June 2023 Paper 1 Q3
3
Find the equation of the curve. [3]
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{f}(x + h) - \mathrm{f}(x) = \left((x + h)^2 + 2(x + h)\right) - \left(x^2 + 2x\right)\) | M1 | 2.1 |
| \(= x^2 + 2xh + h^2 + 2x + 2h - x^2 - 2x\) \(= 2xh + h^2 + 2h\) | M1 | 2.1 |
| \(\dfrac{\mathrm{f}(x + h) - \mathrm{f}(x)}{h} = \dfrac{2xh + h^2 + 2h}{h}\) \(= 2x + h + 2\) | M1 | 2.1 |
| \(\mathrm{f}'(x) = \lim\limits_{h \to 0}(2x + h + 2) = 2x + 2\) | A1 | 2.5 |
| [4] |
Notes
M1: Attempt expression for \(\mathrm{f}(x + h) - \mathrm{f}(x)\)
Allow sign error from no bracket around final term, ie \((x + h)^2 + 2(x + h) - x^2 + 2x\) is M1, but no other errors allowed
If considering \(x^2\) and \(2x\) separately then expressions for both must be seen
M1: Expand and simplify \(\mathrm{f}(x + h) - \mathrm{f}(x)\)
Expand and gather like terms (either separately, or single expression)
Condone sign errors only, so M0 if collecting like terms after an incorrect attempt to divide by \(h\)
Allow BOD if \(2x \ldots + 2x\) becomes 0 rather than \(4x\)
M1: Attempt \(\dfrac{\mathrm{f}(x + h) - \mathrm{f}(x)}{h}\)
Divide all terms by \(h\)
Allow BOD if previous error results in a term with a denominator of \(h\)
A1: Complete proof by considering limit as \(h \to 0\)
www, including correct signs throughout
Must divide by \(h\) before \(h \to 0\)
Must see ‘lim’, ‘\(h \to 0\)’, and \(\mathrm{f}'(x)\) at some point in their solution and not just when quoting the generic formula, but allow BOD for \(\mathrm{f}'(x) = \frac{\mathrm{f}(x+h) - \mathrm{f}(x)}{h}\) followed by =…, =…, =… on subsequent lines
A0 if ‘lim’ still in final answer
Condone \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in place of \(\mathrm{f}'(x)\)
| Scheme | Marks | AO |
|---|---|---|
| \(y = x^2 + 2x + c\) | B1 | 2.2a |
| \(5 = 1 - 2 + c\) \(c = 6\) | M1 | 1.1 |
| \(y = x^2 + 2x + 6\) | A1 | 1.1 |
| [3] |
Notes
B1: State or imply correct equation, including \(+\,c\)
‘\(y =\)’ could be implied by use of 5
\(c\) may be implied by later work
M1: Attempt \(c\) using \((-1, 5)\)
Allow M1 if equation incorrect, as long as from attempt at integrating \(2x + 2\) ie of form \(y = kx^2 + 2x + c\)
\(c\) may be implied by method eg \(y = x^2 + 2x\), followed by \(5 = 1 - 2\) and then an attempt to ‘balance’ the sides
Must use \(x\) and \(y\) the correct way around
As far as attempting a value for \(c\)
A1: Obtain correct equation, including \(y = \ldots\)
Equation must be stated, and not just implied by \(c = 6\) seen
Allow \(\mathrm{f}(x) = \ldots\)
A0 for ‘equation’ \(= x^2 + 2x + 6\)
Just stating \(y = x^2 + 2x + 6\) or \(y = (x + 1)^2 + 5\) gets full marks (may come from observing that \((-1, 5)\) is the minimum point)