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Solution Centre for All Examinations both WAEC/WASSCE/SSCE, NECO, NABTEB, JAMB (UTME), Post-UTME, WAEC GCE, NECO GCE, NABTEB GCE etc

**F/Maths OBJ: 1-10: AACACBDABC 11-20: DAACCDBDAA 21-30: BCCBABABCC 31-40: BCCBCCDDDD**

(1)

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(10)

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(13 CONTINUATION)

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(1)

5^4^(3x/4 -1) + 5^3(x-1)/5^(3x – 2)

5^3x/4 – 4 + 5^3x – 3/5^3x – 2

5^3x – 4 + 5^3x – 3/5^3x – 2

(5^3x ÷ 5^4) + 5^3x ÷ 5^3÷5^3x – 2

5^3x/625 + 5^3x/125÷5^3x- 2

5^3x + 5(5^3x) ÷ 5^3x-3

5^3x + 5(5^3x)/625 ÷ 5^3x/25

Let 5^3x = y

y + 5(y)/625 ÷ y/25

y + 5y/625 × 25/y

=6y/625 × 25/y

=6/25

(2a)

Using y2 – y1/x2 – x1

Where y2 = 7, y1 = -5,

X2 = -2, and X1 = 7

7 – -5/-2 – 7

=7+5/-9

=12/-9

=4/-3

Coordinate points :

-4/3(3 : 2)

=-12/3 : -8/3

= -4 : -2⅔

X = (-4, 2⅔)

(2b)

2/1-√2 – 2/2+√2

=2(2+√2)-2(1-√2)/(1-√2)(2+√2)

=4+2√2 – 2+2√2/2+√2-2√2 – 2

=2 + 4√2/-√2

=(2+4√2)(-√2)/-√2(-√2)

= -2√2 – 4(2)/2

= -8 – 2√2/2

= -4 – √2

(3)

Sn = A/2[2n+(n-1)d]

Where Sn = 165

a = -3, d = 2

165 = A/2[2(-3)+(n-1)2]

165 = n[-6+2n-2]/2

165×2 = n[2n – 8]

330 = 2n² – 8n

2n² – 8n – 330 = 0

n²-4n-165 = 0

Using -b±√b²-4ac/2a

4±√-4²-4(1)(-165)/2(1)

4±√16 + 660/2

4±√676 = 4±26/2

4+26/2 = 30/2

= 15 terms

(4)

Draw the right angled triangle

Using Pythagoras theorem

Third side = √(p+q)² – (p-q)²

=√(p+q+p-q)(p+q-p+q)

Difference of two squares.

=√(2p)(2q)

=√4pq

Adjacent side = 2√pq

Tanx = opp/adj = p – q/2√pq

1 – tan²X = 1-(p-q)²/4pq

=(4pq)-(p²-2pq+q²)/4pq

= -p²+6pq-q²/4pq

= -(p² – 6pq + q²)/4pq

(5)

Draw the diagram

Using cosine law

Cos∅ = 16²+10²-14²/2(16)(10)

Cos∅ = 256 + 100 – 196/320

Cos∅ = 160/320

Cos∅ = 0.5

∅ = cos-¹(0.5)

∅ = 60°

Angle between 10N and 16N

= 180 – ∅ (sum of angles on a straight line)

= 180 – 60

=120°

(6)

Draw the diagram

Taking moment about the pivot,

(T × 25)=(50×10)+(20×45)

25T = 500 + 900

25T = 1400

T = 1400/25

T = 56N

(7)

In a tabular form

Under class interval:

1-5, 6-10, 11-15, 16-20, 21-25, 26-30

Under class mark (X):

3, 8, 13, 18, 23, 28

Under X-Xbar:

-10, -5, 0, 5, 10, 15

Under frequency:

18, 12, 25, 15, 20, 10

Ef = 100

Under f(X – XbarA):

-180, -60, 0, 75, 200, 150

f(X – XbarA) = 185

Where xA = 13

Mean = xA + Ef(X – Xbar)/Ef

=13 + 185/100

=13 + 1.85

=14.85years

(9a)

