Author Topic: MAT1055 Exam Suggested Solutions (Read 2266 times) Tweet Share

Mao · « **on:** June 19, 2008, 03:33:41 pm »

wooo! its over =]

MAT1055 2008 exam suggested solutions

Question 1

(a)
$f(x)=\frac{x+10}{(x+1)(x+4)}$

(i) $\mbox{Domain_f}\in \mathbb{R}\backslash \{-4,-1\}$

(ii) Can f ever equal zero? if so, where?
$x=-10$

(iii) Where are the verticle asymptotoes?
$x=-1$ and $x=-4$ (you should have used limits to check these, but i cbf typing them out here

)

(iv) $f(x)=\frac{3}{x+1}-\frac{2}{x+4}$

Does

demonstrate any asymptotes?

yes. as $x\to -4$ or $x\to -1$ , one of the term becomes constant and the other behaves like an asymptote.
hence asymptotic behaviour.

(v) Are there any holes?

no. all the discontinuities are present as asymptotes. :. no holes

(vi) Where is the function negative/positive?

$\mbox{positive: }-10<x<-4,\; x>-1$
$\mbox{negative: }x<-10,\; -4<x<-1$

(vii) graph cbf'd

(b)
evaluate $\lim_{x\to \infty}f(x)$ and hence state horizontal asymptote.

$\lim_{x\to \infty}f(x)=0$

$\mbox{Horiz. Asympt. at }y=0$

alternatives: C or D

(c)
*will put up on request. i didnt do this one*

(d)
$k(x)=\mbox{tanh }x$
[well, this question is kinda "show this and that and blah", so there really is no point of me typing this up

]

Question 2

(a)
$A=\left[ \begin{array}{cc}1 & -1\\ 1 & 1\\ \end{array} \right]$

(i) $A^2=\left[ \begin{array}{cc}0 & -2\\ 2 & 0\\ \end{array} \right]$

(ii) $|A|=2$

(iii) $A^{-1}=\left[ \begin{array}{cc}1/2 & 1/2\\ -1/2 & 1/2\\ \end{array} \right]$

(iv) $A^{T}=\left[ \begin{array}{cc}1 & 1\\ -1 & 1\\ \end{array} \right]$

(v) solve $\left\{ \begin{align*}x-y &= 1 \\ x+y &= 2 \\ \end{align*}$

$x=\frac{3}{2},\; y=\frac{1}{2}$

(vi) Using the fact that $\left[ \begin{array}{cc}\sqrt{2} & 0\\ 0 & \sqrt{2}\\ \end{array} \right]\cdot \left[ \begin{array}{cc}1/\sqrt{2} & -1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2}\\ \end{array} \right]=A$ , give a geometric argument for why $\left[ \begin{array}{cc}2 & 0\\ 0 & 2\\ \end{array} \right]\cdot \left[ \begin{array}{cc}0 & -1 \\ 1 & 0\\ \end{array} \right]=A^2$

recognising that $\left[ \begin{array}{cc}\sqrt{2} & 0\\ 0 & \sqrt{2}\\ \end{array} \right]$ is a homogenous dilation in both x and y direction and $\left[ \begin{array}{cc}1/\sqrt{2} & -1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2}\\ \end{array} \right]$ is a rotation of 45^o, the matrix A is simply applying these two transformations successively. Hence the matrix $A^2$ will be applying these transformations twice each. as opposed to dilating by a factor of $\sqrt{2}$ , it is now done twice and the dilation factor is 2, hence the
$\left[ \begin{array}{cc}2 & 0\\ 0 & 2\\ \end{array} \right]$ . and the rotation is now 90^o, hence $\left[ \begin{array}{cc}0 & -1 \\ 1 & 0\\ \end{array} \right]$

(b)
$B=\left[ \begin{array}{cccc} 1 & 3 & -3 & 0 \\ 0 & -1 & 1 & 0 \\ -2 & 2 & -2 & 2 \\ 0 & 1 & -1 & 1 \\ \end{array} \right]$

$det(B)=0$ , the second and third column are linearly dependant.

alternatives: c and d

(c)
$\begin{aligh*}<br />x+z &= 5\\<br />2y+z &= 5\\<br />2x+y &= 5\\<br />\end{align*}$

solving with matrices:
$det(B)=-5$

$adj(B)=\left[ \begin{array}{ccc} -1 & 1 & -2 \\ 2 & -2 & -1 \\ -4 & -1 & 2 \\ \end{array}\right]$

$x=2,\; y=1,\; z=3$

(d)
same thing using cramer's rule.

