t)+5,8\sin(-t)+3\right).\]

6. 1. Diverges to \(\infty\)
   2. Diverges to \(\infty\)
   3. Diverges to \(-\infty\)
   4. Diverges to \(-\infty\)
7. 1. \((a_n)\) is given by \(a_n=8-\frac{1}{n}\)
   2. \((a_n)\) is given by \(a_n=4+\frac{1}{n}\)

8. \(5\)

9. Not continuous at \(2\).

10. 1. \(9\)
    2. \(f(2)=9\)

12. \(\mathcal{D}(f)=(-\infty,4)\cup(4,\infty)\) and the continuous extension \(\tilde{f}\) of \(f\) to all of \(\mathbb R\) is given by

\[ \tilde{f}(x) = \begin{cases}\frac{x^2-16}{x-4}&\text{if }x\not=4\\8&\text{if }x=4. \end{cases} \]

13. 1. \(\LO((x-6)^3)\), \(\LO((x-5)^2)\), \(\LO((x+2)^4)\)
    2. \(\LO(x^3)\), \(\LO((x-4)^6)\), \(\LO((x+5)^3)\)
14. 1. \(\LO((x-6)^2)\) and \(\Lo((x-6))\)
    2. \(\LO((x-3)^2)\) and \(\Lo((x-3))\)
    3. \(\LO((x+6)^2)\) and \(\Lo((x+6))\)
    4. \(\LO(x)\) and not \(\Lo(x^n)\) for any natural number \(n\).
15. 1. \(m=32\)
    2. \(m=15\)
    3. \(m=19\);
16. 1. \(m=45\)
    2. \(m=14\)
    3. \(m=4\)
    4. \(m=23\)

17. \(\tfrac{33}{20}\)

18. \(P(0)=2\), \(P(1)=\tfrac{13}{5}\), \(P(2)=\tfrac{16}{5}\), \(P(3)=\tfrac{19}{5}\), \(P(4)=\tfrac{22}{5}\), \(P(5)=5\)

19. \(\tau(1)=2.5\), \(\tau(2)=3.75\), and \(\tau(3)=4.75\)

20. 1. \(-10\)
    2. \(18\pi\)
    3. \(-\tfrac{9}{2}\)
21. 1. \(6\)
    2. \(0\)
    3. \(\sin(5)\)
    4. \(0\)
    5. \(0\)
    6. \(\mathrm{e}^\pi\)
    7. \(\mathrm{e}^4+\ln(5)\)
    8. \(1\)
    9. \(\mathrm{e}-1\)
    10. \(20\)
    11. \(-\tfrac{1}{3}\)
    12. Diverges to \(\infty\)
    13. Diverges to \(-\infty\)
    14. Diverges to \(-\infty\)
    15. Diverges to \(\infty\)
    16. \(0\)
    17. Diverges to \(\infty\)
    18. Diverges to \(-\infty\)
    19. \(1\)
    20. Diverges to \(\infty\)
    21. \(2\)
    22. \(2\)
    23. \(4\)
    24. Diverges to \(-\infty\)
    25. Diverges to\(\infty\)
22. 1. Diverges to \(\infty\)
    2. Diverges to \(-\infty\)
    3. Diverges to \(\infty\)
    4. Does not exist
    5. Does not exist
    6. Does not exist
    7. Diverges to \(\infty\)
    8. Diverges to \(\infty\)
    9. \(0\)
    10. Diverges to \(\infty\)
    11. Diverges to \(-\infty\)
    12. \(\tfrac{\pi}{2}\)
    13. Diverges to \(\infty\)
    14. Diverges to \(\infty\)
    15. Diverges to \(\infty\)
    16. Does not exist
    17. Does not exist
    18. Does not exist
    19. \(0\)
    20. \(0\)
    21. Diverges to \(\infty\)
    22. \(-\tfrac{\pi}{2}\)

23. \(\displaystyle\frac{15+\sqrt{2}}{1001}+\frac{1}{2}\)

24. 1. \(\frac{1}{3}\)
    2. \(\frac{1}{2}\)
    3. \(1\)
    4. \(9\)
    5. \(0\)
    6. \(9\)
    7. \(0\)
    8. \(0\)
25. 1. \(0\)
    2. \(1\)
    3. \(\frac{1-\cos(1)}{6}\)
    4. \(0\)
    5. \(1\)
    6. \(\mathrm{e}-1\)
    7. \(\mathrm{e}^2\)

