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Date	May 2021	Marks available	3	Reference code	21M.1.SL.TZ1.7
Level	Standard Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 1
Command term	Find	Question number	7	Adapted from	N/A

Question

Let $f (x) = m x^{2} - 2 m x$ , where $x \in ℝ$ and $m \in ℝ$ . The line $y = m x - 9$ meets the graph of $f$ at exactly one point.

The function $f$ can be expressed in the form $f (x) = 4 (x - p) (x - q)$ , where $p, q \in ℝ$ .

The function $f$ can also be expressed in the form $f (x) = 4 {(x - h)}^{2} + k$ , where $h, k \in ℝ$ .

Show that $m = 4$ .

[6]

Find the value of $p$ and the value of $q$ .

[2]

Find the value of $h$ and the value of $k$ .

[3]

Hence find the values of $x$ where the graph of $f$ is both negative and increasing.

[3]

Markscheme

METHOD 1 (discriminant)

$m x^{2} - 2 m x = m x - 9$ (M1)

$m x^{2} - 3 m x + 9 = 0$

recognizing $Δ = 0$ (seen anywhere) M1

$Δ = {(- 3 m)}^{2} - 4 (m) (9)$ (do not accept only in quadratic formula for $x$ ) A1

valid approach to solve quadratic for $m$ (M1)

$9 m (m - 4) = 0$ OR $m = \frac{36 \pm \sqrt{36^{2} - 4 \times 9 \times 0}}{2 \times 9}$

both solutions $m = 0, 4$ A1

$m \neq 0$ with a valid reason R1

the two graphs would not intersect OR $0 \neq - 9$

$m = 4$ AG

METHOD 2 (equating slopes)

$m x^{2} - 2 m x = m x - 9$ (seen anywhere) (M1)

$f' (x) = 2 m x - 2 m$ A1

equating slopes, $f' (x) = m$ (seen anywhere) M1

$2 m x - 2 m = m$

$x = \frac{3}{2}$ A1

substituting their $x$ value (M1)

${(\frac{3}{2})}^{2} m - 2 m \times \frac{3}{2} = m \times \frac{3}{2} - 9$

$\frac{9}{4} m - \frac{12}{4} m = \frac{6}{4} m - 9$ A1

$\frac{- 9 m}{4} = - 9$

$m = 4$ AG

METHOD 3 (using $\frac{- b}{2 a}$ )

$m x^{2} - 2 m x = m x - 9$ (M1)

$m x^{2} - 3 m x + 9 = 0$

attempt to find $x$ -coord of vertex using $\frac{- b}{2 a}$ (M1)

$\frac{- (- 3 m)}{2 m}$ A1

$x = \frac{3}{2}$ A1

substituting their $x$ value (M1)

${(\frac{3}{2})}^{2} m - 3 m \times \frac{3}{2} + 9 = 0$

$\frac{9}{4} m - \frac{9}{2} m + 9 = 0$ A1

$- 9 m = - 36$

$m = 4$ AG

[6 marks]

$4 x (x - 2)$ (A1)

$p = 0$ and $q = 2$ OR $p = 2$ and $q = 0$ A1

[2 marks]

attempt to use valid approach (M1)

$\frac{0 + 2}{2}, \frac{- (- 8)}{2 \times 4}, f (1), 8 x - 8 = 0$ OR $4 (x^{2} - 2 x + 1 - 1) (= 4 {(x - 1)}^{2} - 4)$

$h = 1, k = - 4$ A1A1

[3 marks]

EITHER

recognition $x = h$ to $2$ (may be seen on sketch) (M1)

recognition that $f (x) < 0$ and $f' (x) > 0$ (M1)

THEN

$1 < x < 2$ A1A1

Note: Award A1 for two correct values, A1 for correct inequality signs.

[3 marks]

Examiners report

[N/A]

Syllabus sections

Topic 2—Functions » SL 2.6—Quadratic function

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22M.1.SL.TZ2.7a:
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21N.1.SL.TZ0.7b:
Sketch a graph of $v$ against $t$ , clearly showing any points of intersection with the axes.
18M.2.AHL.TZ2.H_10c:
Sketch the graph of $y = g\left( t \right)$ for t ≤ 0. Give the coordinates of any intercepts and the equations of any asymptotes.
18M.2.AHL.TZ2.H_10a.iii:
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21N.1.SL.TZ0.1a.i:
find the $x$ -coordinates of the $x$ -intercepts.
18M.2.AHL.TZ2.H_10a.ii:
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18M.2.AHL.TZ2.H_10a.iv:
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18M.2.AHL.TZ2.H_10a.i:
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21M.1.SL.TZ1.7d:
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18M.2.AHL.TZ2.H_10b:
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22M.1.SL.TZ1.7b.ii:
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22M.1.SL.TZ1.7c:
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EXN.1.SL.TZ0.3:
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22M.1.SL.TZ2.7b.ii:
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22M.1.AHL.TZ2.11c:
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20N.1.SL.TZ0.T_11b:
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16N.1.SL.TZ0.S_1b:
(i) Write down the value of $h$ .

(ii) Find the value of $k$ .
21M.2.SL.TZ2.3b:
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21M.2.SL.TZ2.5a:
Find the range of $f$ .
21M.2.SL.TZ2.5b:
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19N.1.SL.TZ0.S_3b:
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find the coordinates of the vertex.
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Write down the value of $h$ and the value of $k$ .
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20N.1.SL.TZ0.T_11c:
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