Question 3

(a)

(i) $\frac{d}{dx} x^2e^{-3x}+\cos(4x)=e^{-3x}(2x-3x^2)-4\sin(4x)$

(ii) $\frac{d}{dx} ln(x+1)-\frac{2}{x+1}=\frac{x+3}{(x+1)^2}$

(iii) $\frac{d}{dx} 6\mbox{sinh}^2(x)=6\mbox{sinh}(2x)$

(iv) $\frac{d}{dx} \sin (3x^2+2x+1) = (6x+2)\cos (3x^2+2x+1)$

(b)

$n(x)=e^{-x}+\frac{1}{x}$

(i) $\frac{dn}{dx}=-e^{-x}-\frac{1}{x^2}$

(ii) it is self evident that $\frac{dn}{dx}<0$ for $x>0$ , so i shall not type out all the reasonsing

(iii) graph = bleh~

(iv) $n''=e^{-x}+2\frac{1}{x^3}$
this suggests that the gradient of the tangent is always increasing [i.e. becoming less negative]

alternatives: c or d

(c) = bleh~

(d)

$x^2+25y^2=1$

(i) verify that $\left(0,\frac{1}{5}\right)$ and (-1,0) lie on this ellipsis
~bleh~

(ii) implicit differentiation:

$2x+50y\cdot \frac{dy}{dx}=0$

$\implies \frac{dy}{dx}=-\frac{x}{25y}$

(iii) Show that there are exactly 2 points with tangents parallel to the x axis and 2 points with tangents parallel to the y axis

bleh~ $\left(0,\pm \frac{1}{5}\right),\; \left(\pm 1,0\right)$

(iv) find one point where $\frac{dy}{dx}=1$

simultaneously solving both the derivative=1 and the relation yields:

$\left(-\frac{5}{\sqrt{26}},\frac{\sqrt{26}}{130}\right),\; \left(\frac{5}{\sqrt{26}},-\frac{\sqrt{26}}{130}\right)$

Question 4

(a)
$a=5i+3j$ and $b=-i+2j+4k$

(i) which lies in the x-y plane?
a

(ii) Calculate $a\cdot b$ and hence the angle in between them
$a\cdot b=1,\; \theta \apprx 87.86^o$

(iii) Calculate the vector component of a in the direction of b, and a perpendicular to b

$u_{\parallel}=-\frac{1}{21}i+\frac{2}{21}j+\frac{4}{21}k$

$w_{\perp}=5\frac{1}{21}i+2\frac{19}{21}j-\frac{4}{21}k$

(iv) Calculate the area of the triangle of O, A=a and B=b

$A_{triangle}=\frac{1}{2}A_{parallelogram}=\frac{1}{2}a\cdot b=\frac{1}{2}$

(v) Calculate $a\times b$ and hence find the cartesian equation of the plane contaning the vectors a and b

$a\times b=12i-20j+13k=n$

$n\cdot r=n\cdot r_0$ , and since both these vectors originate from the origin, $r_0$ is a zero vector

$\implies 12x-20y+13z=0$

(b) $r(t)=t^2 i + 2t j - 4k$

(i) $r'(t)=2t i + 2j,\; r''(t)=2i,\; r'(t)\times r''(t)=-4k$

(ii) $\kappa = \frac{4\sqrt{2t+2}}{(2t+2)^2}=\frac{\sqrt{2t+2}}{(t+1)^2}$

alternatives: c or d

(c) = bleh~
ps: this question is wrong, the planes are not parallel... so i hope you havent done it!!

(d)
give the 3x3 matrix that:
[0,0,1] -> [1,0,0]
[0,1,0] -> [0,1,0]
[1,0,0] -> [1,1,1]

$A=\left[ \begin{array}{ccc} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0\\ \end{array} \right]$

Question 4

(a)

(i) $\int x^2 e^{x^3}\; dx = \frac{1}{3}e^{x^3}+c$

(ii) $\int \sin(3x)-e^{4x}\; dx = -\frac{\cos(3x)}{3}-\frac{e^{4x}}{4}+c$

(iii) $\int \frac{3x^2+4}{x^3+4x+8}\; dx = log_e |x^3+4x+8| + c$

(b)

$I=\int_0^{\pi /4} \cos(x)-\sin(x)\; dx = \sqrt{2}-1$

graph + justification = bleh~

(c)

$J(x)=\int_0^x e^{-t}\; dt = 1-e^{-x},\; x>0$

$J'(x)=e^{-x}>0$ , hence as x increase, J increases

J(x) is also always positive. as area under the always positive exponential function is always positive.

alternative: d or e

(d) = bleh~

(e)

$V=\pi \int_0^1 \left(3+e^x\right)^2 \; dx=\pi\left(\frac{e^2}{2}+6e+2.5\right)\approx 70.70\; unit^3$

* Mao :*all done!*

Toothpaste · « **Reply #1 on:** June 19, 2008, 03:57:16 pm »

Was fun

Sif refuse food to write answers Mao

Mao · « **Reply #2 on:** June 19, 2008, 04:11:37 pm »

Quote from: Toothpick on June 19, 2008, 03:57:16 pm

Was fun

Sif refuse food to write answers Mao

pffft i came home to get food

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