26. No vertical asymptotes, horizontal asymptote at \(y=9\).

27. Vertical asymptote at \(x=7\), horizontal asymptotes at \(y=2\) and \(y=-\mathrm{e}^2.\)

C-level

1. Use the squeeze theorem to show that the sequence converges to \(\tfrac{1}{10}.\)

2. Rewrite the terms of the sequence like this: \[a_n=\tfrac{1}{\sqrt{36+\frac{1}{n}}+6}.\] Use this to show that sequence converges to \(\tfrac{1}{12}.\)

3. 1. \(\tfrac{2}{5}\)
   2. \(0\)
   3. \(\tfrac{9}{10}\)
   4. \(\tfrac{1}{3}\)
   5. \(\tfrac{1}{3}\)

4. The limit equals \(9\) which does not equal \(f(2)\). So, \(f\) is not continuous at \(2\).

5. 1. Diverges to \(\infty\)
   2. Diverges to \(-\infty\)
   3. Limit does not exist
   4. Diverges to \(\infty\)
   5. Diverges to \(\infty\)
   6. Diverges to \(\infty\)
   7. Diverges to \(-\infty\)
   8. Diverges to \(-\infty\)
   9. Diverges to \(\infty\)
   10. Limit does not exist
   11. Limit does not exist
   12. Limit does not exist
   13. Limit does not exist
6. 1. \(\LO((x-3)^3)\) and \(\Lo((x-3)^2)\)
   2. \(\LO((x+6)^3)\) and \(\Lo((x+6)^2)\)
   3. \(\LO(x^2)\) and \(\Lo((x))\)

7. Rewrite the expression like this: \[\frac{5}{6}\cdot\frac{\sin(5(x-4))}{5(x-4)}.\] Then take the limit as \(x\) goes to \(4\) and use \(\lim\limits_{x\to 4}\tfrac{\sin(x-4)}{x-4}=1\) to obtain that the limit is \(\tfrac{5}{6}.\)

8. \(9\)

9. \(\tfrac{1}{2}\)

10. \(\frac{\sqrt{2}}{{2}}\)

11. \(38\)

12. 1. \(\frac{1}{24}\)
    2. \(\frac{15}{16}+\frac{1}{24}\)
    3. \(\sin(5)\)
    4. \(\ln(2)\)
    5. diverges to \(-\infty\)
    6. diverges to \(\infty\)
    7. \(-\frac{8}{\pi}\)
13. 1. \(28\)
    2. \(\dfrac{3}{16-\frac{\pi}{2}}+28\)
    3. \(\mathrm{e}^2\)
    4. \(30\ln(\tfrac{1}{6})+3^{\pi}\)
    5. \(0\)
    6. \(-1\)
14. 1. \(11\)
    2. \(11\)
    3. Yes and the limit equals \(11\).

15. The function \(f\) is continuous because \(\lim\limits_{x\to 0}f(x)=\tfrac{3}{2}\) and it equals \(f(0)\).

16. The function \(f\) can be rewritten as \[f=\log_3\circ g\quad\text{where}\quad g(x)=\frac{9\sin(x-4)}{x-4}.\] The limit of \(g\) as \(x\) approaches \(4\) is \(9\). Because \(\log_3\) is continuous at \(\tfrac{9}{4}\), use the limit law for composite function to conclude that \[\lim_{x\to 4}\log_3\left(\frac{9\sin(x-4)}{x-4}\right)=\log_3(9)=2.\]

17. Take \(I=[0,1]\). The function \(f\) is continuous on \(I.\) Because \(f(0)=-2\) and \(f(1)=2\) and \(0\) is in the interval \((f(0),f(1)),\) use the intermediate value theorem to conclude that \(f(x)=0\) has a solution in \((0,1)\).

18. 1. Simplify and rewrite the difference quotient so that it looks like this: \(16+3h.\) Take the limit to conclude that \(f^\prime (3)=16.\)
    2. Simplify and rewrite the difference quotient so that it looks like this: \(\tfrac{2h+3}{2(h+2)}.\) Take the limit to conclude that \(f^\prime (2)=\tfrac{3}{4}.\)
    3. Simplify and rewrite the difference quotient so that it looks like this: \(-\tfrac{\sin(2h)}{h}.\) Take the limit to conclude that \(f^\prime \left(\tfrac{\pi}{4}\right)=-2.\)
    4. Simplify and rewrite the difference quotient so that it looks like this: \(\tfrac{1}{\sqrt{16+h}+4}.\) Take the limit to conclude that \(f^\prime \left(17\right)=\tfrac{1}{8}.\)
    5. Simplify and rewrite the difference quotient so that it looks like this: \(\exp(6)\cdot\tfrac{\exp(3h)-1}{h}.\) Take the limit to conclude that \(f^\prime \left(2\right)=3\exp(6).\)
19. 1. \(f'(x)=2\cos(2x+1)\)
    2. \(f'(x)=-2\mathrm{e}x\sin(\mathrm{e}x^2)\)
    3. \(f'(x)=4\sec^2(4x+1)\)
    4. \(f'(x)=(10x+1)\mathrm{e}^{5x^2+x}\)
    5. \(f'(x)=10x+10\)

