Higher November 2021 Unit 2 Q10
10.
(a)
(i) You are given that \(y\) is inversely proportional to \(\sqrt{x}\).
\(y = 65\) when \(x = 51{\cdot}84\).
Find an expression for \(y\) in terms of \(x\). [3]
\(y = 65\) when \(x = 51{\cdot}84\).
Find an expression for \(y\) in terms of \(x\). [3]
(ii) Use the expression you found in part (i) to complete the following table. [2]
| \(x\) | 51·84 | 15·21 | |
| \(y\) | 65 | 78 |
(b) It is known that \(c\) is directly proportional to the square of \(d\).
What happens to \(c\) if \(d\) is doubled?
Circle the correct statement below. [1]
What happens to \(c\) if \(d\) is doubled?
Circle the correct statement below. [1]
| \(c\) is divided by 2 | \(c\) is multiplied by 2 | \(c\) is divided by 4 | \(c\) is multiplied by 4 | \(c\) is squared |
| Scheme | Marks | Comments | ||||||||
|---|---|---|---|---|---|---|---|---|---|---|
| (a)(i) \(y \propto 1/\sqrt{x}\) OR \(y = k/\sqrt{x}\) | B1 | Allow \(y \propto k/\sqrt{x}\) | ||||||||
| \(65 = k/\sqrt{51{\cdot}84}\) OR \(65 = k/7{\cdot}2\) OR \(k = 65 \times \sqrt{51{\cdot}84}\) OR \(k = 65 \times 7{\cdot}2\) OR \(k = 468\) | M1 | M1 implies B1 F.T. for B0 M1 from \(y \propto 1/x^n\) with \(n \gt 0\) and \(n \ne 1/2\) No F.T. from direct proportion | ||||||||
| \((y =)\ 468/\sqrt{x}\) | A1 | May be seen explicitly in part (ii). | ||||||||
(a)(ii)
| B2 | Check working space (if table left blank). B1 for one correct value. F.T. for consistent use of ‘their expression’ for inverse proportion only, but not for \(y = 1/x\) |
| Scheme | Marks | Comments |
|---|---|---|
| \(c\) is multiplied by 4 | B1 |