## 2.** Resolve (3X-10) / (X²-7X+10)**

(3X-10) / (X²-7X+10) =(3X-10) / (X²-5x-2x+10)

(3X-10) /[ x(x-5)-2(x-5) ] = (3X-10)/[X-2][X-5](3X-10)/[X-2][X-5] = A/(X-2) + B(X-5)

(3X-10)/[X-2][X-5] =[ A(X-5)+B(X-2) ]/[X-2][X-5][X-2][X-5] will cancel out

(3X-10) = A(X-5)+B(X-2)

If x-5 = 0

When x=5

(3[5]-10) = A(5-5)+B(5-2)

15-10=A(0) + B(3)

5=3B

B = 5/3

When x =2

(3[2]-10) = A(2-5)+B(2-2)

6-10=-3A

-4=-3A

A = 4/3

Therefore,

[ A(X-5)+B(X-2) ]/[X-2][X-5] = [ 4/3(X-5)+ 5/3(X-2) ]/[X-2][X-5]Or [ 4(X-5)+ 5(X-2) ]/3[X-2][X-5]

### B. **(a) In a class of 40 students, 32 are good in mathematics, 4 are good in physics and 4 do not take either mathematics or physics. How many are good in mathematics and physics as well.**

Solution

Solution

40 = 32-x +x +24-x

40=56-x

X=56-40

X = 16 ( are good in both maths and physics

**Without using table find the value of Sin 75**

Solution

Sin 75 = sin(45⁰+30⁰)

Sin45Cos30 + Sin30Cos45

1√2 × √3/2 + 1/2 × 1/√2

√3/2√2 + 1/2√2

(√3+1)/2√2 (final answer NO!)

Using rationalization method,

(√3+1)/2√2) ×( 2√2×2√2)

[√3×2√2 + 1×2√2]/[2√2 × 2√2][2√6+2√2]/ 8

[√6+√2]/4 (final answer)

a**. If tanA = ¾ (0<a<90) and SinB = 5/13 (90<b<180). Evaluate (I) CosA + secB. (II) TanB**

Solution

TanA = ¾

TanA = 0.75

A = tan–¹(0.75)

A = 36.86⁰

When SinB = 5/13

Sin B = 0.385

B = Sin–¹(0.385)

180-22.64⁰ = 157.36⁰

B = 157.36⁰

**Evaluate CosA + secB.**

Cos(36.86) + sec(157.36)

Cos(36.86) + 1/Cos(157.36)

0.8+(0.923) = 0.8+0.923 = 1.723

Tan B = Tan (157.36)

= 0.417

Or

When tanA = ¾**Using Pythagorean theorem**

|Hyp|² = |opp|² + |Adj|²

|Hyp|² = |3|² + |4|²

9+16 = 25

√25 = 5

**When SinB = 5/13Using Pythagorean theorem,**

|Hyp|² = |opp|² + |Adj|²

|13|² = |5|² + |Adj|²

|Adj|² = 169-25

√144 = 12

Evaluate

(I) CosA + secB.

⅘+cos–¹

⅘+13/12

(48 + 65)/60= 113/60

(II) TanB

Tan B = 5/12

= 0.417

C. **Show that (1+sinA)² + Cos²A = 2(1+sinA)Solution**

(1+sinA)² + Cos²A = 2(1+sinA)

Proving

(1+sinA)(1+SinA) + Cos²A

1 + SinA + SinA + Sin²A + Cos²A

Recall, Sin²A + Cos²A = 1

1 + 2SinA + (Sin²A + Cos²A)

1 + 2SinA + 1 = 2 + 2SinA

2(1 + SinA)

**Expand (2+3x)⁵ in ascending power of x using binomial expansion. Hence solve (2.3)⁵ Solution**

- 1(2⁵3x⁰) + 5(2⁴3x¹) + 10(2³3x²) + 10(2²3x³) + 5(2¹3x⁴) +1(2⁰3x⁵)

32+240x+10×8×9×x² + 10×4×27×x³ + 5×2×81×x⁴ + 243x⁵

32 + 240x + 720x² + 1080x³ + 810x⁴ + 243x⁵

Hence solve (2.3)⁵ = (2+0.3)⁵

1(2⁵0.3⁰) + 5(2⁴0.3¹) + 10(2³0.3²) + 10(2²0.3³) + 5(2¹0.3⁴) +1(2⁰0.3⁵)

32 + 5×16×0.3 + 10×8×0.09 + 10×4×0.027 + 5×2×0.0081 + 0.00243

32+24+7.2+1.08+0.081+0.00243

= 64.36343

64.4(approximately to 1 decimal point)

### B. In how many ways can three men and two women be chosen from six men and 4 women?

Solution

3 men out of 9 can be selected in 9C3, 2 women out of 4 women can be selected in 4C2

9C3 × 4C2. = 9!/(9-3)!3! × 4!/(4-2)!2!

9!/6!3! × 4!/2!2! =( 9×8×7×6!)/6!3! × (4×3×2!)/2!2!

504/6 × 12/2 = 84×6

= 504ways