20. \(c^\prime (t)=\langle 3t^2-1,6\cos(3t)\rangle\)

21. 1. The region is a rectangle and the area is \(52\)
    2. The region decomposes into two right triangles and the total signed area is \(0\)
    3. The region is half a circle with radius \(10\) and the area is \(50\pi\)

22. The total area of the left endpoint rectangles is \(31\). The total area of the right endpoint rectangles is \(61\). The total area of the midpoint rectangles is \(44.5\).

23. The right-endpoint rectangles approximation is \(9.75\). The left-endpoint rectangles approximation is \(7.75\). The midpoint rectangles approximation is \(8.625.\) The pictures of the approximations are given below.

24. The right-endpoint rectangles approximation is \(11.25\). The left-endpoint rectangles approximation is \(6.25\). The midpoint rectangles approximation is \(8.625\). The pictures of the approximations are given below.

25. The partition \(P\) is \(P=(1,2,3,4)\) and the left tagging is \(\tau(1)=1\), \(\tau(2)=2\), and \(\tau(3)=3\). And so, \(\mathcal{R}(f,P,\tau)=20.\)

26. \(-21\)

27. \(28\)

28. \(12\)

29. 1. \(2\leq \displaystyle\int_{2}^{3}f(x)\,\mathrm{d}x\leq 5\)
    2. \(4\leq \displaystyle\int_{1}^{3}f(x)\,\mathrm{d}x\leq 10\)
    3. \(28\leq \displaystyle\int_{1}^{3}7f(x)\,\mathrm{d}x\leq 70\)
    4. \(10\leq \displaystyle\int_{1}^{3}(f(x)+3)\,\mathrm{d}x\leq 16\)
30. 1. \(\frac{3^8}{8}-\frac{2^8}{8}\)
    2. \(\frac{5^3}{3}+\frac{3^3}{3}\)
31. 1. \(6\)
    2. \(1+a\)
    3. \(5\)
32. 1. \(4\)
    2. \(a+5\)
    3. \(-1\)
33. 1. \(23\)
    2. \(a+10\)
    3. \(13\)
    4. No, the function \(f\) is not defined at \(x=2\) so it cannot be continuous

34. \(a=5\) and \(b=\tfrac{1+\sqrt{13}}{2}\)

B-level

1. 1. Translate by \(\langle -1,-2 \rangle\) and rotate by \(\theta=\left(\tfrac{4}{5},-\tfrac{3}{5}\right)\) to get that the new vertices of the translated and rotated rectangle \(R^\ast\) are \(\left(0,0\right)\), \(\left(5,0\right)\), \(\left(5,10\right)\), and \(\left(0,10\right)\). A point \(p=(x(p),y(p))\) is inside the rectangle \(R\) if and only if \(p^\star=R_{\theta}(\langle -1,-2\rangle +p)\) is inside \(R^\ast\) rectangle, meaning that we must have \[0\leq x(p^\ast)\leq 5\quad\text{and}\quad 0\leq y(p^\ast)\leq 10.\]
   2. Rotate and translate \((4,5)\) to get that the transformed point to be \(\left(\tfrac{21}{5},\tfrac{3}{5}\right).\) This is inside the transformed rectangle since the \(x\)-coordinate is between \(0\) and \(5\) and the \(y\)-coordinate is between \(0\) and \(10\) and so it is inside the original rectangle.
2. The function \(f\) given below satisfies all requirements:

\[ f(x)= \begin{cases} 2x+1 &\text{if }x<2\\ \frac{5}{(x-3)^2} &\text{if }2\leq x<3\\ \frac{5}{(x-3)^2} &\text{if }3>x. \end{cases} \]

3. First, rewrite the expression like this \[\frac{4x+1}{\sqrt{36x^2+4x+1}+6x},\] and then like this: \[\frac{4+\frac{1}{x}}{\sqrt{36+\frac{4}{x}+\frac{1}{x^2}}+6},\] Take the limit to get \(\tfrac{1}{3}.\)

4. For horizontal asymptotes, calculate \(\lim\limits_{x\to -\infty}f(x)\) and \(\lim\limits_{x\to \infty}f(x)\) to obtain that \(f\) has a horizontal asymptotes at \(y=\tfrac{5}{3}\) and \(y=2+\tfrac{\pi}{2}.\) For vertical asymptotes determining all \(b\) so the \(f\) diverges to \(\infty\) or \(-\infty\) as \(x\) approaches \(b\) from the left, right, or both sides. So \(f\) has vertical asymptote at \(x=-\tfrac{1}{3}\) and \(x=3\).

