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    Tuesday, 27 September 2016

    Waec Gce 2016 Further Mathematics Obj And Theory Answers – Nov/Dec

    FURTHER MATHS OBJ:
    1-10: BBADDBABDA
    11-20: ACBABDBDAB
    21-30: BDDBCBACCD
    31-40: ACBDACDBAB


    1a)
    8m + 2³m = 1/4
    2²m + 2²m = 2-²
    4³m = 2-²
    Equate the base
    6m = – 2
    m = -2/4 = -1/3

    1b)
    log15 base 4 = x
    x = log15 base/log4 base 10 = 1.1761/0.6020
    = 1.9536
    therefore x approximately 1.954



    ========
    2a)
    (-2) = m(-2)² + n(-2)+2=0
    4m-2n = – – – – – – – – (1)
    F(1) = m(1)²+n(1)+2=3
    m+n+2=3
    m+n=3-2
    m+n=1 – – – – – – – – – – – -(2)
    X Equate (2) by 4
    4m+4n=4 – – – – – – – – – – (3)
    4m-2n= -2
    -4m+4n=4/-6n = – 6
    n=1

    ============
    4)
    Given y= 5/x² + 3
    Y = 5(x² +3)-1
    dy/dx = anxn-1
    y = 5(x² + 3)-1
    dy = 5(-1) (x2 +3)-1^1
    dy/dx = 10x (x² + 3)-²
    dy/dx = – 10/(x² +3)²
    ============
    5a)
    ^nP5 ÷ ^nC4 = 24
    n!/(n-5)! ÷ n!/(n-4)!4! =24
    n!/(n-5)! * (n-4)!4!/n! =24
    (n-4)!4!/(n-5)!=24
    (n-4)(n-5)!4!/(n-5)! = 24
    n-4=1
    n=4+1
    n=5
    b)PR= 5C3 (1/6)³ (5/6)^5-3
    5!/(5-3)!3! (1/6)³ (5/6)²
    5!/2!3! (1/6)³ (5/6)²
    10 (1/216) (25/36)
    =0.03215
    Pr = 1-0.03215
    =0.9678

    ==========
    6) make a table containing
    MARKS,TALLY,F & FX
    UNDER F – 2,9,4,2,2,1
    UNDER FX – 2,18,12,8,10,6
    €F= 20
    €FX = 56

    6b) Mean = €fx / €f
    = 56 / 20
    = 2.8

    ==============
    7)
    Given
    h=15.4t-4.9t
    Velocity V=dh/dt =15.4-9.8t
    At maximum height V=0
    Therefore 15.4-9.8t=0
    9.8t=15.4
    t=15.4/9.8
    t=1.6secs
    Time to reach maximum height is 1.6secs

    7b)
    maximum h=15.4t-9.4t^2
    15.4(1.6)-4.9(1.6)^2
    22.64-12.544
    =12.096
    Max height = 12.1m

    ============
    9a)X+6/(x+1)^2 = A/x+1 + B/(x+1)² + C/(x+1)³
    X+6/ ~(x+1)³~ =A(x+1)² + B(x+1) + C / ~(x+1)³~
    X+6 =A (x+1)² + B(x+1) + C
    Let X+1=0 , X=-1
    -1+6= A (-1+1)² + B(-1+1) +C
    5= C
    Therefore:- C=5
    X+6= A(X2+2x+1) + Bx + B + C
    X+6= Ax²+2Ax+A+Bx+B+C
    comparing the coefficient of X²
    A=0
    Comparing the coefficient of X
    1=2A+B
    1=2(0)+B
    B=1
    x+6/(x+3)² = o/x+1 + 1/(x+1)² + 5(x+1)³
    1/(x+1)² + 5/(x+1)³

    ===========
    14a) m=0.5kg
    F-r=ma
    0.5*10-R=0.5*2
    5-r=1
    R=5-1
    R=4n
    bi)Ra+Rb=80+M+100
    Ra+Rb=180M
    Rb=180+M-90
    Rb=90+m ——-(i)
    Taking moment about B
    A.C.M=80*1.7*2*M
    =136+2m
    C.W.M=100*1.2+90+2*4
    =120+216
    =336
    A.C.W=C.W.M
    136+2M=336
    2M=336-136
    2M=200
    M=200/2
    =100N
    -: M= 100N

    ii)
    Rb=90+M
    =90+100
    =190N
    -: the reaction at B is 190N
    ===========

    10a)
    3x^2+x-2 <= 0 3x^2+3x-2x-2 <= 0 3x(x+1) -2 (x+1) <= 0 (3x-2)(x+1) <= 0 3x-2 <= 0 or 8+1 <= 0 ============== 

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