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    Difficult Questions in Mathematics from Post UTMEs

    Mathematics

    Chima Sunday (Tutor)
    20-05-2016 14:42:00 +0000

    Below are 3 Mathematics questions that between 60-80% 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 m1m2 = -1
    where m1 and m2 are the slopes for the first line and second line respectively
    m1 = (-2 - 4)/(6 - p) = -6/(6 - p)
    m2 = (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 C

    2. x + 2 and x - 1 are factors of the expression Lx3 + 2Kx2 + 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 x-1 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) = Lx3 + 2Kx2 + 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 A

    3. If g(x) = 2x + 3 and L(x) = sin2x, find the dy/dx of g[L(x)]
    A. -2sin2x
    B. 2sin2x
    C. -cos2x + 3
    D. 2sin2x + 3
    (UNILAG 2008)

    Updated Solution
    g(x) = 2x + 3, and l(x) = sin2x
    thus g[l(x)] = 2(sin2x) + 3
    thus the dy/dx = ?
    recall that -2sin2x = cos2x - 1
    so that 2sin2x = -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



    0 4 0

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    Udeaja Benedict

    the answer to number 2 is (A)

    0 20-05-2016 20:03:00 +0000

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    Udeaja Benedict

    school photo

    0 20-05-2016 20:06:00 +0000

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    yhormzee horiyormey

    no 2 is A

    0 20-05-2016 20:53:00 +0000

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    Chima Sunday (Tutor)

    Thanks @ Udeaja Benedict and yhormzee horiyormey for that wonderful and bold attempt. Note that the post has been updated with its solutions and detail explanation. Do check it and report any abnormality you may see. Once again thank you and never allow the fire of learning to quench. cheers!

    0 14-06-2016 16:43:00 +0000