5. The path \(c\) given below satisfies all requirements:

\[ c(t)=\tfrac{t}{t+1}\langle 7, 4 \rangle +(-2,3). \]

6. The path \(c\) given below satisfies all requirements:

\[ c(t)=\tfrac{t}{2t+1}\langle 7, 4 \rangle +(-2,3). \]

7. The solution is in the interval \(\left[\frac{263}{64},\frac{33}{8}\right].\)

8. Simplify and rewrite the difference quotient so that it looks like this: \[\sin(2x^2+1)(2h+4x)\left(\dfrac{1-\cos(h(2h+4x))}{h(2h+4x)}\right)+\cos(2x^2+1)(2h+4x)\left(\dfrac{\sin(h(2h+4x))}{h(2h+4x)}\right).\] Take the limit to conclude that \(f^\prime (x)=4x\cos(2x^2+1).\)

9. Simplify and rewrite the difference quotient so that it looks like this: \[\mathrm{e}^{2x^2+5x}(2h+4x+5)\left(\frac{\mathrm{e}^{h(2h+4x+5)}-1}{h(2h+4x+5)}\right).\] Take the limit to conclude that \(f^\prime (x)=(4x+5)\mathrm{e}^{2x^2+5x}.\)

10. Simplify the Riemann sum so it looks like this: \[ \frac{27(n+1)(2n+1)}{2n^2}+\frac{27(n+1)}{n}+12. \] Take the limit to get \(66\).

11. 1. \(0.3\)
    2. \(4\)
    3. \(2.5\)
    4. Diverges to \(-\infty\)
    5. Diverges to \(\infty\)
    6. \(0.5\)
    7. \(-1\)
    8. \(1\)
    9. \(-4\)
    10. Diverges to \(\infty\)
    11. Diverges to \(-\infty\)
    12. \(-3\)

A-level

1. Here is a path \(c\) that satisfies all requirements:

\[c(t)=\tfrac{2}{\pi}\arctan(t)\langle -2 ,1\rangle +\left(7,3\right).\]

2. 1. Write out \((f\circ g)(x)\) as \[f(g(x))=\begin{cases}10x+16&\text{if }x<-1\\4&\text{if }x=-1\\ 6\ln\left(\mathrm{e}+\tfrac{1}{2}x+\tfrac{1}{2}\right)&\text{if }x>-1\end{cases}\] and take the limit to get \(6.\)
   2. \(4.\)
   3. No, because \(f\) is not continuous at \(1\).
3. The partition is \(P=(0,2,4,6)\), the left endpoint tagging is \(\tau_L(1)=0\), \(\tau_L(2)=2\), and \(\tau_L(3)=4\), the right endpoint tagging is \(\tau_R(1)=2\), \(\tau_R(2)=4\),and \(\tau_R(3)=6\), the midpoint tagging is \(\tau_M(1)=1\), \(\tau_M(2)=3\), and \(\tau_M(3)=5\). So

\[\mathcal{R}(f,P,\tau_L)=0,\quad \mathcal{R}(f,P,\tau_R)=6,\quad \mathcal{R}(f,P,\tau_M)=6.\]

4. The function \(f\) can be written like this: \[f(x)=\begin{cases}2x-1&\text{if }0\leq x\leq 2\\3&\text{if }2<x\leq 4\\ -(x-4)^2+3&\text{if }4<x\leq 6.\end{cases}\] The derivative of \(f\) is given by \[f^{\prime}(x)=\begin{cases}2&\text{if }0<x<2\\ 0&\text{if }2<x<4\\ -2x+8&\text{if }4\leq x<6.\end{cases}\]

5. The function \(f\) given below satisfies all the requirements: \[ f(x)= \begin{cases} \left(\frac{1}{2}\right)^{x-3}+4&\text{if }x<3\\ 10(x-3)^4+7&\text{if }x\geq 3. \end{cases} \]

6. The polygon \(P\) has vertex set \(\{(1,1),(2,3),(5,6),(-2,7),(-1,2)\}\) so its area equals \(22.5.\)

7. The function \(f\) can be written like this: \[f(x)=\begin{cases}2(x-2)^2+1&\text{if }0\leq x\leq 2\\3&\text{if }3<x\leq 6.\end{cases}\] So the area of \(R\) is given by

\[\int_{0}^3 (2(x-2)^2+1)\,\mathrm{d}x+\int_3^6 3\,\mathrm{d}x.\]

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