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

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23 Comments
  1. seun says:

    pls upload furt-maths theory

  2. ddollar says:

    no tym post it nw

  3. Yinka says:

    Please post the answers complete

  4. Anonymous says:

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  10. Abdulrahman says:

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  12. Anonymous says:

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  20. emmanuel says:

    Ghana members
    FURTHER MATHS
    1
    a) 8m + 2³m
    (2³)m + 2³m = 1/4
    2³m + 2³m = 1/4
    2(2³m) = 1/4
    2³m = 1/8
    2³m = 8 – ¹
    2³m = 2-³
    Since bases are the same,
    powers are same
    3m = – 3
    m = – 3/3
    m = -1
    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
    (a)
    Given
    h=15.4t – 4.9t²
    Velocity v = dh/dt = 15.4 –
    9.8t
    At maximum height, V =0
    15.4 – 9.8t = 0
    9.8t = 15.4
    t = 15.4/9.8
    t = 1.6secs
    Time to reach maximum
    height = 1.6secs
    b) maximum height = 15.4 t
    – 4.9t²
    = 15.4(1.6) – 4.9(1.6)²
    =24.64 – 12.544
    = 12.096
    Maximum 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

    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
    9b)
    S²1 1/(x+1)² + 5/(x+1)³ DX
    1n (x+1)³ + 5 1n (X+1)⁴ |²1
    1n 3(3) + 5 1n(3)⁴ – 1n +2³ –
    5 1n +2⁴
    3.2958+21.9722-2.0794-13.
    8629
    =9.3297
    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
    3x <= 2 or x <= -1
    X= x <= 2/3
    Good luck
    NAIJA MEMBER
    FOR NIGERIANS
    1 a)
    g ( x )= y
    y = x + 6
    x = y – 6
    g ^- f( x – 6 )
    = 4- 5 ( x – 6) / 2 = 4 – 5x
    +30/ 2
    = 34-5 x / 2
    1 b)
    coodinate = (x 1+ x 2 / 2 , y
    1+ y 2/ 2 )
    = (7 – 2 / 2 , 7 – 5 / 2 )= ( 5 /
    2 , 2 / 2 )
    = (5 / 2 , 1)
    ======================
    ============
    NAIJA
    10a)
    i ) ( x ^2-1 ) ( x +2 )= 0
    ( x -1 ) ( x +1 ) ( x +2 )
    x =1, or – 1 or -2
    ii ) 2 x -3/( x -1) ( x + 1) ( +2)
    =A /x -1+B /x + 1+C /x +2
    2x – 3=A ( x +1) ( x +2 ) +B
    ( x -1) ( x +2)
    +C ( x -1) ( x + 1)
    let x + 1=0, x =- 1
    2( -1 ) -3=B ( – 1-1) ( – 1+2)
    -5/2 =-2B /- 2 B =5/2
    let x – 1 = 0 x = 1
    2( 1) – 3=A ( 1 +1) ( 1+2 )
    -1= CA , A = -1/6
    Let x +2=0 x =- 2
    2( -2 ) -3=C ( -2- 1) ( -2 +1)
    -7= 3C , C =-7 /3
    =================
    (11a)
    Given:
    f(x)={(4x-x^2)dx
    f(x)=2x^2 – x^3/3 + K
    f(3)=2(3)^2 – (3)^2/3 + K
    =21
    18 – 9 + K=11
    9+K=21
    K=21-9
    K=12
    Therefore
    f(x)= -x^3 + 2x^2 + 12
    ————————————
    ——-
    (14ai)
    SKETCH THE DIAGRAM(14aii)
    Using lami’s theory
    T1/sin60=T2/sin30
    48N/sin60=T2/sin30
    48N/0.8660=T2/0.5
    0.5(48)/0.8660=T2(0.8660)/
    0.8660
    T2=24/0.8660
    T2=27.7N
    (14b)
    Using the equation of
    motion
    H=U^2/2g
    H=(20)^2/2*10
    =20*20/20
    H=20m
    Timetaken to reach the
    maximum height
    S=Ut+1/2at^2
    20=0+1/2(100)t^2
    20/5=5t^2/5
    t^2=4
    t=sqroot4=2S
    5a)
    pr(age)=4/5
    pr(fully)=3/4
    pr(must)=2/3
    pr(age not admitted)=1-4/5
    =1/5
    pr(fully not admitted)=1-3/4
    =1/4
    pr(must not admitted)=1-2/3
    =1/3
    Therefore pr(none
    admitted)=1/5*1/4*1/3
    =1/60
    5b)
    pr(only age and fully gained
    admission)=4/5*3/4*1/3
    =1/5
    4)
    (x^2+5x+1)sqroot
    (2x^3+mx^2+nx+11)=(2x-5)
    remainder:30x+16
    (x^2+5x+1)(2x-5)
    =2x^3+10x^2+2x-5x^2-25x-
    5
    =2x^3+10x^2-5x^2-25x-5
    =2x^3+5x^2-23x+30x+16-5
    =2x^3+5x^2+7x+11
    Therefore m=5, n=7
    12a)
    tabulate
    Marks| 1-10, 11-20, 21-30,
    31-40, 41-50,
    51-60, 61-70, 71-80, 81-90,
    91-100
    F| 3, 17, 41, 85, 97, 115, 101,
    64, 21, 6
    C.B| 0.5-105, 10.5-205,
    20.5-305, 30.5-405,
    40.5-505, 50.5-605,
    60.5-705, 70.5-805,
    80.5-905, 90.5-1005
    C.F| 0+3=3, 3+17=20,
    20+41=61, 61+85=146,
    146+77=243, 243+115=358,
    358+101=459,
    459+64=523, 523+21=544,
    544+6=550
    13ai)
    M=2
    P=5
    C=3
    total=10
    If the books of the same
    subject are to stand
    together
    No of arrangements=2!*5!
    *3!*3!
    =2*120*6*6
    =8640arrangements
    (13aii)
    Only the physics textbook
    must stand together
    No of arrangements=5!*6!
    =120*720
    =86400arrangements
    (13b)
    P=13/20
    q=1-13/20=7/20
    pr(atleast 3 speak E)=1-Pr(2
    speak E)
    =(1-8C1p^1q^7+8C2p^2q^6)
    =1-(8*(13/20)*
    (7/20)^7+28(13/20)^2*
    (7/20)^6
    =1-(0.003346+0.0217467)
    =1-0.0251
    =0.9749
    =0.975(3s.f)

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