NECO 2019 - FURTHER MATHS ANSWER

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F/MATHS OBJ
1-10: DDADAABCDC
11-20: BECEDEEDCA
21-30: DEDDBCECCA
31-40: CCDBEADACC
41-50: BABCEEECAC

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ANSWERS:


(1)
5^2y + 1 = 26(5^y - 1)
Let 5^y = x
X² + 1 = 26(x/5)
5x² + 5 = 26x
5x² - 26x + 5 = 0
5x² - x -25x + 5 = 0
X(5x - 1) -5(5x - 1) = 0
(x - 5)(5x - 1) = 0
X - 5 = 0 OR 5x - 1 = 0
X = 5 OR X = 1/5
Since 5^y = x
If x = 5 OR if x = 1/5 ie 5^-1
5^y = 5¹ OR 5^y = 5^-1
y = 1 OR y = -1



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(4a)
Given: x²+ y² + 2x + 6y + 5 = 0
If (-2, -1) lies in the circle, then f(x, y) = 0
f(-2, -1)=(-2)²+(-1)² + 2(-2)+6(-1) + 5
f(-2, -1) = 4 + 1 - 4 -6 + 5
= 4 + 1 + 5 - 4 - 6
= 10 - 10
= 0
Therefore -2, -1 lies in the circle.

Gradient --> 2x + 2ydy/dx + 2 + 6dy/dx+ 0 =
dy/dx(2y + 6) = -2x - 2
dy/dx = -2x - 2/2y + 6
= -(x +1)/(y + 3)

At(-2, -1)
dy/dx = -(-2+1)/(-1 + 3) = -(-1)/2 = 1/2
Equation is :
y - (-1)/x - (-2) = 1/2
y + 1/x + 2 = 1/2
y + 1 = 1/2(x+2)
y + 1 = 1/2x + 1
y = 1/2x
Or X = 2y

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(9)




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(11a)
1/1-sintita + 1/1+sintita
= 1+sintita+1-sintita/sin²tita
= 2/1-sin²tita
= 2/cos²tita

(11b)
2x + 5≥1
2x ≥ -4
X ≥ -4/2
X ≥ -2
&
X+6 < 12
X < 12 - 6
X < 6
Range --> -2 ≤X<6

(11c)
2tan15°/1 + tan²15° = Cos60°
Cos60°(1+tan²15°)=2 tan15°
½(1+tan²15°) = 2 tan 15°
1+ tan²15° = 4 tan 15°
Tan²15° - 4 tan 15° + 1 = 0
Using Formular method,
tan 15° = -(-4)±√(-4)²-4(1)(1)/2(1)
tan 15° = 4±√16-4/2
tan 15° = 4±√12/2
tan 15° = 4±2√3/2
tan 15° = 2±√3
tan 15° = 2-√3 OR 2+√3
Therefore tan15° = 2-√3



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(15)




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