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FURTHER MATHS, SOLVED ANSWERS 2017 NECO IS AVAILABLE HERE

By | On July 5, 2017 (2 months ago)

Further Mathematics Obj

1-10 EBDCAACBD

11-20 DEAECBECBD

21-30 DDBDEEBEEB

31-40 ACAADEBCED

41-50 ABECBBDEAA

THEORY ANSWERS

(1a)

f(x)=2x^3 + bx -11x +C

if x+3 and x-2 are factors

therefore x=-3, x=2 are solutions

f(2)= 2(2^3)+b(2^2)-11(2)+C=0

2(8)+b(4)-11(2)+C=0

16+4b-22+C=0

=>4b+C=6 eqn(1)

f(-3)=2(-3^3)+ b(-3^2) -11(-3)+C=0

2(-277)+b(9)-11(-3)+C=0

-54+9b+333+C

=>9b+C=21

solving (1) and (2) simultanueosly,

4b+c=6 —–(1)

9b+c=21

(2)-(1)

put b=3 in (1)

4(3)+c= 6

12+c=6

c=6-12

=-6

b=3, c=-6

1b)

f(x)=2x^3 + bx^2 -11x +c

f(x)=2x^3 + 3x^2 -11x -6

(1c)

f(-2)=2(-2^3)+ 3(-2^2) – 11(-2) -6

=2(-8)+ 3(4)-11(-2)-6

=-16+12+22-6

=12

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