ARC 1998TS - 1 SUGGESTED SOLUTIONS 3 Unit MATHEMATICS...

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Unformatted text preview: 1 SUGGESTED SOLUTIONS 3 Unit MATHEMATICS Question 1 (a) y = ln (cot 3 x) = ln (cot x) 3 = 3 ln (cot x) dy = 3 (- cosec 2 x) . 1 dx cot x = - 3 cosec 2 x = 3 sin cos x x cot x (b) Let f(x) = 2x 3- 5x 2 + 3x - 2 f(2) = 16 -20 + 6 -2 = 0 x - 2 is a factor 2x 2- x + 1_________ x - 2 |2x 3-5x 2 +3x - 2 2x 3-4x 2- x 2 +3x - x 2 +2x x - 2 x - 2 f(x) = (x - 2)(2x 2-x + 1) = 0 when x-2 = 0 or 2x 2- x + 1 = 0 x = 2, the real root Consider 2x 2- x + 1 = 0 = b 2- 4ac = 1 - 4(2)(1) = - 7 < 0 ie < 0 so that there are NO real roots f(x) = 0 has only one real root. (c) To find points of intersection - x 2 = x 2 + 11x + 15 2x 2 + 11x + 15 = 0 (x + 3)(2x + 5) = 0 x + 3 = 0 or 2x + 5 = 0 x = -3 or x = -5/2 For y = - x 2 , dy = - 2x = 6 at x = -3 dx ie the curve y = - x 2 has a gradient of 6 at x = -3 For y = x 2 + 11x + 15, dy =2x + 11 = -6+11 = 5 at x = -3 dx ie the curve y = x 2 + 11x + 15 has a gradient of 5 at x = -3 If is the acute angle between the two curves at x = -3 tan = + m m m m 1 2 1 2 1 = 6 5 1 6 5 + ( ) | = 1/31 = 1 ! 51 (d) (i) sin 2x = 2 sin x cox x 1 + cos 2x 1 + (2cos 2 x - 1) = 2 sin x cos x 2 cos 2 x = sin x c o s x = tan x when x=15 ! , tan15 ! = sin cos 30 1 3 + = 1 2 1 3 2 + = 1 2 2 3 2 + = 1 2 3 + (ii) cot 15 ! = __1 __ = 2 + 3 tan 15 ! tan 15 ! + cot 15 ! = 1 2 3 2 3 + + + = 2 3 2 3 2 3 2 3 + + + ( ) ( ) = 2 3 4 3 2 3 + + = 2 3 2 3 + + = 4 2 (e) S T P R Q Join QS TQP = SRQ (corresponding angles, S R | | Q T ) But TSQ = SRQ (angle between tangent and chord) TQP = TSQ PQ is a tangent to the circle through Q, T and S (angle between tangent and chord, PQ being the tangent and TQ the chord). Question 2 (a) Let tan-1 = then tan = , - /2 < < /2 so can be represented as an angle in the first quadrant. By Pythagoras Theorem OA = 5 so that sin (2 tan-1 ) = sin 2 = 2 s i n cos = 2 x 1/ 5 x 2/ 5 = 4 5 (b) d dx [sin-1 (x-1)] = 1 1 1 1 2 ( ) ( ) x = 1 1 2 1 2 + x x = 1 2 2 x x = 1 2 x x ( ) dx x x ( ) . 2 0 5 1 = [sin-1 (x-1)] 1 = sin-1 (0) - sin-1 (-) = 0 + sin-1 () = /6 (c) u = 4 x = (4-x) du = (4-x)- . (-1) dx = dx x 2 4 u = 4 x u 2 = 4-x x = 4 - u 2 Terminals: when x = 0, u = 2 when x = 3, u = 1 x dx x u du ....
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ARC 1998TS - 1 SUGGESTED SOLUTIONS 3 Unit MATHEMATICS...

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