Higher June 2025 Unit 2 Q12
12.
(a)
(i) \(y\) is inversely proportional to \(x^3\).
\(y\) = 0·15 when \(x\) = 6.
Use this information to write an expression for \(y\) in terms of \(x\). [3]
\(y\) = 0·15 when \(x\) = 6.
Use this information to write an expression for \(y\) in terms of \(x\). [3]
(ii) Use the expression you found in part (i) to complete the following table. [2]
| \(x\) | 0·2 | 6 | |
|---|---|---|---|
| \(y\) | 1·2 | 0·15 |
(b) \(c\) is directly proportional to the square root of \(d\).
What happens to \(c\) if \(d\) is multiplied by 4?
Circle the correct statement below. [1]
What happens to \(c\) if \(d\) is multiplied by 4?
Circle the correct statement below. [1]
\(c\) is multiplied by 16 \(c\) is divided by 16 \(c\) is multiplied by 2 \(c\) is divided by 2 \(c\) is squared
| Scheme | Marks | Comments | ||||||||
|---|---|---|---|---|---|---|---|---|---|---|
| (i) \(y \propto \dfrac{1}{x^3}\) OR \(y = \dfrac{k}{x^3}\) | B1 | Allow \(y \propto \dfrac{k}{x^3}\) | ||||||||
| \(0{\cdot}15 = \dfrac{k}{6^3}\) OR \(0{\cdot}15 = k / 216\) OR \(k = 32{\cdot}4\) | M1 | M1 implies B1 F.T. for B0 M1 from \(y \propto \dfrac{1}{x^n}\) with \(n \gt 0\) and \(n\) = integer, with \(n \neq 1\). No FT from direct proportion. | ||||||||
| (\(y\) =) \(\dfrac{32{\cdot}4}{x^3}\) OR (\(y\) =) \(\dfrac{162}{5x^3}\) | A1 | CAO May be seen (explicitly) in part (ii). For reference, use of \(y \propto \dfrac{1}{x^2}\) gives \(y = \dfrac{5{\cdot}4}{x^2}\) (B0M1A0) | ||||||||
(ii)
| B2 | Check working space (if table left blank). B1 for one correct value. F.T. for consistent use, in parts (i) and (ii), of inverse proportion only, including \(1/x\) For reference, use of: \(y \propto \dfrac{1}{x}\), \(\left(y = \dfrac{0.9}{x}\right)\), gives \(y\) = 4·5 and \(x\) = 0·75 \(y \propto \dfrac{1}{x^2}\), \(\left(y = \dfrac{5.4}{x^2}\right)\), gives \(y\) = 135 and \(x = \sqrt{\dfrac{9}{2}} = \dfrac{3\sqrt{2}}{2}\) = 2·12….. |
| Scheme | Marks | Comments |
|---|---|---|
| \(c\) is multiplied by 2 | B1 |
Corrected by Westie's Workshop
- The board's mark scheme printed “c is multiplied by”. Changed to “c is multiplied by 2” because the answer has to name the number, and since c is proportional to the square root of d, multiplying d by 4 multiplies c by \(\sqrt{4} = 2\).
Original: see the board's mark scheme PDF (link above).