Y = 7 – 6/x and y + 2x – 3 = 0

Substitute eqn (1) into eqn(2)

7 – 6/x + 2x – 3 = 0

Multiply through by X

7x – 6 + 2x² – 3x = 0

2x² + 4x – 6 = 0

X² + 2x – 3 = 0

(x² + 3x – x – 3) = 0

X(X+3)-1(x+3)=0

(x-1)(x+3)=0

X – 1 = 0 or X + 3 = 0

X = 1 or X = -3

But y = 7 – 6/x

When X = 1

y = 7 – 6/1

y = 7 – 6

y = 1

When X = -3

y = 7 – 6/-3

y = 7 + 2

y = 9

Coordinates are (1, 1) and (-3, 9)

(9b)

Draw the diagram

Gradient of AB = 9 – 1/-3 -1

=8/-4 = -2

Midpoint of AB = (1+3/2, 1-9/2) = (2, -4)

Gradient of perpendicular = -1/-2 = 1/2

equation of perpendicular is

y-(-4)/x-2 = 1/2

y + 4/x – 2 = 1/2

y = 1/2x – 1 – 4

y = 1/2x – 5

OR

2y = X – 10

(10a) given 4x² – px +1 = 0

For real roots: b² – 4ac >0

(-p) ² – 4(4) (1) > 0

p² – 16 > 0

p² >16

p > ± 4

(10bi) Given: (1 +3x)⁶

Using pascal’s triangle: 1, 6, 15, 20, 15, 6, 1

(1)⁶(3x)º + 6(1)⁵ (3x)¹ + 15 (1)⁴ (3x)₂ + 20(1)³ (3x)³ + (15) (1)²(3x)⁴+6(1)¹(3x)⁵ + 1(1)º(3x)⁶

1 + 6(3x) + 15 (9x²) + 20 (27x³) + 15 (81x⁴) + 6(243 x⁵) 729x⁶

1 + 18x + 135x² + 540x³ + 1215x⁴ 1458x⁵ + 729x⁶

(ii) (1.03)⁶ = (1 + 3(0.01)]

Therefore (1.03)⁶ = 1 + 18(0.01) + 135 (0.01)² + 540(0.01)³ + 1215(0.01)⁴

+ 1458 (0.01)⁵ + 729 (0.01)⁶

+ 1 + 0.18 + 0.0135 + 0.005 + 0.00001215

+ 0.0000001458 + 0.000000000729

= 1.1940523

= 1.194 (4s.f)

(11) u= 5ms-1 u= 6ms-1

A= 1ms-2 a= 3ms-2

p→ ←a

M—————————————N

51m

(11a)

t when particles are 30 meters apart

S= ut + ½ at2

Sp= st + ½ t2

Sa = 6t + 1/2 (3)t2

Where sp and sa are distances covered by particles respectively at

any time t

At time, t= 1.5s

Sp = 5(1.5) + ½(1.5)2 =8.625m

sa = 6(1.5) + ½ (3)(1.5)2 = 12.375m

sp + sa = 8.625 + 12.375

= 21m

Distance between particles = 51 – 21 = 30m

At time t = 1.5s, the particles will be 30 meters apart.

(14a)

Draw the diagram

(14b)

From the diagram

a = v – 20/4

2.5 = v – 20/4

V – 20 = 10

V = 10 + 20 = 30m/s

acceleration/retardation = 3/4

2.5/30/T-12 = 3/4

2.5(T – 12)/30 = 3/4

(T – 12) = 30×3/2.5×4

= 90/10 = 9

T – 12 = 9

T = 9 + 12 = 21

t = T – 12

t = 21 – 12

t = 9secs

(14c)

Total distance of the journey

= Area of BCDI + Area of AFEO + Area of DFI

= 1/2(12+8)10 + 1/2(9×10) + 1/2(21+21)20

= (20/2)10 + 90/2 +(42/2)20

=10(10) + 45 + 21(20)

=100 + 45 + 420

= 565m

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