PRELIMINARY CONTEST 2024
Contest 28
St. John’s Grammar School – 38 pts (Winner)
Accra High School – 32 pts
St. Francis Girls’ SHS – 13 pts
Contest 29
Kumasi Academy – 56 pts (Winner)
Saviour SHS – 16 pts
Battor SHS – 14 pts
Contest 30
Ghana SHS, Koforidua – 49 pts (Winner)
Sefwi Wiawso SHS – 15 pts
Northern School of Business – 09 pts
ROUND 2
SPEED RACE
SECOND QUESTION
Find the values of constants A and B such that $\frac{2x+1}{(x-2)^{2}}=\frac{A}{x-2}+\frac{B}{(x-2)^{2}}$
FORMULA
$\frac{cx+d}{(x-e)^{2}}=\frac{A}{x-e}+\frac{B}{(x-e)^{2}}$
$A=c$
$B=c(e)+d$
SOLUTION
$\frac{2x+1}{(x-2)^{2}}=\frac{A}{x-2}+\frac{B}{(x-2)^{2}}$
$A=2$
$B = 2(2)+1$
$B = 5$
ANSWER: $A=2$ and $B=5$
PRACTICE QUESTIONS
1. Find the values of $A$ and $B$ such that $\dfrac{3x-2}{(x-2)^2}=\dfrac{A}{x-2}+\dfrac{B}{(x-2)^2}$.
ANSWER: $A=3$, $B=4$
SOLUTION 1
For $\dfrac{cx+d}{(x-e)^2}=\dfrac{A}{x-e}+\dfrac{B}{(x-e)^2}$ the constants are:
$A=c,\quad B=ce+d$
Here $c=3$, $d=-2$, $e=2$:
$A=3$
$B=3(2)-2=4$
SOLUTION 2
Multiply both sides by $(x-2)^2$:
$3x-2=A(x-2)+B$
Compare the coefficients of $x$:
$A=3$
Compare the constant terms:
$-2=3(-2)+B$
$B=-2+6=4$
2. Find the values of $A$ and $B$ such that $\dfrac{2x}{(x-3)^2}=\dfrac{A}{x-3}+\dfrac{B}{(x-3)^2}$.
ANSWER: $A=2$, $B=6$
SOLUTION 1
For $\dfrac{cx+d}{(x-e)^2}=\dfrac{A}{x-e}+\dfrac{B}{(x-e)^2}$ the constants are:
$A=c,\quad B=ce+d$
Here $c=2$, $d=0$, $e=3$:
$A=2$
$B=2(3)+0=6$
SOLUTION 2
Multiply both sides by $(x-3)^2$:
$2x=A(x-3)+B$
Compare the coefficients of $x$:
$A=2$
Compare the constant terms:
$0=2(-3)+B$
$B=0+6=6$
3. Find the values of $A$ and $B$ such that $\dfrac{5x+1}{(x-1)^2}=\dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}$.
ANSWER: $A=5$, $B=6$
SOLUTION 1
For $\dfrac{cx+d}{(x-e)^2}=\dfrac{A}{x-e}+\dfrac{B}{(x-e)^2}$ the constants are:
$A=c,\quad B=ce+d$
Here $c=5$, $d=1$, $e=1$:
$A=5$
$B=5(1)+1=6$
SOLUTION 2
Multiply both sides by $(x-1)^2$:
$5x+1=A(x-1)+B$
Compare the coefficients of $x$:
$A=5$
Compare the constant terms:
$1=5(-1)+B$
$B=1+5=6$
4. Find the values of $A$ and $B$ such that $\dfrac{4x-3}{(x+2)^2}=\dfrac{A}{x+2}+\dfrac{B}{(x+2)^2}$.
ANSWER: $A=4$, $B=-11$
SOLUTION 1
For $\dfrac{cx+d}{(x-e)^2}=\dfrac{A}{x-e}+\dfrac{B}{(x-e)^2}$ the constants are:
$A=c,\quad B=ce+d$
Here $c=4$, $d=-3$, $e=-2$:
$A=4$
$B=4(-2)-3=-11$
SOLUTION 2
Multiply both sides by $(x+2)^2$:
$4x-3=A(x+2)+B$
Compare the coefficients of $x$:
$A=4$
Compare the constant terms:
$-3=4(2)+B$
$B=-3-8=-11$
5. Find the values of $A$ and $B$ such that $\dfrac{x+7}{(x-4)^2}=\dfrac{A}{x-4}+\dfrac{B}{(x-4)^2}$.
ANSWER: $A=1$, $B=11$
SOLUTION 1
For $\dfrac{cx+d}{(x-e)^2}=\dfrac{A}{x-e}+\dfrac{B}{(x-e)^2}$ the constants are:
$A=c,\quad B=ce+d$
Here $c=1$, $d=7$, $e=4$:
$A=1$
$B=1(4)+7=11$
SOLUTION 2
Multiply both sides by $(x-4)^2$:
$x+7=A(x-4)+B$
Compare the coefficients of $x$:
$A=1$
Compare the constant terms:
$7=1(-4)+B$
$B=7+4=11$
6. Find the values of $A$ and $B$ such that $\dfrac{2x-5}{(x-3)^2}=\dfrac{A}{x-3}+\dfrac{B}{(x-3)^2}$.
ANSWER: $A=2$, $B=1$
SOLUTION 1
For $\dfrac{cx+d}{(x-e)^2}=\dfrac{A}{x-e}+\dfrac{B}{(x-e)^2}$ the constants are:
$A=c,\quad B=ce+d$
Here $c=2$, $d=-5$, $e=3$:
$A=2$
$B=2(3)-5=1$
SOLUTION 2
Multiply both sides by $(x-3)^2$:
$2x-5=A(x-3)+B$
Compare the coefficients of $x$:
$A=2$
Compare the constant terms:
$-5=2(-3)+B$
$B=-5+6=1$