C3 June 2017 Q3

EdexcelOld spec9 marksFunctions (including |mod|)

Figure 1: sketch of y = g(x), an increasing curve starting at x = -2 and crossing the positive y-axis — Figure 1

Figure 1 shows a sketch of part of the graph of \(y = \mathrm{g}(x)\), where\[\mathrm{g}(x) = 3 + \sqrt{x + 2}, \qquad x \geqslant -2\]

(a) State the range of g. (1)

(b) Find \(\mathrm{g}^{-1}(x)\) and state its domain. (3)

(d) Hence state the value of \(a\) for which\[\mathrm{g}(a) = \mathrm{g}^{-1}(a)\] (1)

Mark scheme (a)

Scheme	Marks
\(y \geqslant 3\)	B1
	(1)

Notes

B1: States the correct range for g Accept \(\mathrm{g}(x) \geqslant 3\), \(g \geqslant 3\), Range \(\geqslant 3\), \([3, \infty)\) Range is greater than or equal to 3
Condone \(\mathrm{f} \geqslant 3\) Do not accept \(\mathrm{g}(x) > 3, x \geqslant 3, (3, \infty)\)

Mark scheme (b)

Scheme	Marks
\(y = 3 + \sqrt{x + 2} \Rightarrow y - 3 = \sqrt{x + 2} \Rightarrow x = (y - 3)^2 - 2\)	M1 A1
\(\Rightarrow \mathrm{g}^{-1}(x) = (x - 3)^2 - 2, \quad \text{with } x \geqslant 3\)	A1
	(3)

Notes

M1: Attempts to make \(x\) or a swapped \(y\) the subject of the formula. The minimum expectation is that the 3 is moved over followed by an attempt to square both sides. Condone for this mark \(\sqrt{x + 2} = y \pm 3 \Rightarrow x + 2 = y^2 \pm 9\)

A1: Achieves \(x = (y - 3)^2 - 2\) or if swapped \(y = (x - 3)^2 - 2\) or equivalent such as \(x = y^2 - 6y + 7\)

A1: Requires a correct function in \(x\) + correct domain or a correct function in \(x\) with a correct follow through on the range in (a) but do not follow through on \(x \in \mathbb{R}\)
Accept for example \(\mathrm{g}^{-1}(x) = (x - 3)^2 - 2, \quad x \geqslant 3\) Condone \(\mathrm{f}^{-1}(x) = (x - 3)^2 - 2, \quad x \geqslant 3\)
or variations such as \(y = (x - 3)^2 - 2, \quad x > 3\) if (a) was \(y > 3\)
Accept expanded versions such as \(\mathrm{g}^{-1}(x) = x^2 - 6x + 7, \quad x \geqslant 3\) but remember to isw after a correct answer
(Condone \(\mathrm{f}^{-1}(x) = x^2 - 6x + 7, \quad x \geqslant 3\))

Mark scheme (c)

Scheme	Marks
\(\mathrm{g}(x) = x \Rightarrow 3 + \sqrt{x + 2} = x\)
\(\Rightarrow x + 2 = (x - 3)^2 \Rightarrow x^2 - 7x + 7 = 0\)	M1, A1
\(\Rightarrow x = \dfrac{7 \pm \sqrt{21}}{2} \Rightarrow x = \dfrac{7 + \sqrt{21}}{2}\) only	M1, A1
	(4)

(c) Alt

Scheme	Marks
Solves \(\mathrm{g}^{-1}(x) = x \Rightarrow (x - 3)^2 - 2 = x\)
\(\Rightarrow x^2 - 7x + 7 = 0\)	M1, A1
\(\Rightarrow x = \dfrac{7 \pm \sqrt{21}}{2} \Rightarrow x = \dfrac{7 + \sqrt{21}}{2}\) only	dM1, A1
	(4)

Notes

M1: Sets \(3 + \sqrt{x + 2} = x\), moves the 3 over and then attempts to square both sides.
Can be scored for \(\sqrt{x + 2} = x - 3 \Rightarrow x + 2 = x^2 \pm 9\)

A1: \(x^2 - 7x + 7 = 0\). The = 0 may be implied by subsequent working

M1: Correct method of solving their 3TQ by the formula/ completing the square. The equation must have real roots.
It is dependent upon them having attempted to set \(3 + \sqrt{x + 2} = x\) and proceeding to a quadratic.
You may just see both roots written down which is fine.
Allow for this mark decimal answers Eg 5.79 and 1.21 for \(x^2 - 7x + 7 = 0\) You may need to check with a calc.

A1: \((x) = \dfrac{7 + \sqrt{21}}{2}\) or exact equivalent only.
This answer following the correct quadratic would imply the previous M
Allow \(x = \dfrac{7}{2} + \sqrt{\dfrac{21}{4}}\) but DO NOT allow \(x = \dfrac{7 \pm \sqrt{21}}{2}\)

(c) can of course be attempted by solving \(3 + \sqrt{x + 2} = \text{"}(x - 3)^2 - 2\text{"} \Rightarrow x^4 - 12x^3 + 44x^2 - 49x + 14 = 0\)
\(\vdots \qquad \Rightarrow \left(x^2 - 7x + 7\right)\left(x^2 - 5x + 2\right) = 0\)

The scheme can be applied to this

Mark scheme (d)

Scheme	Marks
\(a = \dfrac{7 + \sqrt{21}}{2}\)	B1 ft
	(1)
	(9 marks)

Notes

B1ft: \((a) = \dfrac{7 + \sqrt{21}}{2}\) oe. You may condone \(\boldsymbol{x} = \dfrac{7 + \sqrt{21}}{2}\). You may allow this following a re - start.
You may allow the correct decimal answer, awrt 5.79, following exact/decimal work in part (c) or a restart.
Follow through on their root, including decimals, coming from the positive root with the positive sign in (c).
Eg In (c) . \(x^2 - 7x + 11 = 0 \Rightarrow x = \dfrac{7 \pm \sqrt{5}}{2}\) So the correct follow through would be \(x = \dfrac{7 + \sqrt{5}}{2}\)
If they only had one root in (c) then follow through on this as long as it is positive.

SC. If they give the correct roots in parts (c) and (d) without considering the correct answer then award B1 in (d) following the A0 in (c). So \((x) = \dfrac{7 \pm \sqrt{21}}{2}\) as their answer in part (c), allow \((x/a) = \dfrac{7 \pm \sqrt{21}}{2}\) for B1 in (d).