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Mathematics LibreTexts

Test 2(Mock Exam)

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    91392
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    These mock exams are provided to help you prepare for Term/Final tests. The best way to use these practice tests is to try the problems as if you are taking the test. Please don't look at the solution until you have attempted the question(s). Only reading through the answers or studying them, will typically not be helpful in preparing since it is too easy to convince yourself that you understand it.  

    Exercise \(\PageIndex{1}\)

    Solve the following Initial Value Problems:

    1. \(3y^2y^{\prime}-xy^3=x, y(0)=0.\)
    2. \(\displaystyle xy' -y = x , y(1) = -1 \)
    Answer
    1. \( y=(e^{\frac{ x^2}{2}}-1)^{1/3}\)
    2. \( y=x^2-2x.\)
    Solution

    1. \(3y^2y^{\prime}-xy^3=x \implies 3y^2y^{\prime}=xy^3+x=x(y^3+1).\)

    \begin{align*}\displaystyle \int \frac{3y^2 dy}{y^3+1} &= \int xdx\\ ln|y^3+1| &=\displaystyle\frac{ x^2}{2}+C\\ y^3+1 &=Ae^{\frac{ x^2}{2}}\\ y &=(Ae^{\frac{ x^2}{2}}-1)^{1/3}\\ \end{align*}

    Now, \( x=0 , y=0 \implies A=1.\)

    Hence \(y=(e^{\frac{ x^2}{2}}-1)^{1/3}.\)

    2. 

    \( \displaystyle xy' -y = x , y(1) = -1 \implies \displaystyle y' - \frac{1}{x}y = 1 , y(1) = -1.\)

    Hence the integrating factor \(= e^{\int  \frac{-1}{x} dx}=e^{-ln(x)}=e^{ln(x^{-1})}=x^{-1}.\)

    \begin{align*} \displaystyle x^{-1} y' -x^{-1}\frac{1}{x}y & = 1 \\\displaystyle \frac{d}{dx}(x^{-1}y) &= 1 \\\displaystyle x^{-1}y &= \int dx \\\displaystyle x^{-1}y & = x+C \\\displaystyle y & = x^2+Cx \\\end{align*}

    Since \(y(1)=-1,-1=(1)^2+C.\) Hence \(C=-2.\)Thus the solution is \( y=x^2-2x.\)

    Exercise \(\PageIndex{2}\)

    Use an appropriate test to determine if the following series converges or diverges. Justify your answer.

    1. \( \sum_{k=2}^{\infty}   \displaystyle \frac{k}{7^k}\)
    2. \( \sum_{k=1}^{\infty}   \displaystyle \frac{k+2}{(2k+1)}\)
    3. \( \sum_{k=2}^{\infty}  \displaystyle \frac{k^k}{(ln(k))^k}\)
    Answer

    Converges, diverges, diverges.

    Solution

    1. By Ratio Test,

    \(\rho=\lim_{k\to \infty} \displaystyle \frac{ (\frac{k+1}{7^{k+1}})}{(\frac{k}{7^k})}= \lim_{k\to \infty} \displaystyle \frac{k+1}{k} \frac{1}{7}= \displaystyle \frac{1}{7} <1.\) Hence the series converges.

    2. By divergence test \(\lim_{k\to \infty} \frac{k+2}{(2k+1)} =\frac{1}{2}\ne 0\), the series diverges.

    3. \begin{align*} \rho &=\lim_{k\to \infty} \displaystyle \sqrt[k]{\frac{ k^k}{(ln(k))^k}} \\&= \lim_{k\to \infty} \displaystyle \frac{ k}{ln(k)} \\&= \lim_{k\to \infty} \displaystyle \frac{1}{ \frac{1}{k} }\mbox{( by L'H\^{o}pital's rule)}\\&= \infty.\end{align*}

    By the Root test,  the series diverges.

     

     

    Exercise \(\PageIndex{3}\)

    Determine whether each of the following series converges or diverges, and if it converges, find its sum.  Justify your answer.

    1. \(\sum_{k=1}^{\infty}  \displaystyle \frac{1}{(k+1)(k+2)}\)
    2. \( 1+ \displaystyle  \frac{\pi}{e}+ \displaystyle \frac{\pi^2}{e^2}+ \displaystyle  \frac{\pi^3}{e^3}+\cdots\)
    Answer
    1.  Convergent telescoping series with sum is \(\frac{1}{2}\).
    2. Divergent geometric series.
    Solution

    1. By using partial fraction decomposition, we get  \( \displaystyle \frac{1}{(k+1)(k+2)}= \displaystyle \frac{1}{k+1}-\frac{1}{k+2}.\)

    Thus \(\sum_{k=1}^{\infty}  \displaystyle \frac{1}{(k+1)(k+2)} = \sum_{k=1}^{\infty}( \displaystyle \frac{1}{k+1}- \displaystyle \frac{1}{k+2}).\) The second part of each term cancels with the first part of the  succeeding term, hence \(S_n= \displaystyle \frac{1}{2}- \displaystyle \frac{1}{n+2}.\) Therefore, \(\lim_{n\to \infty} S_n= \displaystyle  \frac{1}{2}\).

