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2015 JUPEB MATHEMATICS – Essay Questions

JUPEB: MATHEMATICS 2015 JUPEB MATHEMATICS – Essay Questions

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(1a) A= (-5, -3, -1, 0, 2, 3,)

B = (-4, -3, 0, 3, 5, 8)

Find A n B

A n B = (-3, 0, 3)

(1b) Show that A U (B U C) = (A U B) U C

(1c) Vote for party A = 190

Vote for party B = 200

Vote for party C = 250

Vote for A & B = 50

Vote for A & C = 60

Vote for B only = 40

Total number of voters = 500

(i) Number that voted for A only =

= 190 – 80 – 60

= 190 – 140

= 50

Number that voted for C only

= 250 – 160 – 100

= 250 – 160 = 90

Number that voted for B only = 40

Total number of votes

Total number of = 50 + 90 + 40 + voter for the 50 + 60 + 100 = 420

(ii) Number of voters who voted for A and B but not C

= number that voted B only

= 50 +40 = 90

(iii) Number that did not vote for any party

= 500 – 420 = 80

(ai)

A = +–+123456789A = 15689–24679+34567A=148–45–242–36+335–32A =13–26+33A=3–12+9=0

(ii) The matrix is a singular matrix

(iii) Resolve x3–1x+3x+12Into partial fractionSolutionx3–1x+3x+12= Ax+3+Bx +Cx+12x3–1=Ax2+2x+1+x+3Bx+cx3–1=Ax2+2x+1x+3Bx+cx3–1=Ax2+2Ax+A+Bx2+x+3Bx+3crearrangex3–1=Ax2+Bx2+2Ax+x+3Bx+A+3Cx3–1=AA+Bx2+2Ax+x+3Bx+A+3CEquating x– term x3 has no terms on the RHSx3– termsA+B=0______1x– terms2A+C+3B=0______2cons tan t termsA+3C=–1______3A=–Bfrom equ1A=B–4from equ 3A =–3c–1______5

Equate 4 & 5A= – B ______4A=–3c–1____5–B=–3c–1–B=–3c+1B=3c+1_____6Put 6 into 22A+c+3B=0_____2B=3c+1______62A+C+33c+1=02A+C+9C+3=02A+3+10C=0____ 7FromA=–3x–1______5Put 5 into 72–3x–1+3+10c=0–6c–2+3+10c=04c+1=0C=–14

From 5A=–3c–1A=––14–1A=34–1A=3–44A=–14From 4A=–B–14=–B∴B=14x2–1x+3x+12=14x+3+14x–14x+12=–14x+3+14x–1x+12=–14x+3+x–14x+12

2b cos x=α=sinx+βind tan x in form of α &βcos x+α = cons x cosα – sin α xsin x+β= sin β cos x+sin x cos βsin ce: cos x+α = sin β cos x + sin x cos βDivide through by cos xcos x cos αcos x– sin α sin xcos x=sin β cos xcos x+ sin cos βcos xcos α –tan x sin  =sin β + tan x cos β collect like termscos α –sin β = tan x sin α + tan x cos βcos α – cos β = tan x sin α + cos βtan = cos α – sin βsin α + cos 

2c sin A =45, cos β = 1213Find i sin A+BSin A+B = sin A cos B + sin B cos A =

SIn A =45 , Cos B=1213Cos A =35 , Sin B =513Tan = 43 , Tan B =512Sin A+B= SIn A Cos B + Sin B Cos A=45+1315+51235=4865+14=48x4 +65x165x4=192+65260=257260

2ciiTan A–B–tan B–tan B1+tan A Tan B= 43–531+4351243–512=16–512=11121+43512=1+59=1491112149=1112×914=11×34×14=3356

Limx →01x– cos xcos x = cos ecx = 1sin xlimx →01x–1sinxlimx →01x– limx →01sinx = 0∴ ∞–∞=0

b loge1+x= xx22+x33–x44+...xnn

Note: A taylor series is a series expansion of a function about a point. A one dimensional taylor series an expansion of a reel function f(x) about a point x=a is given by f(x)

f(x)=f(a)+f1(a)(x–a)+f11(a)2!(x–a)2+f111(a)3!(x–a)3+...+f(n)(a)n!(x–a)2If a=0, the expansion is known as aMaclaurin series for the unction loge(1+x)f(x)=loge(1+x) , f(0)=log1e=0

f‘(x)=1(1+x)–1 , f‘(0)=1f‘‘(x)=–1(1+x)–2 , f‘‘(0)–1f‘‘‘(x)=(–1)(–2)(+x)–3 , f‘‘‘(0)=2!f‘‘‘‘(x)=(–1)(–2)(–3)(1+x)–4 ,f‘‘‘‘(0)=–3!∴fn(x)=(–1)(–2)...(–n–1)[1+x]–n=(–1)n–1(n–1)(1+x)–n

Thusloge1+x=0+x.1+x2–12!+x3.2!3!+x4–3!4!+–––loge1+x=x–x22!+x32!3×2!+x4–3!4×3!+loge1+x= x–x22+x33=x44+–––for kx ≤1Hence, evaluate loge1.1

(C) x= 1.0, 1.4, 1.8, 2.2, 2.6, 3.0

Evaluate ∫131x+1dx using trapezoidal rule

h= interval i.eh= x2–x1=x2–x1=x3–x2=x6–x5h=x5–x4h=2.6 – 2.2h=0.4∫131x+1dx=12hfirst term+last term+hsum ofall other terms=120.41+0.25+0.40.4167+0.3571+0.3125+0.2778=0.21.25+0.41.3641=0.25+0.54564=0.79564

(ai)

Moment of inertial of a body about an axis is that point of the body that causes it to resist a change in its angular velocity about that axis.

