RI revision programme calculus (hard)

RAFFLES INSTITUTION H2 Mathematics 9758 2025 Year 6 Term 3 Revision 1 (Zeta) Topic(s): Calculus I (Differentiation, Applications, Maclaurin and Graphing (Solutions) 1 HCI Prelim 9758/2023/01/Q11 [The volume of a right circular cone of radius r and height h is 21 π3 rh .] A student makes a printer nozzle for his self -built 3D printer. The printer nozzle consists of a hollow inverted right circular cone of negligible thickness with radius r mm where 02 r , height h mm and slanted edge 2 mm joined to a hollow cylinder of negligible thickness with radius r mm and height r mm. A height of 1 4 h mm is cut off from the vertex of the cone for the nozzle opening (see diagram). The volume of the printer nozzle is V mm3 . (a) Show that 3 2 2 21π π 4 64V r r r= + − . [3] The student wants V to be a maximum. (b) It is given that 1rr= gives the maximum value of V . Show that 1r satisfies the equation 424537 18736 3136 0rr− + = . [4] (c) Show that one of the positive roots of the equation in part (b) does not give a stationary value of V . Suggest a reason why this value is a solution to the equation in part (b) even though it does not give a stationary value of V . [3] (d) Given that 1r is the largest positive root of the equation in part (b), state the value of 1r and find the corresponding value of h. (You need not show that your answer gives a maximum.) [1] (e) With reference to the value of 1r found in part (d), comment on the practicality of having a maximum volume for the printer nozzle. [1] r mm mm h mm 2 mm mm

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ ______________________________________ Y6 H2 Math Term 3 Revision Lecture 1: Calculus I Page 2 of 16 Solution: (a) By Pythagoras’ Theorem, 2 2 2 22 2 4 (reject 4 since 0) rh h r h r h +=  = − =− −  Let 0r be radius of cut off cone. Using similar s , 0 0 1 44 r rrr =  = 2 2 2 0 3 2 2 2 2 3 2 2 1 1 1π ( ) π π ( )3 3 4 11π π 4 π 43 192 21π π 4 (shown)64 V r r r h r h r r r r r r r r  = + − = + − − − = + − (b) 2 2 2 2 2 22 2 2 2 2 d 21 13π π 2 4 ( )( 2 )d 64 24 213π π 2 464 4 21 8 33ππ 64 4 V r r r r rr r rr r r r rrr r = + − + −  − = + − −  − −=+ − At maximum value of V, d 0d V r = , since 02 r 2 2 2 2 2 2 2 2 2 2 2 21 8 33π π 0 64 4 7 8 33π0 64 4 7 8 3 0 6 (since 0 2)64 4 7 3 8 64 4 64 4 21 5 rrr r rrr r rr r r r rr r rr − + = −   −+=  −  −+ =   −  −=  − − −  = 2 2 2 24096 (4 ) (21 56)r r r− = − 2 4 4 2 42 16384 4096 441 2352 3136 4537 18736 3136 0 (shown) r r r r rr − = − +  − + =

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ ______________________________________ Y6 H2 Math Term 3 Revision Lecture 1: Calculus I Page 3 of 16 (c) Using GC, since 0r  , 0.41806129085388 or 1.988674386383 4rr== When 0.41806129085388r= , d 3.294435715 0d V r = Hence 0.41806129085388r= does not give a stationary value of V . The squaring of the equation to remove the square root created additional roots to the equation. These additional roots may not give rise to stationary values of V . (d) 1 1.99 (3 s.f.)r = 2 14 0.2125421957 0.213 (3 s.f.)hr= − = = (e) The value of 1 1.99r = is almost the same length as the slant height 2 of the inverted cone. This means that the inverted cone is essentially flat and non -existent, leaving the printer nozzle in the shape of a cylinder only. Hence it is not realistic to have a maximum volume for the printer nozzle. Ideal measurements/values/conditions based on theoretical calculations may not be practical or feasible in real life. 2 NYJC Prelim 9758/2022/01/Q7 A curve is defined parametrically by 2 ,11 at atxy tt==++ , where a is a non-zero constant and ,1tt − . (i) Show that the equation of the normal to the curve at the point T with parameter t is given by 32( 1)( 2) ( 1) ( 2 1)t t t y t x at t t+ + + + = + + . [4] (ii) The normal to the curve at the point P 1, 2aa− meets the curve again at point Q. Find the exact coordinates of Q in terms of a. [4] (iii) Let M be the mid-point of PQ. Find a cartesian equation of the curve traced by M if a varies. [2]

