575 Answers Chapter 4

AnswersSelected answers are given here. For a full set of worked solutions, see Specialist Maths Dimensions Units 3 & 4 Solutions and Study Guide.g 2x + 6yy′ h 2xy + x 2 y′ + 12y 2 y′iy – xy′ – 3yk2 y′------------------------------------y 2l 6xy + 3x 2 y′ – 6xyy′ – 3y 2mopy + 22 a --------------b6y – xceg8xiy 4 + ( x 2 – 3y 2 ) 2 y------------------------------------------------------28x 4 y 3 + ( x 2 – 3y 2 ) 2 x 23 a –y′siny b 2y′cos2yc y′sec 2 y d 2y′sinycosye2y – 2xy′----------------------y 22y – 2xy′----------------------( x + y) 24xy 2 – 4x 2 yy′----------------------------------( x 2 + y 2 ) 2jn2x( 1 – y 2 ) + 2yy′ ( x 2 – 1)---------------------------------------------------------------( 1 – y 2 ) 22x – 3xy 2 – y-----------------------------------33x 2 y + 3xy 2y 5 – 4x 3 y-----------------------xy 4 – 6x 41 – ( x + y) ---------------------------2( x + y) 2 – 1y′---ydfhf10x( y – xy′ )------------------------------y 312xy′ – 12y----------------------------( 2x – 3y) 23x 2 – 4xy-----------------------2x 2 – 3y 23y 2 + 4y – 3x 2 y-----------------------------------------5( 3x 3 y 4 + 4x)2x 4 y + 2y--------------------------------------22yx 3 + 2yx – x 5–2y-------------------------------------------3( x 2 – y) 2 y 2 – 2xy′-----------------1 – y 29 a –1 b –11 – y10 a ------------------------------------------------------- 2b 2+2( 1 – x 2 )( 1 – 1 – y 2 )11 1Revision QuestionsMultiple choice1 D 2 D 3 D 4 D5 B 6 B 7 E 8 C9 C 10 B 11 B 12 C13 A 14 A 15 DShort answer1 m = 3, –12 y = – 1 4 --x 3 π+ ------ – --16 63 644 a minimum = ⎛3-- ,37 ----- ⎞⎝216⎠stationary point of inflection = (0, 4)b C(1, 3) Non-stationary point of inflection15 --96 ay = f(x)f(0) = 0a b c d e x586gy′---yh y′e yi (1 + y′)e x + y j 2e x2 + y 2( x + yy′ )k y′cosye siny l 2cos(x 2 + y 2 ) × (x + yy′)4 a1525----- b -----892cd13–----- e --102f5 28 7+− ±---------12215 --------- 67 There is a stationary point at ⎛ 10,ln-- ⎞ .⎝ 2⎠8 At ⎛ 56,–-- ⎞ 2the gradient is –--.⎝ 3⎠3At ⎛6,11 ----- ⎞ 2the gradient is --.⎝ 3 ⎠31365-----------3811-----4Specialist Maths Dimensions Units 3 & 4b a: Slope = 0 and local maximumb: Non-stationary point of inflectionc: Slope = 0 and local minimumd: Non-stationary point of inflectione: Slope = 0 and local maximumApplication Tasks1 a For f (x) = tan – 1 ( --------------- x – 1)x + 1,b 1⎛⎞⎜( x + 1) f ′(x) =– 1 – ( x – 1) ( x + 1) – 2 ⎟⎜-------------------------------------------------------------------⎛ x – 11 + -----------⎝ ⎝⎛ ⎟⎜x + 1⎠⎞2 ⎞ ⎟⎝⎠ ⎠x + 1 – x + 1 ( x + 1) = -----------------------------( x + 1) 2 × --------------------------------------------2( x + 1) 2 + ( x – 1) 22 1= ---------------------- = QED( 2x 2 -------------------+ 2)( x 2 + 1)