# Neco 2019 | 2020 Further Mathematics Questions and Answers

FURTHER MATHS OBJ

11-20: BECEDEEDCA

21-30: DEDDBCECCA

41-50: BABCEEECAC

(1)

(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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(9)

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(11)

(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