O-LevelAdditional MathematicsDifferentiation and integration4037/22 · Oct/Nov 2022 · Q810 Marks
In this question all lengths are in centimetres. The volume of a cylinder with radius r and height h is πr²h and its curved surface area is 2πrh. The volume of a sphere with radius r is 4/3 πr³ and its surface area is 4πr². The diagram shows a solid object in the shape of a cylinder of base radius r and height h, with a hemisphere of radius r on top. The total surface area of the object is 300cm². [Figure 8.1]
- (a)Find an expression for h in terms of r.[2]
- (b)Show that the volume, V, of the object is 150r - 5/6 πr³.[3]
- (c)Find the maximum volume of the object as r varies.[5]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2022). The mark or mark type is shown in brackets.
- [M1] 2πrh + 2πr² + πr² [= 300] oe
- [A1] h = (300 - 3πr²) / (2πr) oe, isw
- [M2] V = πr² (their (300 - 3πr²) / (2πr)) + 2/3 πr³ oe (M1 for V = πr²(their h) + kπr³, k≠2/3 oe)
- [A1] Correct completion to given answer: 150r - 5/6 πr³ nfww
- [B1] Derivative of V: 150 - 5/2 πr² oe, soi, isw
- [M1] their 150 - 5/2 πr² = 0 and solves as far as r = ...
- [A1] r = √(300/(5π)) oe or 4.37 (cm)
- [M1] 150(their 4.37) - 5/6 π(their 4.37)³
- [A1] 437 or awrt 437 (cm³) isw
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →