TMJC Chp 8 Applications of Differentiation Extra Practice Solutions 2024 (1)

Chapter 8 Applications of Differentiation Extra Practice Solutions TMJC 2024 Page 1 of 15 H2 Mathematics (9758) Chapter 8 Applications of Differentiation Extra Practice Solutions Q1 2018/MI Promo/1/5(a) (i) ( ) ( ) ( ) ( ) 2 22 2 2 2 2 f e , for , f 2 e 2 e 2 e 1 x xx x x x x x x x x xx =  =+ =+ For the function to be increasing, ( ) ( ) 2 2f 2 e 1 0 xx x x = +  Method 1: By GC, From the sketch, for ( ) 2 22 e 1 0, 0xx x x +    Method 2: 22Since 1 0 and e 0, for all , 0 xxx x +     (ii) When ( ) ( )1, f 1 2(1)e(2) 4e, f 1 ex = = = = Equation of tangent at 1x= : ( )e 4e 1 4e 3e yx yx − = −  = − y x O ( ) 2 22 e 1xy x x=+

Chapter 8 Applications of Differentiation Extra Practice Solutions TMJC 2024 Page 2 of 15 Q2 2018/DHS Prelim/2/5(a) (i) ( ) ( ) 2 4e dd2 1 4e dd d 2 e d 2 e xy xy xy xy xy yyx y x y xx y y y x x y x x +=    + + = +       −−= +− (ii) When x = 0, ( ) ( ) ( ) 2 0 0 4e 20 y y yy += = When at (0, 2), ( )2 2 2d 1d2 y x −== Equation of the tangent to the curve at (0, 2) is ( )2 1 0 2 yx yx − = − =+ (iii) Substitute y = x + 2 into ( ) 2 4exyxy+= , ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 4e 2 2 4e 1e Using G.C., 2 or 0 (reject it's point ) xx xx xx xx x x x x A + + + + + = += += =− = ( )2,0B−

Chapter 8 Applications of Differentiation Extra Practice Solutions TMJC 2024 Page 3 of 15 Q3 2018/HCI Prelim/1/6 (i) d 2cosd d sind d sin tan d 2cos 2 x tt y tt y t t xt = =− −= =− At P, t = p Equation of tangent at P: 2 22 tan1 cos ( 2sin ) 2 tan sin1 cos 2 cos cos sin tan1 cos 2 tan1 sec 2 py p x p ppy p x p p p p xp ppx − − =− − = + − + += + − = + − 2 tan 2(1 sec )y x p p+ = + (shown) (ii) When y = 0, tan 1 sec2 2 2sec tan p xp px p =+ += When x = 0, 1 secyp=+ Method 1: Coordinates of 1 sec 1 sec,tan 2 ppM p  ++= 1 sec sec 2 12 py p y+=  = − 1 sec 2 tan tan pyx pp +== Using 221 tan secpp+= , 2 2 2 2 2 22 22 2 2 21 (2 1) 41 4 4 1 ( 1) 1 y yx y yyx y yx x x y x xy x + = − + = − + =− =− = −

Chapter 8 Applications of Differentiation Extra Practice Solutions TMJC 2024 Page 4 of 15 Method 2: Coordinates of 1 seccot cosec , 2 pM p p +=+ 1 sec sec 2 12 py p y+=  = − cos 1cot cosec sin px p p p += + = Using 22sin cos 1pp+= , 22 2 22 2 2 2 2 2 2 2 2 2 2 2 2 22 2 2 cos 1 1 121 1 121 1 1(2 1) (2 ) 1 1(2 1) (2 1) 4 (4 4 1) ( 1) 1 p xy y xy y y x y y x y y x y y x yx x y x xy x + += −   + − += − +=−− + = − + =− =− = −

Chapter 8 Applications of Differentiation Extra Practice Solutions TMJC 2024 Page 5 of 15 Q4 2018/MJC Promo/1/6 (i) d 2d x t = , 2 d4 d y tt=− 2 d d d 2 d d d y y x x t t t=  =− Gradient of normal = 2 2 t At point M , 1t = , gradient of normal = 1 2 Equation