Revesion week 3 t2 ahahahahahahaha aa a ha ha ha a ha ha hahaa haahaah PDF

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1)

2) 3)

a.

Write down, in surd form, the values of sin 45, cos 45, sin 30, cos 30. 1 3 b. Hence show that sin 75  .¤ 2 2 Find all values of  with 0     such that 2 sin  + cos  = 1.¤   cos x  sin x Given that 0  x  , prove that tan (  x)  .¤ cos x  sin x 4 4

4)

h 12

11

A 1000 m

5) 6) 7) 8) 9) 10)

The angle of elevation of a tower PQ of height h metres at a point A due east of it is 12. From another point B, the bearing of the tower is 051T and the angle of elevation is 11. The points A and B are 1000 metres apart and on the same level as the base Q of the tower. i. Show that AQB = 141. ii. Consider the triangle APQ and show that AQ = h tan 78. iii. Find a similar expression for BQ. iv. Use the cosine rule in the triangle AQB to calculate h to the nearest metre.¤ Find all the angles with 0    2 for which sin  + cos  = 1.¤ Solve the equation 2 sin2 = sin 2 for 0    2 .¤ sin A sin A Prove the following identity:  tan 2 A .¤  cos A  sin A cos A  sin A sin3 cos3 Prove that   2 (for sin   0, cos   0).¤ sin cos sin 3 x Evaluate Lim .¤ x0 5x i. Express sin 4t  3 cos 4t in the form R sin(4t + ), where  is in radians. ii.

Hence, or otherwise, find the general solution of the equation sin 4t  3 cos 4t  0 in exact form.¤ Express cos x – sin x in the form Rcos(x + ), where  is in radians. Hence, or otherwise, sketch the graph of y = cos x – sin x for 0  x  2  .¤

11)

i. ii.

12)

By making the substitution t  tan

13) 14) 15) 16)

17)



, or otherwise, show that cos ec   cot   cot



.¤ 2 2 Use the principle of Mathematical Induction to prove that 7n + 2 is divisible by 3 for all positive integers n.† Use the Principle of Mathematical Induction to show that 9n + 2  4n is divisible by 5 for all positive integers n.† Use mathematical induction to prove that 2 + 4 + 6 + ⋯ + (2𝑛 ) = 2𝑛 (𝑛 + 1) , for all positive integers n. A school council at a co-ed school consists of 7 girls and 6 boys. In how many ways can a subcommittee of 4 girls and 3 boys be chosen from this council so as to exclude a particular girl, Mary, but include a particular boy, John?† In the English alphabet of 26 letters, the five letter a, e, i, o, u are known as ‘vowels’ and the remaining twenty one as ‘consonants’. i. The letters of the word FACETIOUS are arranged in a line in all possible ways. How many ways are there, if there are no restrictions at all?

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ii.

Out of all the possible arrangements found in (i) above, how many will contain: a. all the vowels together in the order AEIOU? b. all the vowels separated by consonants?† 18) In a tennis club, there are five married couples available to play a (mixed doubles) match, that is, a match in which a combination of one man and one woman play against a combination of another man and another woman. In how many ways can a group of four persons be chosen for this match if: i. a man and his wife play in a match but not as partners; ii. a man and his wife may not play in the match either as partners or as opponents.† 19) Find the number of ways in which the letters of the word EXTENSION can be arranged in a straight line so that no two vowels are next to each other.† 20) Find the number of ways in which the letters of the word EXTENSION can be arranged in a straight line so that no two vowels are next to each other.† [[End Of Qns]]

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[Solutions] 1

1

1 3 b) Proof » , 2 2 2 2 «2)  = 0, 2·214 (to 4 sig. figs) » «3) Proof » «4) i) ii) Proof iii) BQ = h tan 79 iv) 108 m » «1) a)

,

,

«5)  = 0,  , 2 » 2  «6)  = 0, , , 5 , 2 » 4 4 «7) Proof » «8) Proof » 3 «9) » 5 «10) i) 2sin(4t +

 3

) ii) t =

 12

( 3n 1 ) » y 1



2

x



«11) i) cos x  sin x  2 cos ( x  ) ii) » 4 «12) Proof » «13) Step (I): For n = 0, 7 + 2 = 3 which is divisible by 3 Step (II): Assume for n = k, where k is a positive integer, the statement is true i.e. 7 + 2 = 3𝑀, where 𝑀 is an integer Step(III): Prove that, for n = k + 1, the statement is true 7 + 2 = 7 × 7 + 2 = 7(3𝑀 − 2) + 2 from (II) 21𝑀 − 12 = 3(7𝑀 − 4) which is divisible by 3 Hence by the axiom of Mathematical Induction the statement is true » «14) Step (I): For n = 0, 9 − 4 = 81 − 1 = 80 which is divisible by 5 Step (II): Assume for n = k, where k is a positive integer, the statement is true i.e. 9 − 4 = 5𝑀, where 𝑀 is an integer Step(III): Prove that, for n = k + 1, the statement is true 9() − 4 = 9 × 9 − 4 × 4 = 95𝑀 + 4  − 4 × 4 from (II)   = 9 × 5𝑀 + 54  = 5(9𝑀 + 4 ) which is divisible by 5 Hence by the axiom of Mathematical Induction the statement is true » «15) Step (I): For n = 1, LHS = 23 = 8 RHS = 21222 = 8  LHS = RHS Step (II): Assume for n = k, where k is a positive integer, the statement is true i.e. 2 + 4 + 6 + ⋯ + (2𝑘) = 2𝑘  (𝑘 + 1) Step(III): Prove that, for n = k + 1, the statement is true 2 + 4 + 6 + ⋯ + (2𝑘) + (2𝑘 + 2) = 2𝑘  (𝑘 + 1) + (2𝑘 + 2) from (II) 2𝑘 (𝑘 + 1) + 8(𝑘 + 1) = 2(𝑘 + 1) (𝑘  + 4𝑘 + 4) = 2(𝑘 + 1) (𝑘 + 2) Hence by the axiom of Mathematical Induction the statement is true » «16) 6C4 × 5C2 = 150 » «17) i) 9 = 362880 ii) a) 5 = 120 ©EDUDATA SOFTWARE PTY LTD

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b) 5 × 4 = 2880 » «18) i) 10C4 ‒ 5C1 × 4C1 × 4C1 = 130 ii) 5 × 4 × 3C2 = 60 » «19) «20)

   

× ×

   

= 720 » = 720 »

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