# ARC 1998TS - dy =2x 11 =-6 11 = 5 at x =-3 dx ie the curve...

This preview shows pages 1–3. Sign up to view the full content.

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 Π is the acute angle between the two curves at x = -3 tan θ = + m m m m 1 2 1 2 1 = 6 5 1 6 5 + ( ) | = 1/31 Π = 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 cos x = tan x when x=15 ! , tan15 ! = sin cos 30 1 30 + = 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document