Past papers › WAEC › 2017 Paper WAEC 2017 Further Maths Theory
Theory paper · 2 questions · partial
WAEC · 2017 · May/June · Further Maths · Paper 2 Topics include Factor theorem, Polynomials, Trigonometric identities, Stationary points, Coordinate geometry.
Our copy of this paper is missing questions 1, 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15.
Sit this paper Answer every question in order, timed if you like (suggested 30 min). You're marked when you hand in, then you see where to focus and the working for each question.
Or read it here: every question below has a worked solution.
4 9 If ( x + 1 ) (x + 1) ( x + 1 ) ( x − 2 ) (x - 2) ( x − 2 ) g ( x ) = x 4 + a x 3 + b x 2 − 16 x − 12 g(x) = x^4 + ax^3 + bx^2 - 16x - 12 g ( x ) = x 4 + a x 3 + b x 2 − 16 x − 12 a a a b b b
(a) (b) Try it on a graph Move a and b until the curve passes through both marked points. Which values work?
Open the interactive graph Worked solution (try it first) ( x + 1 ) (x + 1) ( x + 1 ) is a factor, so
g ( − 1 ) = 0 g(-1) = 0 g ( − 1 ) = 0 :
1 − a + b + 16 − 12 = 0 1 - a + b + 16 - 12 = 0 1 − a + b + 16 − 12 = 0 , which gives
a − b = 5 a - b = 5 a − b = 5 .
( x − 2 ) (x - 2) ( x − 2 ) is a factor, so
g ( 2 ) = 0 g(2) = 0 g ( 2 ) = 0 :
16 + 8 a + 4 b − 32 − 12 = 0 16 + 8a + 4b - 32 - 12 = 0 16 + 8 a + 4 b − 32 − 12 = 0 , which gives
8 a + 4 b = 28 8a + 4b = 28 8 a + 4 b = 28 .
Divide the second equation by 4:
2 a + b = 7 2a + b = 7 2 a + b = 7 .
Add it to
a − b = 5 a - b = 5 a − b = 5 :
3 a = 12 3a = 12 3 a = 12 , so
a = 4 a = 4 a = 4 .
Then
b = a − 5 = − 1 b = a - 5 = -1 b = a − 5 = − 1 .
Check:
g ( x ) = x 4 + 4 x 3 − x 2 − 16 x − 12 g(x) = x^4 + 4x^3 - x^2 - 16x - 12 g ( x ) = x 4 + 4 x 3 − x 2 − 16 x − 12 = ( x + 1 ) ( x − 2 ) ( x + 2 ) ( x + 3 ) = (x + 1)(x - 2)(x + 2)(x + 3) = ( x + 1 ) ( x − 2 ) ( x + 2 ) ( x + 3 ) ✓.
Watch out
Put − 1 -1 − 1 ( − 1 ) 4 = 1 (-1)^4 = 1 ( − 1 ) 4 = 1 a ( − 1 ) 3 = − a a(-1)^3 = -a a ( − 1 ) 3 = − a − 16 ( − 1 ) = + 16 -16(-1) = +16 − 16 ( − 1 ) = + 16 Simplify each equation (divide by 4) before solving them together. Report a problem with this question
(a) Simplify 1 1 − cos θ + 1 1 + cos θ \dfrac{1}{1 - \cos\theta} + \dfrac{1}{1 + \cos\theta} 1 − cos θ 1 + 1 + cos θ 1 sin θ \sin\theta sin θ
(b) Find the equation of the line joining the stationary points of y = x 2 ( x − 3 ) y = x^2(x - 3) y = x 2 ( x − 3 )
Try it on a graph Plot the curves, move them, and read values off the graph.
Open the interactive graph Worked solution (try it first) (a) The common denominator is
( 1 − cos θ ) ( 1 + cos θ ) = 1 − cos 2 θ (1 - \cos\theta)(1 + \cos\theta) = 1 - \cos^2\theta ( 1 − cos θ ) ( 1 + cos θ ) = 1 − cos 2 θ = sin 2 θ = \sin^2\theta = sin 2 θ .
The top is
( 1 + cos θ ) + ( 1 − cos θ ) = 2 (1 + \cos\theta) + (1 - \cos\theta) = 2 ( 1 + cos θ ) + ( 1 − cos θ ) = 2 .
So the sum is
2 sin 2 θ \dfrac{2}{\sin^2\theta} sin 2 θ 2 .
(b) y = x 3 − 3 x 2 y = x^3 - 3x^2 y = x 3 − 3 x 2 , so
d y d x = 3 x 2 − 6 x = 3 x ( x − 2 ) \dfrac{dy}{dx} = 3x^2 - 6x = 3x(x - 2) d x d y = 3 x 2 − 6 x = 3 x ( x − 2 ) .
It is zero at
x = 0 x = 0 x = 0 and
x = 2 x = 2 x = 2 .
The stationary points are
( 0 , 0 ) (0, 0) ( 0 , 0 ) and
( 2 , 8 − 12 ) = ( 2 , − 4 ) (2, 8 - 12) = (2, -4) ( 2 , 8 − 12 ) = ( 2 , − 4 ) .
The gradient of the line joining them is
− 4 − 0 2 − 0 = − 2 \dfrac{-4 - 0}{2 - 0} = -2 2 − 0 − 4 − 0 = − 2 , so the line is
y = − 2 x y = -2x y = − 2 x , that is
y + 2 x = 0 y + 2x = 0 y + 2 x = 0 .
The distance between them is
2 2 + 4 2 = 20 ≈ 4.47 \sqrt{2^2 + 4^2} = \sqrt{20} \approx 4.47 2 2 + 4 2 = 20 ≈ 4.47 .
Watch out
In (a), 1 − cos 2 θ = sin 2 θ 1 - \cos^2\theta = \sin^2\theta 1 − cos 2 θ = sin 2 θ In (b), find each y y y ( 0 , 0 ) (0, 0) ( 0 , 0 ) ( 2 , − 4 ) (2, -4) ( 2 , − 4 ) Report a problem with this question