MTH1301 Test Solutions
(1) Let the universal set be \begin{align*}
&\bigcup=\{x:x\ \text{is a natural number less than} 10\}\\
&A=\{x:x\ \text{is an odd number}\}\ \text{and}\
B=\{x:x\ \text{is a prime number}\}\\
\end{align*}
find (i)$A\Delta{B}$  (ii)$A^c-B^c$

Solution:
firstly, we rewrite the sets from set builder to the builder method.
\begin{align*}
&\bigcup=\{1,2,3,4,5,6,7,8,9\}\\
&A=\{1,3,5,7,9\}\\
&B=\{2,3,5,7,9\}\\
&A^c=\{2,4,6,8\}\\
&B^c=\{1,4,6,8,9\}
\end{align*}
(i)the symmetric difference $A\Delta{B}$ of sets is given by $A\Delta{B}=(A\backslash{B})\bigcup(B\backslash{A})$,
$A\Delta{B}=(A\bigcap{B^c})\bigcup(B\bigcap{A^c})$,
$A\bigcap{B^c}=\{1,9\}$, and $B\bigcap{A^c}=\{2\}$
$A\Delta{B}=\{1,2,9\}$
(ii)$A^c-B^c=\{2\}$ hence the order of $A^c-B^c$ is 1
(2) The $(2m)^th$ term of an A.P is 50,more than the $mth$ term, and $(m+1)^th$ is 56, find the first term.
Solution: Let x be the $mth$ term so that
\begin{align*}
&x=a+(m-1)d..............(i)\\
&x+50=a+(2m-1)d..........(ii)\\
&\text{since the (2m)th term is 50 more than the $mth$ term}\\
&56=a+((m+1)-1)d\\
&56=a+md...........(iii)\\
&\text{substitute (i) in (ii)}\\
&a+(m+1)d+50=a+(2m-1)d\\
&a+md-d+50=a+2md-d\\
&md+50=2md\\
&md=50\\
&\text{substitute md=50 into (iii) so that}\\
&56=a+50\\
&a=6
\end{align*}
(3) find the remainder when $f(x)=3x^3-4x^2+5x-1$ is divided by $x-3$
Solution:
using the polynomial long division method you will get
$\sqrt[x-3]{3x^3-4x^2+5x-1}=59$ or better still you can use the simple short method, let $x=3$ and substitute into $f(x)$ so that $f(3)=3(3)^3-4(3)^2+5(3)-1=59$
(4) Obtain the first 4 terms of the expansion of $(2x-y)^5$
Solution: with out wasting any time,lets use the pascal triangle to expand for $n=5$, the pascal triangle is $(1 5101051) $
\begin{align*}
&(2x-y)^5=(2x)^5+5(2x)^4(-y)+10(2x)^3(-y)^2\\
&+10(2x)^2(-y)^3+5(2x)(-y)^4+(-y)^5\\
&=32x^5-80x^4y+80x^3y^2-40x^2y^3+10xy^4-y^5\\
&\text{hence the first four terms of the expansion is}\\
&=32x^5-80x^4y+80x^3y^2-40x^2y^3
\end{align*}
(5) If $\alpha$ and $\beta$ are the roots of the equation $mx^2+nx+q=0$. find the value of
(i)$\alpha^2+\beta^2$   (ii)$\alpha^2\beta+\beta^2\alpha$
Solution: first we find the sum and product of the equation which is $\alpha+\beta=\frac{-b}{a}=\frac{-n}{m}$ and $\alpha\beta=\frac{c}{a}=\frac{q}{m}$
(i)  \begin{align*}
&\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\\
&=(\frac{-n}{m})^2-2(\frac{q}{m})=\frac{n^2}{m^2}-\frac{2q}{m}\\
&=\frac{n^2-2mq}{m^2}
\end{align*}
(ii)   \begin{align*}
&\alpha^2\beta+\beta^2\alpha\\
&\text{factorize the given expression to sum and product form}\\
&=\alpha(\alpha\beta)+\beta(\alpha\beta)\\
&=(\alpha+\beta)(\alpha\beta)\\
&=(\frac{-n}{m})(\frac{q}{m})=\frac{-nq}{m^2}
\end{align*}