    2. This is a geometric series with first term \(a=1\) and the ratio \(r= \displaystyle  \frac{\pi}{e} >1\). Thus this series diverges.

    Exercise \(\PageIndex{4}\)

    A glass of juice with a temperature of \(40^{\circ} F\) is placed in a room with a constant temperature of \(75^{\circ}F,\) and 1 hour later its temperature is \(55^{\circ}F.\) Using Newton’s Law of Cooling, show that \(t\) hours after the juice is placed in the room its temperature is approximated by \(T=75-35e^{-.56t}.\)

    Solution

    The ambient temperature (surrounding temperature) is \(75°F\), so \(T_s=75\). The temperature of the glass of juice when it placed in a room  is \(40°F\), which is the initial temperature (i.e., initial value), so \(T_0=40\). Therefore Equation becomes

    \[\dfrac{dT}{dt}=k(T−75) \nonumber\]

    with \(T(0)=40.\)

    Rewrite the differential equation by multiplying both sides by \(dt\) and dividing both sides by \(T−75\):

    \[\dfrac{dT}{T−75}=k\,dt. \nonumber\]

    Integrate both sides:

    \[\begin{align*} ∫\dfrac{dT}{T−75} &=∫k\,dt \\ \ln|T−75| &=kt+C.\end{align*} \]

    Solve for \(T\) by first exponentiating both sides:

    \[\begin{align*}e^{\ln|T−75|} &=e^{kt+C} \\ |T−75| &=C_1e^{kt}, & & \text{where } C_1 = e^C. \\ T−75 &=\pm C_1e^{kt} \\ T−75 &=Ce^{kt}, & & \text{where } C = \pm C_1\text{ or } C = 0.\\ T(t) &=75+Ce^{kt}. \end{align*} \]

    Solve for \(C\) by using the initial condition \(T(0)=40:\)

    \[\begin{align*}T(t) &=75+Ce^{kt}\\ T(0) &=75+Ce^{k(0)} \\ 40 &=75+C \\ C &=-35.\end{align*} \]

    Therefore the solution to the initial-value problem is

    \[T(t)=75-35e^{kt}.\nonumber\]

    To determine the value of \(k\), we need to use the fact that after \(1\) hour the temperature of the juice is \(52°F\). Therefore \(T(1)=55.\) Substituting this information into the solution to the initial-value problem, we have

    \[T(t)=75-35e^{kt}\nonumber\]

    \[T(1)=55=75-35e^{k}\nonumber\]

    \[-20=-35e^{k}\nonumber\]

    \[e^{k}=\dfrac{4}{7}\nonumber\]

    \[\ln e^{k}=\ln(\dfrac{4}{7})\nonumber\]

    \[k=\ln(\dfrac{4}{7})\nonumber\]

    \[k≈−0.56.\nonumber\]

    So now we have \(T(t)=75-35e^{−0.56t}.\)

    Exercise \(\PageIndex{5}\)

    For each of the following series determine whether the series converges absolutely, converges conditionally, or diverges.  Justify your answer.

    1. \(\sum_{k=1}^\infty \displaystyle (-1)^{k+1}{3^k \over k!} \)
    2. \(\sum_{k=1}^\infty \displaystyle (-1)^{k+1}{\sqrt{k+3} \over k}\)
    3. \(\sum_{k=1}^\infty \displaystyle (-1)^{k+1}  \left( \frac{k+1}{k} \right)^k\)
    Answer

    Converges absolutely, Converges conditionally, diverges

    Solution

    1. \begin{align*}\rho & = \lim_{k\to\infty}\left|a_{k+1}\over a_k\right|\\&=\lim_{k\to\infty} \dfrac{\left|(-1)^{k+2} {3^{k+1}\over (k+1)!} \right|}{\left|(-1)^{k+1} {3^k \over k!} \right|}\\&= 3 \lim_{k\to\infty} \dfrac{1}{(k+1)}\\&= 0<1. \end{align*}

    Hence by the Ratio test, the series converges absolutely.

    2. Converges conditionally by alternating series test, since \(\displaystyle \sqrt{k+3}/k\) is decreasing. Does not converge absolutely by comparison with p-series, \(\displaystyle p=1/2\)

    3. Diverges by divergence test since \(\displaystyle \lim_{k→∞}|a_k|=e\).


    Test 2(Mock Exam) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Pamini Thangarajah.

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