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 of normal: ( )143 2yx− = − 25yx=+ (ii) At point N : ( )41 2 1 52 tt  = + + 2 3 4 0tt + − = ( ) 1 reject this is point t M = or 4t =− Coordinates of N is ( )7, 1−− (iii) Equation of tangent at P : ( )2 42 21y x p pp  −− = − − At point Q : 0y=  ( )2 420 2 1 xppp  −− = − − 41xp=+ At point R : 0x=  ( )2 42 21yp pp  −− = − − ( ) 2 2 4 1py p += Area of triangle OQR = ( )2 1 2(4 1) 412 p pp  + + 2 41p p  +=

Chapter 8 Applications of Differentiation Extra Practice Solutions TMJC 2024 Page 6 of 15 Q5 2018/CJC Promo/1/7 (i) ( ) 22 2 3 5 d d2 2 2 6 0 dd d 2 2 d 6 2 shown3 kx xy y yykx x y y xx y kx y x y x kx y yx + − = + + − = += − += − (ii) For tangents parallel to x-axis, d 0d y x = , 0kx y+= y kx=− or yx k=− Method 1: Substitute y kx=− into C, ( ) ( ) ( ) 22 2 2 2 2 2 3 5 35 5 3 kx x kx kx x k k x kk + − − − = − − = −= + Since k is a non-zero constant, ( ) 230 1 3 0 1 03 kk kk k + + −   Method 2: Substitute yx k=− into C, ( ) 2 2 2 22 2 2 2 2 2 2 2 2 2 3 5 2 3 5 35 5 3 5 13 yyk y ykk ky ky k y k y k k k ky kk k k    − + − − =       − − = + =− −= + −= + Since k is a non-zero constant, 5 013 1 03 k k k + −   Method 3: [Discriminant] 0 + + – 0 + + –

Chapter 8 Applications of Differentiation Extra Practice Solutions TMJC 2024 Page 7 of 15 Substitute y kx=− into C, ( ) ( ) ( ) ( ) ( )( ) 22 22 2 2 2 2 2 2 2 3 5 535 3 3 5 0 40 0 4 3 5 0 kx x kx kx x k k x kk k k x b ac kk + − − − = −− − =  = + + + = − − +  ( ) 230 1 3 0 1 03 kk kk k + + −   Since 10, 3kk − , 1 03 k−   Method 4: [Discriminant] Substitute yx k=− into C, ( ) ( )( ) ( ) 2 2 22 2 2 2 2 2 2 22 2 2 22 2 2 3 5 2 3 5 53 5 0 3 40 0 4 3 5 0 30 1 3 0 1 03 yyk y ykk ky ky k y k ky k k k y kk b ac k k k kk kk k    − + − − =       − − = −+ + =  = + − − +  + + −   Since 10, 3kk − , 1 03 k−   (iii) 13,k = x = 1 and y = 2, d 5dt x = ( ) ( ) ( ) ( ) 13 1 2d d 3 2 1 y x += − =3 d d d dt d dt y y x x= =15 units per second.

Chapter 8 Applications of Differentiation Extra Practice Solutions TMJC 2024 Page 8 of 15 Q6 2018/NJC Promo/1/12 Let V be the volume of water in the pool when the depth of water is x m. By similar triangles, 5 5 2 2 ax ax=  = . ( )( ) 2 2 15 45 45 20 900 2522 d d d d d d 1 100900 50 10 90 5 V x x x x x x V t V t xx x  = + + = + = = + = + When x = 1.6, ( ) d 10 5 m/min d 90 5 1.6 49 x t == + 2 m 45 m 5 m 50 m x a x 45 45 + 5 2 𝑥 Width of pool is 20m

Chapter 8 Applications of Differentiation Extra Practice Solutions TMJC 2024 Page 9 of 15 Q7 2020/SAJC Promo/1/11 (i) 2 2 2 32 2 1123 6 2 4 13 24 13 24 rv r r r h r r h vrh r    =  + +  =+ =− 22 22 2 2 122 4 1 1322 4 24 17 2 12 A r r rh vrr r r r vr r     = + +    = + + −       =+ (ii) 2 d 17 2 d6 Av rrr =− For stationary values, let d 0d A r = 2 3 3 17 2 06 17 12 0 12 17 vr r rv vr    −= −= = 3 12 17 vr = Using second derivative test to check for minimum, 2 23 d 17 4 +d6 Av rr = When 3 12 ,17 vr = 2 2 d 17 17 17 +0d 6 3 2 A r   = =  Therefore, A is minimum when 3 12 .17 vr =

Chapter 8 Applications of Differentiation Extra Practice Solutions TMJC 2024 Page 10 of 15 (iii) 2 2 3 1 3 2 13 024 2 13 24 48 13 48 1.0613 rh r r r r r     = −    = But 0,r so 0 1.06.r (iv) Q8 2018/AJC Promo/1/9(b) APC and CQB are similar triangles because: 90APC CQB ACP CBQ   = =  = = (3 interior angles are equal) By similar triangles, AC AP CB CQ= 12x y CQ= 22 12 12 24 144(12 ) x y yyy  = = −−− 12 6 2 6( 6) 6( 6) yyx yy  = = −− x y 24 – x 12 – y x y L P Q A B C α 90o – α α