PRELIMINARY CONTEST 2024
Contest 34
T.I. AMASS, Kumasi – 71 pts (Winner)
Amenfiman SHS –40 pts
Tuobodom SHTS – 21 pts
Contest 35
Accra Girls’ SHS – 39 pts (Winner)
Notre Dame Girls’ SHS – 34 pts
Aburi Girls’ SHS –27 pts
Contest 36
Anlo SHS – 69 pts (Winner)
Tamale Islamic Science SHS –34 pts
Yendi SHS –28 pts
ROUND 2
SPEED RACE
QUESTION
Find the value of $k$ if $\displaystyle\int_0^1(3x^2-4kx+5)\,dx=0$.
ANSWER: $k=3$
SOLUTION 1
Apply $\int_0^1 x^n\,dx=\dfrac{1}{n+1}$ directly to each term:
$\dfrac{3}{3}-\dfrac{4k}{2}+5=0$
$1-2k+5=0$
$6-2k=0$
$2k=6$
$k=3$
SOLUTION 2
$\int_0^1(3x^2-4kx+5)\,dx=0$
$\left[x^3-2kx^2+5x\right]_0^1=0$
$(1-2k+5)-(0)=0$
$6-2k=0$
$2k=6$
$k=3$
PRACTICE QUESTIONS
Find the value of $k$.
1. $\displaystyle\int_0^1(6x^2-2kx+3)\,dx=0$
ANSWER: $k=5$
SOLUTION 1
Apply $\int_0^1 x^n\,dx=\dfrac{1}{n+1}$ directly to each term:
$\dfrac{6}{3}-\dfrac{2k}{2}+3=0$
$2-k+3=0$
$5=k$
$k=5$
SOLUTION 2
$\int_0^1(6x^2-2kx+3)\,dx=\left[2x^3-kx^2+3x\right]_0^1=0$
$(2-k+3)-(0)=0$
$5-k=0$
$k=5$
2. $\displaystyle\int_0^1(3x^2-6kx+8)\,dx=0$
ANSWER: $k=3$
SOLUTION 1
Apply $\int_0^1 x^n\,dx=\dfrac{1}{n+1}$ directly to each term:
$\dfrac{3}{3}-\dfrac{6k}{2}+8=0$
$1-3k+8=0$
$9=3k$
$k=3$
SOLUTION 2
$\int_0^1(3x^2-6kx+8)\,dx=\left[x^3-3kx^2+8x\right]_0^1=0$
$(1-3k+8)-(0)=0$
$9-3k=0$
$k=3$
3. $\displaystyle\int_0^1(3x^2-2kx+4)\,dx=0$
ANSWER: $k=5$
SOLUTION 1
Apply $\int_0^1 x^n\,dx=\dfrac{1}{n+1}$ directly to each term:
$\dfrac{3}{3}-\dfrac{2k}{2}+4=0$
$1-k+4=0$
$5=k$
$k=5$
SOLUTION 2
$\int_0^1(3x^2-2kx+4)\,dx=\left[x^3-kx^2+4x\right]_0^1=0$
$(1-k+4)-(0)=0$
$5-k=0$
$k=5$
4. $\displaystyle\int_0^1(3x^2-2kx+1)\,dx=0$
ANSWER: $k=2$
SOLUTION 1
Apply $\int_0^1 x^n\,dx=\dfrac{1}{n+1}$ directly to each term:
$\dfrac{3}{3}-\dfrac{2k}{2}+1=0$
$1-k+1=0$
$2=k$
$k=2$
SOLUTION 2
$\int_0^1(3x^2-2kx+1)\,dx=\left[x^3-kx^2+1x\right]_0^1=0$
$(1-k+1)-(0)=0$
$2-k=0$
$k=2$
5. $\displaystyle\int_0^1(3x^2-2kx+2)\,dx=0$
ANSWER: $k=3$
SOLUTION 1
Apply $\int_0^1 x^n\,dx=\dfrac{1}{n+1}$ directly to each term:
$\dfrac{3}{3}-\dfrac{2k}{2}+2=0$
$1-k+2=0$
$3=k$
$k=3$
SOLUTION 2
$\int_0^1(3x^2-2kx+2)\,dx=\left[x^3-kx^2+2x\right]_0^1=0$
$(1-k+2)-(0)=0$
$3-k=0$
$k=3$
6. $\displaystyle\int_0^1(6x^2-2kx+1)\,dx=0$
ANSWER: $k=3$
SOLUTION 1
Apply $\int_0^1 x^n\,dx=\dfrac{1}{n+1}$ directly to each term:
$\dfrac{6}{3}-\dfrac{2k}{2}+1=0$
$2-k+1=0$
$3=k$
$k=3$
SOLUTION 2
$\int_0^1(6x^2-2kx+1)\,dx=\left[2x^3-kx^2+1x\right]_0^1=0$
$(2-k+1)-(0)=0$
$3-k=0$
$k=3$