{[userprofile.firstName]} {[userprofile.lastName]}
{[userprofile.totalFollowing]} Followers
{[userprofile.totalFollowed]}
{[userprofile.bio]}
{[userprofile.gender]}
{[userprofile.town]}
{[userprofile.state]}
{[userprofile.realNumber]}
{[userprofile.website]}
-
Difficult Problems in mathematics From post utme (ABU 2008)
Mathematics
Chima Sunday (Tutor)
17-05-2016 12:54:00 +0000Hey friends! Checkout this mathematics questions, about 70% of students that attempted it got it wrong. Try it and see if you can get it right
1. X*Y = XY - Y2; Evaluate (2x + y)*(y + 2x)
A. 4xy
B. 4x2
C. 0
D. 1Updated Solution
X*Y = XY - Y2
here X = (2x + y), and Y = (y + 2x) = (2x + y)
thus X*Y = (2x + y)(2x + y) - (2x + y)2
= (2x + y)2 - (2x + y)2
= 0, so answer is Option C2. Evaluate x2 + 1/x2 if x - 1/x = 5
A. 24
B. 25
C. 26
D. 27Updated Solution
x2 + 1/x2 = ?
but x - 1/x = 5
now sqauring both side
(x - 1/x)2 = 52
(x - 1/x)(x - 1/x) = 25
x2 + 1/x2 - 2 = 25
x2 + 1/x2 = 25 + 2 = 27
So answer is Option D3. If (2x - 1)(2x + 3) + (2x + 1) = ax2 + bx + c; Find a,b,c
A. 0,6,6
B. 8,0,0
C. 8,0,6
D. 4,6,-2Updated Solution
(2x - 1)(2x + 3) + (2x + 1) = a2 + bx + c
opening the bracket
4x2 + 6x - 2 - 3 + 2x + 1 = ax2 + bx + c 4x2 + 6x - 2 = ax2 + bx + c
so by comparison
a = 4
b = 6
c = -2
so answer is Option D
-
Chima Sunday (Tutor)
Solutions to the questions above. Keep the fire burning.
0 14-06-2016 13:17:00 +0000
-
{[reply.name]}
{[reply.voteCount]} {[reply.voteCount]} {[reply.created]}
{[reply.voteCount]} {[reply.created]}