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NAME: OFODI MARTINS EMMANUEL
LEVEL: 100L
COURE: MAT 101
DEPARTMENT: M.B.B.S
1. Find the middle term of (2p – 1/2q)10
(a+x)n = n(n-1) (n-2)(n-3)(n-4)an-5x5
(2p – 1/2q)10 = a=2p, x= -1/2q, n=10
10(10-1)(10-2)(10-3)(10-4)2p10-5(-1/2q)5 = 10*9*8*7*6 2p5 (-1/2q)5
5!
5!
30240*32p5*-1/32q5
= -252p5q5 ans
5!
(b). 3*[n+1 C3] = 7*[n C2]
= 3*(n+1)!
=
7*n!
=
3* (n+1)(n+1-1)(n+1-2)(n+1-3) =
7* n(n-1)(n-2)
(n+1-3)!3!
(n-2)!2!
3* n(n+1)(n-1) = 7* n(n-1)
6
2
(n+1-3)!3!
=
(n-2)!2!
n(n+1)(n-1) = 7* n(n-1) = n(n2-1) = 7n(n-1)
2
2
2
2
= 2n(n2-1) = 14n(n-1)
Divide both sides by 2n
n2-1 = 7(n-1)
= n2-1 = 7n-7
= n2 -7n +6 =0
(n-1) (n-6) =0
n = 1 or n = 6
n = 6 is valid because the number of objects n has to be greater than the number of ways in which a set of r
members can be selected that is nCr.
2. Expand (c – 1/c)5 using the binomial series.
(a+x)n = an + nan-1x + n(n-1)an-2x2 + n(n-1)(n-2)an-3x3 + n(n-1)(n-2)(n-3)an-4x4 + X5
2!
3!
4!
= C5 + 5C5-1(-1/c) + 5(5-1)C5-2(-1/c)2 + 5(5-1)(5-2)C5-3(-1/c)3 + 5(5-1)(5-2)(5-3)C5-4(-1/c)4 + (-1/c)5
2!
3!
4!
= C5 – 5C3 + 10C - 10C-1 +5C-3 - C-5 ans
3. Expand ¼-x)2 in ascending powers of x up to term in x3, using the binomial theorem.
(a+x)n = an + nan-1x + n(n-1)an-2x2 + n(n-1)(n-2)an-3x3
2!
3!
-2
(-2-1)
(-2-2)
2
4 + (-2) 4
–x + (-2)(-2-1) 4
(-x) + (-2)(-2-1)(-2-2)4(-2-3)(-x)3
2!
3!
= 1/16 + x/32 + 3x2/256 + x3/256 ans
4. Expand 1/√(1 – 2t) in ascending powers of t up to the term in t3, using the binomial theorem. State the
limits of t for which the expression is valid.
1/√(1-2t) = 1/(1-2t)1/2 = (1-2t)-1/2
(a+x)n = an + nan-1x + n(n-1)an-2x2 + n(n-1)(n-2)an-3x3
2!
3!
-1/2
-1/2-1
= 1 + (-1/2)(1
)(2t) + (-1/2)(-1/2-1)(1-1/2-2)(-2t)2 + (-1/2)(-1/2-1)(-1/2-2)(1-1/2-3)(-2t)3
2!
3!
2
3
= 1 + t + 3t /2 + 5t /2 ans
5. Prove that cos(y-π) + sin(y+π/2) = 0
Cos(y-π) = (cosy(cosπ) – siny(sinπ))
Where π = 180
= cosy(cos180) – siny(sin180) = cosy(-1) –siny(0)
= -cosy
Sin(y+π/2) = siny(cosπ/2) + cosy(sinπ/2)
= siny(cos90) + cosy(sin90)
= siny(0) + cosy(1) = cosy
Therefore
-cosy +cosy = 0
6. Show that tan(x+π/4) tan(x-π/4) = -1
Where π = 180
Tan(x+π/4) = tanx + tan(45) = tanx + 1
1- tanxtan45
1 – tanx
tan(x-π/4) = tanx + tan(-45) = tanx – 1
1 – tanxtan(-45)
1 + tanx
= tan2x2 – tanx + tanx - 1
= tanx + 1 * tanx – 1
1 – tanx
= (tan2x2 -1)
( 1 - tan2x2)
1 + tanx
= 1(tan2x2 – 1)
-1(-1 + tan2x2)
1 + tanx – tanx –tan2x2
= 1/-1
= -1 ans
7. solve the equation 4sin(x-20°) = 5cosx
= 4sinx * 4cos20 – 4cosx * 4sin20 = 5cosx
= 16sinx * cos20 – 16cosx sin20 = 5cosx
Divide through by cosx
=(16sinx * cos20) / (cosx) – (16cosxsin20)/(cosx) = 5
(16tanx * cos20) = (5 + 16sin20)
= (16 * 0.9397)tanx = 5 + (16 * 0.3420)
Tanx = 10.472/15.0352; tanx = 0.6965
X= tan-1 (0.6965); x= 34.86 ans
8. Given cosA = 0.42 and sinB = 0.73, evaluate
(a) Sin(A-B)
cosA = 0.42
A = cos-1(0.42) = 65.1654
sinB = 0.73
B = sin-1 (0.73) = 46.8864
= sin(A-B) = sinAcosB – cosAsinB
= (sin65.1654cos46.8864) – (cos65.1654sin46.8864)
= (0.9075*0.6834) – (0.4200*0.7300)
= 0.6201855 – 0.3066 = 0.3135855; to 4 decimal places = 0.3136 ans
(b) Cos (A-B) = cosAcosB + SinAsinB
= (cos65.1654cos46.8864) + (sin65.1654sin46.8864)
= (0.4200*0.6834) + (0.9075*0.7300) = 0.287028 + 0.662475
= 0.949503; to 4 decimal places = 0.9495
(c). tan(A+B) = tanA + tanB = tan65.1654 + tan46.8864
1- tanAtanB
=
3.228885
1 – 2.30795
1- tan65.1654tan46.8864
= 2.16077 + 1.068115
1- 2.16077*1.068115
= 3.228885 = -2.4687 ans
-1.30795
9. An oil company bores a hole 80m deep. Estimate the cost of boring if the cost is £30 for drilling the first
metre with an increase in cost of £2 per metre for each succeeding metre.
Cost of 80m deep = ?
Cost of I metre deep = £30
Cost of 2 metre deep = £32
Cost of 3 metre deep = £34
Therefore: £30, £32, £34……….
a=30 d=2 n=80
Sn = n/2 (2a + (n-1)d)
S80 = 80/2 (2*30 + (80-1) 2) = 40 (60 +(79)2)
S80 = 8720
The cost of 80m deep is £8720
10.Find the sum of all the numbers between 5 and 250 which are exactly divisible by 4
Numbers divisible by 4 = 8,12,16,………………..248.
A=8; d=4; l=248
Tn = a + (n-1)d
248 = 8 + (n-1)4;
248 = 8 + 4n – 4
= 248 – 8 + 4 = 4n;
244 = 4n
N = 244/4 = 61.
Sn = n/2 (2a + (n-1) d) = 61/2 (2*8 + (61-1)4)
= 61/2 (16 + 60(4)) = 30.5 (16 + 240)
S61 = 7808
Sum of number between 5 and 250 divisible by 4 = 7808