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ ______________________________________ Y6 H2 Math Term 3 Revision Lecture 1: Calculus I Page 4 of 16 Solution: (i) 1 atx t= + 2 1 aty t= + 22 22 2 22 d ( 1) d 2 ( 1) 2 d ( 1) ( 1) d ( 1) ( 1) d 2d x a t at a y at t at at at t t t t t t y ttx + − + − += = = =+ + + +  = + Equation of normal at point T is 2 1 1 ( 2) 1 at atyx t t t t − =− − + + +  2 ( 2) ( 2) 11 at att t y t t x tt+ −  + =− +++ 3( 1)( 2) ( 2) ( 1)t t t y at t t x at+ + − + =− + + 3 2( 1)( 2) ( 1) ( 2 1)t t t y t x at t t+ + + + = + + (shown) (ii) When 1 2t =− , xa=− , 1 2ya= and d3 d4 y x =− Equation of normal at P is ( )14 23y a x a− = + 4 11 36y x a = + OR : At P 1, 2aa− , 1 12 at att =−  =−+ . Using result in (i), Equation of normal at P is 3 1 11 8 2 16y x a− + =− 6 8 11y x a− + =− Given that this normal meets the curve again at Q, 2 2 2 6 8 1111 6 8 11( 1) 6 19 11 0 (2 1)(3 11) 0 1 11 1 ( is rejected as this corresponds t o point )r 3o22 at at att t t t tt tt t t P  − + =− ++   − = +  − − =  + − =  =− =−

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ ______________________________________ Y6 H2 Math Term 3 Revision Lecture 1: Calculus I Page 5 of 16 When 11,3t = 2 11 11 11 12133 ,11 11 14 421133 aa x a y a          = = = =   ++        The coordinates of Q are 11 121,14 42aa  (iii) Mid-point of PQ, M is 3 71,28 42aa− 142 , 0 since is non-zero 3 71 and 42 9 28x a y ayx a x =−  =− = 3 ACJC Prelim 9758/2021/01/Q1 (modified) (a) A function is defined as 2 1ef ( ) e x xx − −= . Describe a sequence of transformations that transforms the graph of exy= onto the graph of f ( )yx= . [5] (b) A function is defined as g( ) cos 2xx= . (i) Describe a sequence of transformations that transforms the graph of sin 2yx= onto the graph of g( )yx= . [2] (ii) Describe a sequence of transformations that transforms the graph of 2sinyx= onto the graph of g( )yx= . [3] Solution: (a) 2 22 22 1ef ( ) e 1e ee ee x x x xx x x − −− −+ −= =− =− 1. Translation of 2 units in the negative x-axis direction.

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ ______________________________________ Y6 H2 Math Term 3 Revision Lecture 1: Calculus I Page 6 of 16 2. Reflection in the y-axis. 3. Translation of 2e units in the negative y-axis direction. Alternative 1: 1. Reflection in the y-axis. 2. Translation of 2 units in the positive x-axis direction. 3. Translation of 2e units in the negative y-axis direction. Alternative 2: 2 2 2 2f ( ) e e e e exxx − + −= − =  − 1. Reflection in the y-axis. 2. Scaling parallel to the y-axis by the scale factor 2e . 3. Translation of 2e units in the negative y-axis direction. Alternative 3: ( ) 2 2 2f ( ) e e e e 1xxx − + −= − = − 1. Reflection in the y-axis. 2. Transl