
Difficult Questions in Mathematics from Post UTMEs
Mathematics
Chima Sunday (Tutor)
20052016 14:42:00 +0000Below are 3 Mathematics questions that between 6080% of students got wrong.
Some of these questions are among the most challenging ones we've seen in any school. Try it and see if you can get it right
1. Find the value of p if the line joining (p, 4) and (6, 2) is perpendicular to the line joining (2, p) and (1, 3)
A. 0
B. 3
C. 4
D. 6
(UNIPORT 2008/2009)Updated Solution
When two lines are perpendicular to each other, the product of their slopes = 1
slope = change in y/change in x
so m_{1}m_{2} = 1
where m_{1} and m_{2} are the slopes for the first line and second line respectively
m_{1} = (2  4)/(6  p) = 6/(6  p)
m_{2} = (3  p)/(1 2) = (3  p)/3
so (3  p)/3 × 6/(6  p) = 1
6(3  p)/3(6  p) = 1
(18 + 6p)/(18 + 3p) = 1
cross multiplying
18 + 6p = 18  3p
6p + 3p = 18 + 18 = 36
so 9p = 36
p = 4
answer is Option C2. x + 2 and x  1 are factors of the expression Lx^{3} + 2Kx^{2} + 24, find the values of L and K
A. L = 6, K = 9
B. L = 2, K = 1
C. L = 2, K = 2
D. L = 0, K = 1
(FUTO 2008/2009)Updated Solution
If x+2 and x1 are factors of a polynimial, i.e x = 2 and 1 are part of the roots of the polynomial, then p(2) and p(1) are zero respectively
so for P(x) = Lx^{3} + 2Kx^{2} + 24
then P(2) = L(2)^{3} + 2K(2)^{2} + 24 = 0
L(8) + 2K(4) + 24 = 0
8L + 8K = 24
dividing both side by 8
L + K = 3 (1)
also P(1) = L(1)^{3} + 2K(1)^{2} + 24 = 0
L + 2K = 24 (2)
adding (2) and (1)
L + K + (L + 2K) = 3  24
3K = 27
dividing both side by 3
K = 9
now recall from equation (1) L + K = 3
but K = 9
so by subject formular L = 3 + 9 = 6
L = 6
So answer is Option A3. If g(x) = 2x + 3 and L(x) = sin^{2}x, find the dy/dx of g[L(x)]
A. 2sin2x
B. 2sin2x
C. cos2x + 3
D. 2sin^{2}x + 3
(UNILAG 2008)Updated Solution
g(x) = 2x + 3, and l(x) = sin^{2}x
thus g[l(x)] = 2(sin^{2}x) + 3
thus the dy/dx = ?
recall that 2sin^{2}x = cos2x  1
so that 2sin^{2}x = cos2x + 1
so that g[l(x)] = cos2x + 1 + 3 = cos2x + 4
dy/dx = (2sin2x) + 0 = 2sinx2x
n/b: dy/dx of cos2x = 2sin2x
answer is Option B

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