RI Differentiation C6B Add Prac (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 _______________________________________________________________ Additional Practice Questions for Chapter 6B: Applications of Differentiation Page 1 of 20 Additional Practice Questions for Chapter 6B: Applications of Differentiation (Solutions) 1 9758/2017/01/Q5 (modified) A curve has equation f( )yx where 32 33f( ) 7 22xx x x   . (i) Show that the gradient of the curve is always positive. Hence explain why the equation f( ) 0x  has only one real root and find this root. [3] (ii) Find the x-coordinates of the points where the tangent to the curve is parallel to the line 23yx . [3] [(ii) 0.145x or 1.15x ] 1(i) 32 33f( ) 7 22xx x x   22 31 3f' ( ) 3 3 3 ( ) 22 4xxx x   Since 213( ) 0 for all .2xx Therefore f' ( ) 0x  for all x.    f as and f as . Since gradient of curve is always positive, curve is always increasing and thus will cut -axis at only one point. So, f 0 has only one root. xx x x x x      From GC, 1.33x (ii) 213f' ( ) 3 ( ) 2 24xx  From GC, 0.145x or 1.15x

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ _______________________________________________________________ Additional Practice Questions for Chapter 6B: Applications of Differentiation Page 2 of 20 2 IJC Prelim 9740/2007//01/Q9 The equation of a closed curve is 22(2 ) 3 ( )2 7xy x y   . (i) Show, by differentiation, that the gradient at the point (x, y) on the curve may be expressed in the form d4 d7 yy x x yx   . (ii) Find the equations of the tangents to the curve that are parallel to (a) the x-axis, (b) the y-axis. [(ii)(a) y = 2 (b) 7x ] 2(i) --- (1) Differentiating both sides with respect to x: (Shown) 2(ii) (a) To find tangents parallel to x-axis, set , From (2), equation of tangents: y = 2 (b) To find tangents parallel to y-axis, is not defined: From (3), equation of tangents: 22(2 ) 3 ( )2 7xy x y   dd2( 2 )(1 2 ) 6( )(1 ) 0dd yyxy x y xx    d(233 ) ( 2433 ) 0 d yxyxy xyxy x    d(4 ) ( 7 ) 0 d yxy x y x    d4 d7 yy x x yx   d 0d y x  22 4 0 4 ...(2) Substitute (2) into (1), ( 8 ) 3( 4 ) 27 yx y x xx xx      22 2 81 27 27 11 42 xx xx     d d y x 22 7 0 7 ...(3) Substitute (3) into (1), (9 ) 3(6 ) 27 yx x y yy   2189 27 1 7 y y   7x

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ _______________________________________________________________ Additional Practice Questions for Chapter 6B: Applications of Differentiation Page 3 of 20 3 NYJC JC2 CT1 9758/2018/01/Q6 (modified) The diagram shows the curve C with parametric equations x = 2a cot t , y = 2a sin 2 t, where 0 t  and a is a positive constant. (i) Show that 3d 2sin cosd y ttx  . [1] (ii) Hence find, in terms of a, the equation of the tangent l to C which is parallel to the x-axis. [2] (iii) A point P on C has parameter 3t  . Given that the tangent to C at P meets l at the point Q and the point R is the foot of the perpendicular from P to l, find the exact area of triangle PQR in terms of a. [5] [(ii) 2ya (iii) 2 33 a ] (i) x = 2a cot t 2d 2c o s e cd x att  y = 2a sin2 t d 4s i nc o sd y at tt  3 2 d4 s i n c o s 2sin cosd 2c o s e c ya tt ttx at  (ii) 3d 2sin cos 0d y ttx  sin 0 or cos 0 0, or 2 tt tt    Since 0 t  , 2t  . Equation of the tangent parallel to x-axis is y = 2a. (iii) At P, 3t  , 3d3 3 2sin cosd3 3 8 y x      22c o t 3 3 axa     and 2 3 2 sin 32 aya   Equation of tangent at P is 33 3 2 28 3 aayx    33 3 3 33 9 84 2 84 aa ayx x   x y (0,2a) y = 0

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ _______________________________________________________________ Additional Practice Questions for Chapter 6B: Applications of Differentiation Page 4 of 20 At Q, y = 2a, 33 9 8 9 222 84 4 33 33 aa aax x a        23 , 23 aaP  2 ,2 33 aQa  2 ,2 3 aRa  Area of triangle PQR = 1 2 PR QR 2 12 2 22 33 3 14 22 33 33 aa a aa a            y (0,2a) P R Q l

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ _______________________________________________________________ Additional Practice Questions for Chapter 6B: Applications of Differentiation Page 5 of 20 4 9740/2012/01/Q10 [It is given that a sphere of radius r has surface area 24 r and volume 34 .3 r ] A model of a concert hall is made up of three parts.  The roof is modelled by the curved surface area of a hemisphere of radius r cm.  The walls are modelled by the curved surface of a cylinder of radius r cm and height h cm.  The floor is modelled by a circular disc of radius r cm. The three parts are joined together as show n in the diagram. The model is made of material of negligible thickness. (i) It is given that the volume of the model is a fixed value k cm3, and the external surface area is a minimum. Use differ entiation to find the values of r and h in terms of k. Simplify your answers. [7] (ii) It is given instead that the volume of the model is 200 cm 3 and its external surface area is 180 cm2. Show that there are two possible values of r. Given also that r < h, find the value of r and the value of h. [5] [(i) 3 3 5 khr  (ii) 3.04r , 4.88h ] (i) Considering the volume of the model gives 32 2 4 23 21 .3 khr rkr r h       The external surface of the model is given by  2 2 2 2 2 25 22. ( * ) 2 3 42 3 1 3 Sr kkr r r r rh rr r            For minimum S, d 0,d S r  giving 3 2 d1 0 .d5 03 32Sr k r r kr    

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ _______________________________________________________________ Additional Practice Questions for Chapter 6B: Applications of Differentiation Page 6 of 20 Since 2 2 ,3 khr r we have 3 12 133 5 .3 2 k k hk rr     Therefore 3 3 .5 khr  2 23 d1 0 4 3d Sk rr  Since r > 0 and k > 0, 2 2 d 0d S r  for all r > 0 3 3 5 khr  gives minimum external area. (ii) Given that k = 200, from (*) in (i), 2 3 3 5 540 12 52 ( 2 00 0 0 108 2 0) 18 40 0. 03 rr r r rr           Using GC, we get 3.0372, 3.7215 or 6.7587.r Since r is clearly positive, 3.04r  (3 s.f.) or 3.72r  (3 s.f.) The corresponding values of h are 2 200 2 (3.0372) 4.883(3.0372) h    (3 s.f.) or 2 200 2 (3.7215) 2.12.3(3.7215) h    (3 s.f.) Since ,rh we have 3.04r and 4.88h to 3 s.f.

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ _______________________________________________________________ Additional Practice Questions for Chapter 6B: Applications of Differentiation Page 7 of 20 5 9740/2013/02/Q2 Fig. 1 shows a piece of card, ABC, in the form of an equilateral triangle of side a. A kite shape is cut from each corner, to give the shape shown in Fig. 2. The remain