(ii) Radius of gyration of a system: For all bodies rotating through any axis, the moment of inertial (I) can be written as

I = mk2
k2 = 1/M
K=1MK= radius of gyration of the body.(b) I =?, K=?length = 2a, density= pI=ML23= I=mx2a22I =4a2m3radius of gyration K=K=1m=K=4a2m3mK= 4a2m3mK=2a3radius of gyration

(a) Newtons law of cooling = This states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature (i.e the temperature of the surroundings).

Differential equation of Newton’s law of cooling dT/dt=ky
dT/dt= -k(T-Ta)
(b) Temperature of room 15o
0o = 100oc
t1= 2 min
02=83oC
3o=?
t2= 4min)
04
t=?
15min,100oc -85oc
Room temperature = 15oc
dy/dt=k(y-15o)
dy/dt-k(y-15)
u(t) = y(t)
u(t)= y(t) – 15
du/dt = dy/dt
du/dt = ku
from
u(t) = U(o)ekt
U(t) =10ekt
y(t) = 10ekt
y(15) = 85
But
U(t) = 10ekt
10ekt =70 (final temperature – room temperature i.e 85-15=70)
ekt = 70/10
add loge 0.7
kt= 0.35667
kt = 0.357
k= 0.357/t
k= 0.357/15
k= -0.024
U(t) = 10ekt
y(t) = 15 + 10ekt
y(4) = 15+ 10ekt-(0.024)(4)
15 + 10e-0.096
15+10(0.90846)
15+9.08
24.08oc
Time taken for the water to reach y (t)= 40o
15+10e-0.024t = 40
(c) a=i+3j-7k
b= 7i +6j + 5k
c= 9i + 7j + Bk
Find (i) /a+b/

ii β if A, B & C are collinearSolutiona+b=i+3j–7k+7i+6j+5ka+b=8i+9j–2ka+b =82+92+–22=64+81+4 =149=12.206

(iii) For collinear vectors

A x  B x C= 0

a = i +3j -7k

b = 7i +6j + 5k

c = 9i +7j +Bk

a x b x c = 0

+–+13–776597β1657β–375Gβ–77697= 0

s = ∑fx2∑f–∑fx∑f2=24890550–3495502= 4978.1 – 4886.01=92.09=9.596≈9.6∑f = 50∑x = 894∑fx2=248905∑fx = 3495Mean x= ∑fx∑f=349550=69.9

(ii) Cv =5x×100coeficient of variation = 5x×1009.669.9×100 =13.73%

(iii) It represent population

(b) Mode is the measure of location associated with frequency. To find the mean, add up the values in the data set and then divide by the number of values that you added. To find the median, list the values of the data set in numerical order and identify which value appears in the middle o the list. To find the mode, identify which vale in the data set occurs most often.

 

∑x=15 , n=5∑y=20 , ∑x2=55∑y2=114 , ∑xy=78r=n∑xy–∑x∑yn∑x2–∑x2.n∑y2–∑y2r=578 – 1520555 – 152. 5114 – 202r= 390 –300275 – 225. 570 – 400r = 9050. 170r =907.071 ×13.04r = 0.976

N= 5 , ∑D2=14rk=1–6∑D2NN–1rk=184552–1 = rk= 84524=1–24120=120 – 84120=36120 = rk = 0.3

The five student results in mathematics and physics are highly correlated.

(b) A discrete random variable has a countable number of possible values while a continuous random variable takes all values in a given interval of numbers. Or

A discrete variable is a variable whose value is obtained by counting while a continuous random variable is a variable whose value is obtained by measuring.

c % of defective = 5% % of non–deective = 95%=95100 = 1920∴Probability p of non–defectivep1920q=11920 = q= 20 – 1920q = 120n = 8px=2 = nCn, prqn–1=8C2192021202= 280.90251.5625×10–83.38 ×10–7=3.948×10–7

(ii) p(at least 2)= p(x≥2)p(x≥x)=p(x=2) + p(x=3)p(x=2)= 8C2(1920)2(120)2p(x=2)=28(0.9025)(1.5625×10–8)p(x=2)=3.948 ×10–7p(x=3)= 8C3(1920)3(120)8–3=56(0.95)3(120)5=56(0.8574)(0.05)5=56(0.8574)(3.125 ×10–7)=1.50045 ×10–5

(iii) Variance𝞉2=npq𝞉2=8(1920)(120)𝞉2=8(0.95)(0.05)𝞉2=0.38

 

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