MTH1301 Test Solutions

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)   $$\begin{align*} &\alpha^2\beta+\beta^2\alpha\\ &\text{factorize the given expression to sum and product form}\\ &=\alpha(\alpha\beta)+\beta(\alpha\beta)\\ &=(\alpha+\beta)(\alpha\beta)\\ &=(\frac{-n}{m})(\frac{q}{m})=\frac{-nq}{m^2} \end{align*}$$