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# week3c2 - x = 1-2 sin 2 x = 2 cos 2 x-1 Substituting x = π...

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Week 3 Version 2 Class Time allowed: 15 minutes. 1. (2 marks) On the same axes, sketch the graphs of y = | 2 x - 7 | and y = 3 x - 2. Hence, or otherwise, solve | 2 x - 7 | = 3 x - 2. 2. (2 marks) Show that f ( x ) = x 3 + 6 x 2 + 15 x + 3 has an inverse function. 3. (4 marks) Determine a, b, c, d such that sin π 8 = p a + b 2 and cos π 8 = p c + d 2. 4. (2 marks) Find d dx cosh(sin x ).

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Week 3 Version 2 Class 1. | 2 x - 7 | = 3 x - 2 2 x - 7 = 3 x - 2 or - (2 x - 7) = 3 x - 2 2 x - 5 = 3 x or 5 x = 9 x = - 5 or x = 9 5 . However, when x = - 5 , | 2 x - 7 | = 17 and 3 x - 2 = - 17, so x = 9 5 is the only solution. Figure 1: y = | 2 x - 7 | and y = 3 x - 2 2. f ( x ) = x 3 + 6 x 2 + 15 x + 3 f 0 ( x ) = 3 x 2 + 12 x + 15 = 3( x 2 + 4 x + 5) = 3(( x + 2) 2 + 1) 0 for all x R . Since the curve is monotonic increasing, it is one-to-one. [Also, since f ( x ) is a cubic polynomial, it is onto.] Thus, it has an inverse function.
Week 3 Version 2 Class 3. According to the double angle formulae, cos 2
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Unformatted text preview: x = 1-2 sin 2 x = 2 cos 2 x-1. Substituting x = π 8 gives: cos π 4 = 1-2 sin 2 π 8 ⇒ sin 2 π 8 = 1 2 1-√ 2 2 ! ⇒ sin π 8 = r 1 2-√ 2 4 (taking the positive square root, since π 8 is in quadrant 1). ∴ a = 1 2 and b =-1 4 . Substituting x = π 8 gives: cos π 4 = 2 cos 2 π 8-1 ⇒ cos 2 π 8 = 1 2 √ 2 2 + 1 ! ⇒ cos π 8 = r √ 2 4 + 1 2 (taking the positive square root, since π 8 is in quadrant 1). ∴ c = 1 2 and d = 1 4 . 4. d dx cosh(sin x ) = d dx cosh( u ) where u = sin x = du dx ± d du cosh( u ) ² = cos x × sinh( u ) = cos x sinh(sin x ) ....
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