K H S U R GXF W P RP H QWF RUH O D WL RQ FRH I L FL H QW E HW Z Q xD Q G y LVFO

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K H S U R GXF W P RP H QWF RUH O D WL RQ FRH I L FL H QW E HW Z Q xD Q G y LVFO RV HU WRDV F R P S D U H G W R K D E H Z H Q x D Q G y ±L y x y Z KH Q x ±Y ²0D \QR WEH YD OL GD VF RU HOD WLR QG RHV QR W QHF HVD ULO \ LP SO \FDXVD WLRQ ± E ²
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1 H2 M a t hem a t ic s2 0 1 7 P r e lim Ex am pe 2 So lut io n 1 ( i) y = ( ) l n 1 s i n x + e 1 sin y x = + d d y x = cos 1 sin x x + [B1] 2 2 d d y x = ( )( ) ( )( ) ( ) 2 1 sin sin cos cos 1 sin x x x x x + - - + = ( ) 2 2 2 sin sin cos 1 sin x x x x - - - + = ( ) 2 sin 1 1 sin x x - - + [A1] = ( ) ( ) 2 sin 1 1 sin x x - + + = 1 1 sinx - + = 1 e y - = e y - - (Shown) (ii) 3 3 d d y x = d e d y y x - - - = d e d y y x - 4 4 d d y x = 2 2 d d d e e d d d y y y y y x x x - -  + -   = ( ) 2 d e e e d y y y y x - - - - - (from (i) ) = ( ) 2 2 d e e d y y y x - - - - or 2 d e e d y y y x - - - + (iii) When 0 x= , y = ln 1 = 0 d d y x = cos0 1 1 sin 0 = + 2 2 d d y x = 0 e 1 - = - 3 3 d d y x = 1 4 4 d d y x = 1 1 2 - - = -
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2 ( ) ln 1 sinx \ + = ( ) ( ) 2 3 4 1 2 1 0 ... 2 3! 4! x x x x - - + + + + + = 2 3 4 1 1 1 ... 2 6 12 x x x x - + - + 2 (i) d d cos , sin 10 , d d d sin 10 d cos 10 tan cos 10 tan sec (Shown) x y u u t t t y u t x u t u t u q q - q - q q q q q = = = = - = - (ii) When 30 u = and 1 , 2 t= 5 15cos , 15sin 4 x y q q = = - , d 1 tan sec d 6 y x q q = - Equation of tangent is 5 15sin 4 y q - + = ( ) 1 tan sec 15cos 6 x q q q - - = 1 5 tan sec 15sin 6 2 x q q q - - + \ y = 1 5 tan sec 6 4 x q q - + 3 (i) A(3, 0, 2), B (1, 0, 3), C(2, -3, 5) 1 3 2 0 0 0 3 2 1 AB - = - = uuur 2 3 1 3 0 3 5 2 3 AC - = - - = - uuur 1 1 2 1 3 0 3 5 1 3 6 3 1 3 Take 5 , = 0 5 3 0 18 21 6 3 6 AB AC - - · = · - = = = + + = n a n uuur uuuur 1 3 A vector equation of i 6 21 s 5 H = r
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3 ( i) Equa tionof 2 H is 2 x y kz - + =14. Sub. A ( 3,0,2)intoe qua tionof 2 H , 2(3 ) 0 (2) k - + =14 \ k =4 ( Shown) Sub. B ( 1,0,3)into L HS ofe qua tionof 2 H , L HS = 2 4 x y z - + = 2( 1 ) 0 4(3 )- + =14=RHS \ B isa lsoin 2 H . Sinc e B isinboth 1 H a nd 2 H , \ B isonther ive r . ( De duce d) (i) Re c al A B = uuur 2 0 1 - , using A(3, 0, 2) or B (1, 0, 3), a cartesian equation of the river (line AB) is 3 2, 0 2 x z y - = - = - or 1 3, 0 2 x z y - = - = - (iv) Since 1 2 3 0 1( 2) ( 3)(0) 2(1) 0 2 1 BC AB -     = - = - + - + =       uuur uuur , BC is perpendicular to AB. \ B is the point on the river that is nearest to C . Exact distance from C to the river = 1 3 1 9 4 14 2 BC = - = + + = uuur (v) Acute angle between BC and 2 H 1 1 1 2 3 1 2 4 13 sin sin 14 21 14 21 49.3 or 0.861 rad q - -     - -       = = = ° 4 (i) ( ) d 100 d A kA A t = - (ii) ( ) 1 d 100- A A A = k dt By partial fractions, ( ) 1 100- A A = ( ) 1 1 100 100 100 + - A A
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4 1 1 1 d 100 100 \ + - A A A = + kt c ( ) 1 ln ln 100 100 - - A A = + kt c ( 0 > Q A and 100 0 - > A ) ( ) 1 ln ln 100 100 A A - - = + kt c ln 100- A A = ( ) 100 kt c + 100- A A = ( ) 1 100 100 100 e e e e kt c k t kt c D + = = where 1 100 k k = and 100 e c D = . When 0, 20 = = t A , 20 100 20 - = D D = 1 4 When 5, 40 = = t A , 40 100 40 - = 1 5 1 e 4 k 1 5 1 e 4 k = 2 3 1 5 e k = 8 3 1 5k = 8 ln 3 1 k = 1 8 ln 5 3 \ 100- A A = ( ) ln 8 1 5 3 1 e 4 t When 0.5 100 50 = · = A , 50 100 50 - = ( ) ln 8 1 5 3 1 e 4 t 1 = ( ) ln 8 1 5 3 1 e 4 t ( ) ln 8 1 5 3 e t = 4 1 8 ln 5 3 t = ln 4 t = 8 1 5 3 ln 4 7.07 ln = (2 dp) The required time is 7.07 days.
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5 (i) Whe n ( ) 1 4 d a y s = t , 10 - A A = ( ) 1 8 l n 1 4 5 3 1 e 4 Method 1 Solve algebraically 100- A A = 3.8963 (5 sf) A = ( )( ) 100 3.8963 - A = 389.63 3.8963 - A 4.8963 A = 389.63 A = 79.58 (2 dp) For the bread to be deemed safe for human consumption in terms of preservatives present, 79.58
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