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Revision No 3

Revision No 3. 1. Graph is in form y = log a (x – b) Find values of a and b. (13,2). (5,0). 2. A recurrence relation is in the form U n+1 = mU n + c Given that U 1 = 7, U 2 = 32 and U 3 = 153, find the values of m and c. 3. B divides AC in ratio 1:2.

calvin-holcomb
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Revision No 3

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  1. Revision No 3

  2. 1. Graph is in form y = loga(x – b) Find values of a and b. (13,2) (5,0) 2. A recurrence relation is in the form Un+1 = mUn + c Given that U1 = 7, U2 = 32 and U3 = 153, find the values of m and c

  3. 3. B divides AC in ratio 1:2. A(1,0,1) and C(7,3,13) Find coordinates of B 4. Solve 5cos(x – 20) = 2 0<x<3600

  4. 1. Graph is in form y = loga(x – b) Find values of a and b. (13,2) using (5,0) → 0 = loga(5 – b) = 1 (5,0) →b = 4 using (13,2) → 2 = loga(13 – 4) → 2 = loga9 → 9 = a2 → a = 3

  5. 2. A recurrence relation is in the form Un+1 = mUn + c Given that U1 = 7, U2 = 32 and U3 = 157, find the values of m and c U2 = mU1+ c U3 = mU2+ c 125 = 25m m = 5 Substituting 32 = 7(5) + c c = -3 → 32 = 7m + c →157 = 32m + c

  6. A B C 3. B divides AC in ratio 1:2. A(1,0,1) and C(7,3,13) Find coordinates of B 1 2 2AB = BC 2(b – a) = c – b 2b – 2a = c – b ( ) 5 2 9 b = 3b = c + 2a ( ) ( ) 1 0 1 = 7 3 13 + 2 B(5 , 2 , 9) ( ) 15 6 27 =

  7. 4. Solve 5cos(x – 20) = 2 0<x<3600 Let 5cosA = 2, where A = x – 20 cosA = 0.4 cos-1(0.4) = 66.40 A = 66.40 or 360 – 66.40 = 293.60 x – 20 = 66.40 OR x – 20 = 293.60 x =86.40 x = 313.60

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