
Common Questions Ask in Post UTME Mathematics Exams. Day2
Mathematics
Chima Sunday (Tutor)
22062016 15:34:00 +0000Try your best to solve these questions. Show working on paper when necessary and include a picture of your hand written Solution in your response. Wish you the best!
1. If m is the gradient of the line pq  px  qy = 0; and q ≠ 0. Find 1/m
A. q/p
B. p/q
C. q/p
D. p/q
(oau)Updated Solution
equation of a straight line, y = mx + c; were m = slope or gradient
so for pq  px  qy = 0
making y subject formular
y = (p/q)x + p
so by comparison with equation of straight line, m = (p/q)
so 1/m = (q/p)
Answer is Option C2. If α and β are the roots of the quadratic equation, x^{2}  10x + 2 = 0. Find the value of 1/α^{2} + 1/β^{2}.
A. 26
B. 24
C. 3/2
D. 3
(uniben)Updated Solution
If α and β are the roots of the quadratic equation, then we can construct the equation as x^{2}  (α + β)x + αβ = 0
then comparing this with x^{2}  10x + 2 = 0
it seen that (α + β) = 10; and αβ = 2
so 1/α^{2} + 1/β^{2} = (α^{2} + β^{2})/α^{2}β^{2} = [(α + β)^{2}  2αβ]/(αβ)^{2} (please try and convince yourself that this is true by working it out)
now substituting we that
1/α^{2} + 1/β^{2} = [(10)^{2}  2(2)]/(2)^{2}
= (100  4)/4 = 96/4 = 24
Answer is Option B3. If sec^{2}x + tan^{2}x = 3, find the value of x
A. 30^{o}
B. 45^{o}
C. 60^{o}
D. 90^{o}
(abu)Updated Solution
recall that sin^{2}x + cos^{2}x = 1
divide through by cos^{2}x to have
tan^{2}x + 1 = sec^{2}x
were sec^{2}x = 1/cos^{2}x
substituting sec^{2}x into the question, we have
tan^{2}x + 1 + tan^{2}x = 3
so 2tan^{2}x = 3 1 = 2
tan^{2}x = 1
square rooting both side to have
tanx = 1
thus x = tan^{1}(1) = 45
So answer is Option B

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