
Difficult Question for Math JAMB 2006
Mathematics
Admin Admin
29012016 15:39:00 +0000This is another questions from jamb 2006, in which over 90% of the participant fail,now give a try and see if you can have it correct.
Question 4: Find the value of x for which the function 3x^{3}= 9x^{2} is minimum
A. 3 B. 2 C. 0 D. 5
Question 24: If the mean of five consecutive integers is 30, find the largest of the numbers
A. 30 B. 28 C. 34 D. 32
UPDATED_ANSWERS TO QUESTION ABOVE
Question 4. Find the value of x for which the function 3x^{3}=9x^{2} is minimium
Solution
Correct option is C
3x^{3}9x^{2}=0
Factorizing yield x^{2}(3x9)=0
x=0 or x=2.
∴ x=0 is the minimum value.
Question 24. If the mean of five consecutive integers is 30,find the largest of the numbers.
Solution
let the consecutive numbers be x, x+1, x+2,x+3, x+4.
x + x + 1 + x + 2 + x + 3 + x + 4/5 = 30
5x + 10 = 30 x 5
simplifying for x, we have
x = ^{140}/_{5} = 28
x + 4 = 28 + 4 = 32 which is the largest number.

KUFRE IKPE
C. 34+32+30+28+26 =150, 150/5 = 30 so the largest number is 34 c. pls let me know if I got it right.
0 30012016 18:36:00 +0000

Sandra Asagade
(x+x+1+x+2+x+3+x+4)/5=30 (5x+10)/5=30 X+2=30 X=28 The largest of the numbers x+4=28+4=32 The answer is 32D
0 30012016 18:43:00 +0000

Joce Ekeanya
(x)+(x+1)+(x+2)+(x+3)+(x+4)/5=30 5x+10=30*5 5x=15010 X=28 ..therefore d largest number is 32, D
0 30012016 19:00:00 +0000

Cyprian Udebuani
No 1: 3X*3=9X*2 Collect like times 3X*3  9X*2=0 (3X  9X)*3  2=0 (3x)*1=0 3X=0 3X/3=0/3 X=0
0 30012016 19:25:00 +0000

OAA Oke
3x*3=9x*2 Collct lyk term 3x*39x*2=0 Dy/dx=9x*218x=0 Divide through by 9 X*22x=0 X[x2]=0 X=0, x2=0 X=0, x=2 For which the function is minimum therefore ur highest value is your minimum I.e x=2
0 31012016 07:30:00 +0000

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