\def\section{{\bf Class Time:}}

%\marknumbers{1,20}

\input perm3
\normalafterproblemskip=1cm

\title{Math 105 Test III}
\date{November 26, 1991}


\setproblempermutations{\version 1&2:3,5,10,11,15,16,1,2,4,6,7,8,9,12,13,14,17,18,19,20}


\lastversion=0

\gdef\scriptfrac#1#2{{\scriptscriptstyle{#1}\over\scriptscriptstyle{#2}}}
\let\ss=\scriptstyle

\problem %%  1 %%

If the equation $5x^2y-y^3=2$ defines $y$ implicitly as a 
function of  $x$, then $\frac{dy}{dx}\Bigg|{\vtop{\null\vskip 4pt
\hsize=20pt$\ss x=1$\par\vskip 3pt$\ss y=2$}}=$
\correct$\frac{20}7$
\wrong$\frac{18}7$
\wrong$0$
\wrong$\frac 53$
\wrong$-\frac85$

\problem %%  2 %%
A point is moving along the graph $x^2y^3=4$.
When the point is at $(2,1)$ its $x$ coordinate is changing
at the rate of $-3$ units per minute.
At what rate is the $y$ coordinate changing at this point?
\correct$1$ unit per minute
\wrong$\frac43$ units per minute
\wrong$3$ units per minute
\wrong$4$ units per minute
\wrong$0$ units per minute

\problem %%  3 %%
$\frac{12^{2x}}{6^{-3x}}=$
\correct$2^{7x}3^{5x}$
\wrong$2\cdot3^{4x}$
\wrong$2^{5x}3^{5x}$
\wrong$4^x3^{3x}$
\wrong$\frac{2^{4x}}{3^{2x}}$

\problem %%  4 %%
If $f(x)=(3x^2+2)e^x$, then $f^{\prime}(x)=$
\correct$(3x^2+6x+2)e^x$
\wrong$6xe^x$
\wrong$6x^2e^{x-1}$
\wrong$6x$
\wrong$6x\ln x$

\problem  %%  5 %%
If $2^{5x-3}=4^x$, then $x=$
\correct$1$
\wrong$\ln 2$
\wrong$3\ln2$
\wrong$-1$
\wrong$\frac37$

\gdef\xx{e^{x^2+2x^3}}
\problem %%  6 %%
$\frac d{dx}\xx = $
\correct$(2x+6x^2)\xx$
\wrong$\frac{\xx}{x^2+2x^3}$
\wrong$e^{2x+6x^2}$
\wrong$(x^2+2x^3)e^{x^2+2x^3-1}$
\wrong$\frac{2x+6x^2}{\xx}$

\problem %% 7 %%
The function $f(x)=(3x+1)e^{5x-1}$ has a relative minimum 
whose $x$ coordinate is
\correct$-\frac8{15}$
\wrong$\frac35$
\wrong$\frac53$
\wrong$\frac{15}8$
\wrong$0$

\problem %%  8 %%
Suppose that the function $y=f(x)$ satisfies $y\>^{\prime}=-3y$
and has $f(0)=4$.
Then $f(x) =$
\correct$4e^{-3x}$
\wrong$-3e^{4x}$
\wrong$\frac 34 e^x$
\wrong$e^{\scriptfrac43x}$
\wrong$-\frac 43e^x$

\problem %%  9 %%
$\frac d{dx} (3+e^{x^2})^5 = $
\correct$10xe^{x^2}(3+e^{x^2})^4$
\wrong$5e^{x^2}(3+e^{x^2})^4$
\wrong$5(2xe^{x^2})^4$
\wrong$10xe^{5x^2}$
\wrong$15+5e^{x^2}$

\problem %% 10 %%
$\frac{e^{3\ln 2+1}}{e^2}=$
\correct$\frac 8e$
\wrong$3e$
\wrong$9e^2$
\wrong $\frac e9$
\wrong$\frac9{e^2}$

\problem%% 11%%
If $\frac{e^x(e^{3x-2})}{(e^3)^x}=5$, then $x=$
\correct $2+\ln5$
\wrong$\ln3$
\wrong$1+3\ln2$
\wrong$\frac 13\ln5$
\wrong$2+3\ln5$

\problem %% 12 %%
The derivative of $\ln(3x^2+1)$ is
\correct$\frac{6x}{3x^2+1}$
\wrong$\frac1{3x^2+1}$
\wrong$\frac 2x$
\wrong$6xe^{3x^2+1}$
\wrong$6x\ln(3x^2+1)$

\problem %%  13 %%
The derivative of $\ln(e^{x^2})$ is
\correct$2x$
\wrong$\frac {x^2}e$
\wrong$e^{x^2}$
\wrong$x^2$
\wrong$\frac1{e^{x^2}}$


\problem %%14%%
If $f(x)=(2x-1)(x+27)^3(x-81)$, then $f^{\prime}(x)=$
\correct $(2x-1)(x+27)^3(x-81)\left[\frac 2{2x-1}
+\frac3{x+27}+\frac1{x-81}\right]$
\wrong$\ln\left[(2x-1)(x+27)^3)(x-81)\right]$
\wrong$\frac 2{2x-1}+\frac3{x+27}+\frac1{x-81}$
\wrong $0$
\wrong$2(x+27)^4(x-9)+3(2x-1)(x+27)^2(x+9)$

\problem %% 15 %%
If $3-2\ln x=0$ then $x=$
\correct $e^{3/2}$
\wrong$\root 3 \of{e^2}$
\wrong$-\frac23 e$
\wrong$2e^3$
\wrong $\frac{e^2}3$

 
\problem %% 16 %%
The solution of the equation $(3x+2)5^{4x}-(x-4)5^{4x}=0$
is $x=$
\correct$-3$
\wrong$4\ln 5$
\wrong$1$
\wrong$-\frac12$
\wrong$-\frac 54$

\problem %% 17 %%
The function $y=3\ln(x+1) -(x+1)^3$ has its only maximum at
\correct $(0,-1)$
\wrong$(0,1)$
\wrong$(1,0)$
\wrong$(-1,0)$
\wrong$(1,3\ln2-8)$

\problem %% 18 %%
$\frac{d}{dx}\left[-(\ln x)^3\right]=$
\correct$-3\frac{(\ln x)^2}{x}$
\wrong $\frac{-1}{x^3}$
\wrong$-3(\ln x)^2$
\wrong$\frac 1{e^3}$
\wrong $-\frac{3x}{\ln x}$

\problem %% 19 %%
Mr.~M.~Bags invested \$25,000 in a certificate of deposit at his bank.
If the certificate pays 8\% annual interest compounded continuously,
how much was it worth when it matured 3 years later?
\correct $25,000e^{.24}$ ( or approximately \$31,781.23)
\wrong $25,000e^{2.4}$ ( or approximately \$275,579.41)
\wrong $25,000+.24(25,000)$ ( or \$31,000)
\wrong $25,000(1.08)^{3}$ ( or approximately \$31,492.80)
\wrong $25,000e^{.024}$ ( or approximately \$25,607.26)


\problem %% 20 %%
Ten grams of a certain radioactive material decays to 9 grams after 15 years.
After how many years will 5 grams remain?
\correct $\frac{15\ln 2}{\ln(10/9)}$
\wrong $75$
\wrong $\frac{2\ln(1/2)}{\ln(3/5)}$
\wrong$\frac{15\ln(1/2)}{\ln(10/9)}$
\wrong$\frac{\ln(15)}{2\ln(9)}$

\end
