Neco 2016 FURTHER MATHEMATICS (Obj And Theory) Answers – May/July Expo

1a) A=(2 -1)(1 3) |A|=(2*3--1) |A|=6+1 |A|=7 since the determinant of A is not 0 the A is not non singular matrix To find A^-1 A^-1=(3 1)(-1 2) then 1/7(3 1)(-1 2) A^-1=(3/7 1/7)(-1/7 2/7) ====================== 2a) 3root2-3root5/5root2+2roo5 conjugate=5root2-2root5 =(3root2-3root5)(5root2-2root5) / (5root2+2root5)(5root2-2root5) =(15*2-6root10-15root10-6*5) / (25*2-10root10+10root10-4*5) =(30-31root10-30) / (50-20) =-31root10/30 2b) cosx=adj/hyp=12/13 then tanx=opp/adj=5/12 siny=3/5 tany=3/4 tan(x+y)=tanx+tany/1-tanxtany =(5/12+3/4)/(1-(5/12)(3/4) =(14/12)/(1-15/48) =36/125+18/125+16/125 =70/125 =14/25 ====================== 3a) Pr(A passed)=4/5,pr(A failed)=1/5 Pr(B passed)=3/5,Pr(B failed)=2/5 Pr(C passed)-2/5,Pr(C failed)=3/5 Pr(atleast one passed) =4/5*2/5*3/5+3/5*1/5*3/5+2/5*1/5*2/5 =24/125+9/125+4/125 =47/125 3b) Pr(atleast one failed) =4/5*3/5*3/5+3/5*2/5*1/5+4/5*2/5*2/5 =101/125 ====================== 4) (3+2y)^5 with (a+b)^5 shows a=3,b=2y Using pascals triangle the coefficient of (a +b)^5 are 1,5,10,10,5 and 1 (3+2y)^5 =(3)^5+5(3)^4(2y) + 10(3)^3(2y)^2 + 10(3)^2(2y)^3 + 5(3)(2y)^4 + (2y)^5 =243+810y+1080y^2+720y^3+240y^4+32y^5 (3.04)^5=(3+0.04)^5 =3^5+5(3)^4(0.04) + 10(3)^2(0.04)^2 +10(3)^2(0.04)^3 + 5(3)(0.04)^4 + (0.04)^5 =243 + 5*81(0.04) + 10*27(0.0016) + 10*9(0.000064) + 15(0.0000026) + 0.0000001 =243 +16.2 + 0.432 + 0.00576 + 0.000039 + 0.0000001 =259.63783 ====================== 11a) 5x^2-2x+3=0 Divide both sides by 5 x^2-2/5x+3/5=0--eq1 Compare eq1 above with x^2-(alpha+Beta)x+alphabeta=o alpha+beta=2/5 alphabeta=3/5 11ai) alpha^3+beta^3 = (alpha+beta) (alpha +beta)^2 - 3alphabeta =2/5(2/5)^2-3(3/5) =2/5(4/25-9/5) =2/5(-41/25) =-82/125 11aii) Sum of alpha +1/beta and beta+1/alpha =(alphabeta+1)/beta + (alpha beta + 1)/alpha =alphabeta^2+alpha^2beta+beta/alphabeta =alphabeta(alpha+beta) + (alpha+beta)/ alphabeta product root (alpha +1/beta)(beta+1/alpha) =alpha(beta+1/alpha)=1/beta(beta+1/alpha) =alphabeta+1+1+1q/alphabeta sum=alphabeta(alpha+beta) + (alpha+beta)/ alphabeta =3/5(2/5)+(2/5) =6/25+2/5 =(6+10)/25 =16/25 Product alphabeta+2+1/alphabeta =3/5+2+1/(3/5) =3/5+2/1+5/3 =64/15 Recall x^2-(alpha+beta)+alphabeta=0 x^2-16/25x+64/15=0 11bi) f(x)=x^3+2x-kx-6 x-2=0 x=+2 f(2)=2^3+2(2)^2-k(2)-6=0 8+8-2k-6=0 16-6-2k0 10-2k=0 2k=10 k=10/2 k=5 11bii) (x^2+4x+3) (x-2)rootx^3+2x^2-5x-6 =x^3-2x^2 =4x^2-5x =4x^2-8x =3x-6 =3x-6 =0 =>x^2+4x+3 =x^2+3x+x+3 =x(x+3)+1(x+3) =(x+1)(x+3) the remaining factors of f(x) are (x+1) and (x+3) 11biii) Zero of f(x) x-2=0,then x=2 x+1=0, then x=-1 x+3=0, then x=-3 Zeros of f(x) are 2,-1 and -3 ====================== 10a) T4=ar^4=162---(1) T8=ar^7=4374---(2) Divide (2) by (1) 4374/162=ar^7/ar^4 2187/81=r^3 27=r^3 r=3root27 r=3 Recall(1) T5=ar^4=162 162=a(3)^4 a=162/81=2 10ai)a=1st,T2=ar,T3=ar^2 a=2,T2=2*3=6,T3=2*3^2=18 therefore the three terms are 2,6 and 18 10aii) Sn=a(r^n-1)/(r-1) =2(3^10-1)/(3-1) Sn=2(3^10-1)/2 =59049-1 =59048 10b) a=6,c=60,Sn=330 Sn=n/2(a+c) 330=n/2(6+60) 300*2=n(66) 660=n(60) n=660/66 =60 L=a+(n-1)d 60=6+(10-1)d 60=6+9d 60-6=9d 54=9d d=54/9 d=6 10c) T2=ar=6---(1) T4=ar^3=54---(2) divide 2 by 1 54/6=ar^3/ar 9=r^2 r=root9 r=3 Recall(1) ar=6 a(3)=6 a=6/3 a=2 Tn=ar^n-1 Tn=2(3)^n-1 ====================== U Like It? Kindly Drop Your Comments Below& Hit Share Continue Reading Headlines: