Get better study experience with the jambite for andriod app! jambite app download
  • Common Questions Ask in Post UTME Mathematics Exams. Day2


    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

    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

    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

    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

    2 5 0

  • Olabode Olayiwola

    Aii... Back in a jiff

    2 22-06-2016 15:42:00 +0000

  • Olabode Olayiwola

    Been trying.... Don't understand

    2 22-06-2016 16:25:00 +0000

  • ahmad muhd Amg

    MODIBBO ADAMA UNIVERSITY OF TECHNOLOGY YOLA Screening Exercise for 2016/2017 Academic session for admission into first degree programme in MAUTECH Screening of candidates for admission into undergraduate programmes of the university schedule as follows The platform will be ready for online registration from monday 20th june to friday 8th july 2016 while the screening exercise for 100 level candi's will start from 11th and for 200 level is 14th july 2016 Candidates who chose mautech as their most preferred first choice are advise to report at the univer'y on designated dates for the screening exercise ELIGIBILITY cand's who scored a mini 180point in utme are to present themselves physically for the screening exercise into 100 level Cand'ts with a mini of lower credit in diploma or merit grade in NCE from a recognize tertiary institution are to go for screening into 200 level Your to come along with both the originals and photocopies of your credential one pasport and sratch card for 0 level result verification REG procedure log on click on 2016/17 You will be required to pay 2300 for registration NOTE after successful registration, you will later recieve a text massage indicating date time and venue of your screening within the scheduled period SIGNED REGISTRAR

    2 22-06-2016 17:05:00 +0000

  • Jonathanowo Akinlabi

    1 c 2 a 3 b

    2 22-06-2016 17:20:00 +0000

  • Chima Sunday (Tutor)

    Thanks to you @Jonathan for such a fearless attempt keep it up and also to you Olabode you are most respectfully welcome. Please @ all do know that this post has been updated with its detailed solution, so try and check it up. Equally, you are free to ask any question. Once again you are welcome and thank you.

    2 23-06-2016 14:05:00 +0000