Further Maths Revision ANSWERS 1* 2* Simplify by rationalising Rationalise 1 √3 √3 3 2√3 5+√2 3 Factorise x² - 64 4 Factorise 3x² + 8x - 3 10√3 − 2√6 23 (x+8)(x-8) (3x-1)(x+3) 5 Factorise fully x³ - 4x X(x²-4) = x(x+2)(x-2) 6 7 Make h the subject A = πr²+2πrh Make x the subject h= y= 10 11 2−π¦ 3π¦−1 Prove that (y+6)(y+3) - y² is a multiple of 9 When y> 0 9 2ππ π₯+2 1+3π₯ X= 8 π΄−ππ² Simplify 3π²π 2π x 9y+18 = 9(y+2) 4π³ 9ππ Find a and b such that x² + 8x + 10 ≡ (π₯ + π)² + b f(x) = x² + 5x -1 work out i) f(-3) ii)f(3x) iii)f(x-2) 2 ππ² 3 (x+4)² - 6 a=4 b=-6 i) ii) iii) -7 9x²+15x-1 x²+x-7 12 13 Work out the Range, where Domain is −2 ≤ π₯ ≤ 2 f(x) = x² 0 ≤ π(π₯) ≤ 4 Calculate i) the gradient between A(-3,7) B(2,-6) ii) the equation of this line iii)the gradient of the normal −13 i) ii) π¦= iii) 14 Find the gradient & intercept of 2x + y – 4 = 0 15 Draw the graph of y = f(x) where f(x) = 4 −4 ≤ π₯ ≤ −2 = x² −2 < π₯ < 2 = 8 – 2x 2≤π₯ ≤4 16 Solve the following, using −π ± √π 2 − 4ππ π₯= 2π 5 −13 5 5 x- 13 m = -2 c = 4 a=3 b=5 c=1 3x² + 5x + 1 = 0 x = -0.23 or -1.4 17 Solve 2x + y = 8 5x + 2y = 21 X = 5 y = -2 4 5 18 Expand (π₯ − 1)(π₯ + 1)(π₯ + 3) x³ + 3x² - x -3 19 20 21* 22 Given that f(x) = x³+x²-4x-4 i) show (x+1) is a factor ii) factorise f(x) iii) solve f(x) = 0 i) ii) iii) x=-1 f(x)=0 (x+1)(x-2)(x+2) X=-1, 2, -2 Solve Y – 4 > 3y – 12 Y<4 3 Solve π₯ 2 = 27 3 (√27)² = 9 Find the nth term of 3 6 13 24 39 2n²-3n+4 23 The nth term of a sequence is i) ii) π+1 2π+1 work out the first 3 terms find the limiting value as n→ ∞ i) ii) 2 3 1 2 , 3 4 , 5 7 24 A straight line DEF has points D(4,3) E(8,5) DE:EF is 2:3 Find co-ordinates of F F ( 14, 8 ) 25 (x-3)² + (y+2)² = 36 find the Centre & Radius of the circle Centre (3, -2) r = 6 26 Work out x Use angle in a semi-circle = 90° So is 90-x Opp angle = 112° ∴ 2π₯ − 60 + 112 + (90 − π₯) = 180° AB is a diameter X = 38° 27 Prove that triangle ABD is isosceles Using alt. Segment theory ABD = BCD As AB=BC (isosceles) BAD = BCD ∴ π΅π΄π· = π΄π΅π· (ππ ππ πππππ ) AB is a tangent at B ADC is a straight line AB = BC 28* Demonstrate cos 60 = 0.5 sin 60 = √3 2 2 √3 60° 1 cos 60 = 29 1 2 sin 60 = √3 2 Find x X = 4.6 Use π π π = = ππππ΄ ππππ΅ ππππΆ 30 Find θ Angle A = 41.4° Use π2 = π 2 + π 2 − 2πππΆππ π΄ 31 Solve for 0≤ π₯ ≤ 360 i)tan x = √3 ii)3Cos π₯ = 2 iii)√2 πΆππ π₯ − 1 = 0 32 X= 60°, 240° X = 48°, 312° X = 45°, 315° Solve for 0≤ π ≤ 360 2Sinπ + Cosπ = 0 ππππ Use [πΆππ π = Tanπ] 33 i) ii) iii) p.v = -26.6° θ = 333.4° , 153.4° Re-write the equation as a quadratic with only one trigonometric function Sin²π − 2πΆππ π + 1 = 0 Use [πππ2 π + πΆππ 2 π = 1] Cos²θ + 2Cosθ – 2 = 0 34 Prove the identity 1 − πΆππ 2 π ≡ πππ2 π 1 − πππ2 π π ππ2 π πππ 2 π Use 35 37 38 π πππ πππ π )² = (tanθ)² = tan²θ [πππ2 π + πΆππ 2 π = 1] Differentiate i) y=3xβ΄ ii)y=4x³ + 2x – 5 i) ii) 36 =( Find the gradient of Y=x³+2x²+4 at a tangent at A(3,2) ππ¦ ππ₯ ππ¦ ππ₯ ππ¦ ππ₯ = 12π₯³ = 12x² + 2 = 3π₯² + 4π₯ when x =3 Is 39 Y=x²-7 find the co-ordinates when gradient = 6 (3,2) The sketch shows y=5x-x² Find ππ¦ i)the gradient function ππ₯ ii)the gradient of curve at P iii)the equation of the normal at P i) ii) iii) 5-2x Gradient = -1 Y = -x + 9 39 2 π΄= ( 1 3 2 ) B=( ) 1 −1 i) Find i) 4A ii) AB iii) Write the Identity matrix ii) 8 12 ( ) 4 4 1 ( ) 1 I= ( 40 Sketch the graph y=Sin π 1 0 0 1)