WAEC 2016 Further Mathematics OBJ and Theory

FURTHER MATHS OBJ:
1-10 ABACCDCCCD
11-20 CDDDDABDBC
21-30 BDADABACBC
31-40 CCBBABABDD
15a)at max heightV=0m/sg=10m/s^2V^2=u^2-2gh0^2=30^2-2*10H0=900-20H20H=900H=900/20H=45m
15b)time taken to get to max heightV=u-gt0=30-10t10t=30t=30/10t=3secsTimetaken to return=2t=2*3=6secs15c)H=40mH=ut-1/2gt^2=4030t-1/210t^2=4030t-5t^2=405t^2-30t+40=0t^2-2t-4t+8=0(t^2-2t)-(4t+8)=0t(t-2)-4(t-2)=0(t-2)(t-4)=0t-2=0 or t-4=0t=2secs or t=4secs

11a)Kp2=72K!/(k-2)!=72K(k-1)(K-2)!/(K-2)!=72K^2-K=72K^2-K-72=0K^2-9k+8k-72=0K(K-9)+8(k-9)=0(K+8)(K-9)=0k=-8,K=9We consider positive value of K=9
11b)The equation 2cos^2tita-5costita=3Let cos tita=x2x^2-5x=3using quadratic formulara=2,b=-5,c=-35+_root(25+24)/4=5+_root(49)/4=(5+_7)/4=(5+7)4=3 or (5-7)/4=-2/4=-1/2since x cos titacos tita=-0.5tita=cos^-1(-0.5)tita=120degrees——————
10a)(1+x)^77Co(1)^7(x)^0 + 7C1(1)^6(x) + 7C2(1)^5(x)^2 + 7C3(1)^4(x)^3 + 7C4(1)^3(x)^4 + 7C5(1)^2(x)^5 + 7C6(1)(x)^6 + 7C7(1)^0(x)^7= 1+7x + 21x^2+35x^3 + 35x^4+21x^5 + 7x^6+x^7

(10b)35 21 7a=35d=T2-T1=21-35d=-14=================

=6a)1+4+k+k+4+11/5=k+120+2k/5=k+1 cros multiply20+2k=5(k+1)20+2k=5k+520-5=5k-2k15/3 = 3k/3k=15/3 =5the numbers are 1,4,5,9,11 the mean X=6

6b)tabulatex| 1, 4, 5, 9, 11x-x| -5, -2, -1, 3, 5(x-x)^2| 25, 4, 1, 9, 25total| 64standard deviation=sqr £(x-x)^2/n=sqr64/5=sqr12.8=3.58====================2)(5,2)(-4,k)(2,1)(y3-y2)/(x3-x2)=(y2-y1)/(x2-x1)(1-k)/2-(-4)=(k-2)/(-4-5)(1-k)/(2+4)=(k-2)/-9(1-k)/6=(k-2)/-9-9(1-k)=6(k-2)-9+9k=6k-129k-6k=-12+93k=-3k=-1—————7)m1=3u1=8m/sm2=?u2=5m/sv=6m/sm1u1+m1u2=(m1+m2)v3*8+m2*5=(3+m2)624+5m2=18+6m224-18=6m2-5m2m2=6(7b)m2u2-m1u1=V(m1+m2)6*5-3*8=V(3+6)30-24=9v9v=6v=6/9v=0.67m/s—————

-5a)pr(age)=4/5pr(fully)=3/4pr(must)=2/3pr(age not admitted)=1-4/5=1/5pr(fully not admitted)=1-3/4=1/4pr(must not admitted)=1-2/3=1/3Therefore 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+11Therefore m=5, n=7

3a)If f(x+2)=6x^2+5x-8)To find f(5)Therefore f(x+2)=f(5)where x+2=5x=5-2x=3therefore f(5)=6(3)^2+5(3)-8=6(9)+15-8=54+7=61

3b)(7root2+3root3)/(4root2-2roo3)*(4root2+2root3)/(4root2+2root3)(24*2+14root6+12root6+6*3)/(16*2+8root6-8root6-4*3)(48+26root6+18)/(32-12)=(66+26root6)/20=66/20+(26root6/20)=33/10+(13root6/10)=3.3+1.3root6====================

12a)tabulateMarks| 1-10, 11-20, 21-30, 31-40, 41-50, 51-60, 61-70, 71-80, 81-90, 91-100F| 3, 17, 41, 85, 97, 115, 101, 64, 21, 6C.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, 90.5-1005C.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

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(13ai)M=2P=5C=3total=10If the books of the same subject are to stand togetherNo of arrangements=2!*5!*3!*3!=2*120*6*6=8640arrangements(13aii)Only the physics textbook must stand togetherNo of arrangements=5!*6!=120*720=86400arrangements

(13b)P=13/20q=1-13/20=7/20pr(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)