Practice NAB Unit 3

( ) ( ) ( ) ( ) ( ) ( ) Outcome 1 ii) BC = c – b i) AC = c – a – – = = 1. (a) A, B and C have coordinates (1, 2, 3), (4, –4, 12) and (5, –6, 15). (i) Write down the components of AC (ii) Hence show that the points A, B and C are collinear. ii) BC = c – b i) AC = c – a ( ) ( ) ( ) ( ) 5 -6 15 4 -4 12 5 -6 15 1 2 3 – – ( ) ( ) = 1 -2 3 = 4 -8 12 Thus AC = 4BC with common point C, therefore collinear (could also have used AB)

( ) ( ) ( ) ( ) KL = 2LM l – k = 2(m – l) l – k = 2m – 2l 3l = 2m + k 5 -3 6 -1 0 3 = 2 + ( ) 9 -6 15 3l = coordinates L (3,-2,5) ( ) 3 -2 5 l =

a) 2(3) + -1(0) + 2(4) = 14 b) cosϴ = PQ . PR PQ = √22 + (-1)2 + 22 PQ PR = √9 = 3 cosϴ = 14 PQ = √32 + 02 + 42 3(5) = √25 = 5 ϴ = cos-1(14/15) ϴ = 21.00 Threshold 9 out of 12

→ -3cosx dy/dx = -½sinx Outcome 2 3. (a) Differentiate –3sin x with respect to x. → -3cosx dy/dx = -½sinx (b) Given y = ½cosx, find dy/dx

4. Find f/(x) when, f(x) = (x + 3)-5 let y = U-5 where U = x + 3 dy/dU = -5U-6 dU/dx = 1 dy/dx = -5U-6.1 = -5(x + 3)-6

b) Integrate –¾sinx with respect to x → ¾cosx + c 5) a) Find ∫4cosx dx → 4sinx + c b) Integrate –¾sinx with respect to x → ¾cosx + c

[ ] ∫ = (x – 1)5 1(5) = (3 – 1)5 – (2 – 1)5 5 5 = 31/5 units2 c) Evaluate (x – 1)4 dx ∫ 2 [ ] 3 = (x – 1)5 1(5) 2 = (3 – 1)5 – (2 – 1)5 5 5 = 31/5 units2 Threshold 8 out of 11

Outcome 3 loga24 = log825 – log84 = log832 – log84 = log88 = 1 6. (a) Simplify loga12 + loga2. loga24 (b) Simplify 5 log82 – log8 4. = log825 – log84 = log832 – log84 = log88 = 1

(c) If y = 102.9, find an approximation for y. 7) a) If x = loge7 find an approximation for x x = 1.40 (using calculator) (b) Given that log10y = 3.4, write down an expression for the exact value of y. y = 103.4 (c) If y = 102.9, find an approximation for y. 794.3 (using calculator) loge4 Threshold 5 out of 8

kcos(x – α) = kcosxcosα + ksinx sinα Outcome 4 8) Express 2cosx + 3sinx in the form kcos(x-α) where k>0 and 0≤α<3600 2cosx + 3sinx kcos(x – α) = kcosxcosα + ksinx sinα equating coefficients kcosα = 2, ksinα = 3 k = √(22 + 32) tanα = 3/2 tan-1(3/2) = 56.30 Quad i → α = 56.30 → √13cos(x – 56.30) = √13 Threshold 3 out of 5