• Common Questions Ask in Post UTME Mathematics Exams. Day2

    Mathematics

    Chima Sunday (Tutor)
    22-06-2016 15:34:00 +0000

    Try 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 C

    2. If α and β are the roots of the quadratic equation, x2 - 10x + 2 = 0. Find the value of 1/α2 + 1/β2.
    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 x2 - (α + β)x + αβ = 0
    then comparing this with x2 - 10x + 2 = 0
    it seen that (α + β) = 10; and αβ = 2
    so 1/α2 + 1/β2 = (α2 + β2)/α2β2 = [(α + β)2 - 2αβ]/(αβ)2 (please try and convince yourself that this is true by working it out)
    now substituting we that
    1/α2 + 1/β2 = [(10)2 - 2(2)]/(2)2
    = (100 - 4)/4 = 96/4 = 24
    Answer is Option B

    3. If sec2x + tan2x = 3, find the value of x
    A. 30o
    B. 45o
    C. 60o
    D. 90o
    (abu)

    Updated Solution
    recall that sin2x + cos2x = 1
    divide through by cos2x to have
    tan2x + 1 = sec2x
    were sec2x = 1/cos2x
    substituting sec2x into the question, we have
    tan2x + 1 + tan2x = 3
    so 2tan2x = 3 -1 = 2
    tan2x = 1
    square rooting both side to have
    tanx = 1
    thus x = tan-1(1) = 45
    So answer is Option B



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