IB Math – Standard Level Year 1 – Quadratics Practice: MarkScheme
Alei - Desert Academy
Quadratics – Practice Problems - Markscheme
1.
(a)
(b)
2
x – 3x – 10 = (x – 5)(x + 2)
x2 – 3x – 10 = 0  (x – 5)(x + 2) = 0
 x = 5 or x = –2
(M1)(A1) (C2)
(M1)
(A1)(C2)
[4]
2.
(a)
(b)
3.
4.
5.
6.
7.
1
,q=2
2
or vice versa
By symmetry C is midway between p, q
Note: This (M1) may be gained by implication.
1 2 3
2 
 x-coordinate is
2
4
p=–
(7 – x)(1 + x) = 0
 x = 7 or x = –1
7  1
B: x =
= 3;
2
y = (7 – 3)(l + 3) = 16
(A1)(A1) (C2)
(M1)
(A1)(C2)
(M1)
(A1)(C1)(C1)
(A1)
(A1)(C2)
[4]
2
4x + 4kx + 9 = 0
Only one solution  b2 – 4ac = 0
16k2 – 4(4)(9) = 0
k2 = 9
k = 3
But given k > 0, k = 3
OR
One solution  (4x2 + 4kx + 9) is a perfect square
4x2 + 4kx + 9 = (2x  3)2 by inspection
given k > 0, k = 3
Graph of quadratic function.
Expression
+
–
a
c
b2 – 4ac
b
(a)
(M1)
(A1)
(A1)
(A1)(C4)
(M1)
(A2)
(A1)(C4)
[4]
0
(A1)
(A1)
(A1)
(A1)
(C1)
(C1)
(C1)
(C1)
[4]
2
f (x) = x – 6x + 14
f (x) = x2 – 6x + 9 – 9 + 14
f (x) = (x – 3)2 + 5
(M1)
(M1)
(b)
Vertex is (3, 5)
(A1)(A1)
(a)
When t = 0,
h = 2 + 20 × 0 – 5 × 02 = 2
(M1)
(A1) 2
(b)
[4]
When t = 1,
h = 2 + 20 × 1 – 5 × 12
= 17
h=2
[4]
(M1)
(A1)
(AG) 2
C:\Users\Bob\Documents\Dropbox\Desert\SL\1Algebra&Functions\TestsQuizzesPractice\SLA1QuadraticsPractice.docx on 11/30/14 at 7:54 AM
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IB Math – Standard Level Year 1 – Quadratics Practice: MarkScheme
(c)
(d)
(i)
(ii)
(i)
(ii)
(iii)
(iv)
8.
Alei - Desert Academy
h = 17  17 = 2 + 20t – 5t2
5t2 – 20t + 15 = 0
 5(t2 – 4t + 3) = 0
 (t – 3)(t – 1) = 0
Note: Award (M1) for factorizing or using the formula
 t = 3 or 1
Note: Award (A1) for t = 3
h = 2 + 20t – 5t2
dh

= 0 + 20 – 10t
dt
= 20 – 10t
(M1)
(M1)
t=0
dh

= 20 – 10 × 0 = 20
dt
(M0)
(M1)
(A1) 4
(A1)(A1)
(A1)
dh
=0
dt
 20 – 10t = 0  t = 2
(M1)
(A1)
t=2
 h = 2 + 20 × 2 – 5 × 22 = 22  h = 22
(M1)
(A1) 7
y = (x +2)(x – 3)
= x2 – x – 6
Therefore, 0 = 4 – 2p + q
OR
y = x2 – x – 6
OR
0 = 4 – 2p + q
0 = 9 + 3p + q
p = –1, q = –6
(M1)
(A1)
(A1)(A1)
[15]
(C2)(C2)
(C3)
(A1)
(A1)
(A1)(A1)
(C2)(C2)
[4]
9.
8
y =x2
6
2
y =5–3(x–4)
4
2
–2
q=5
k = 3, p = 4
0
2
4
6
(A1)(C1)
(A3)(C3)
[4]
C:\Users\Bob\Documents\Dropbox\Desert\SL\1Algebra&Functions\TestsQuizzesPractice\SLA1QuadraticsPractice.docx on 11/30/14 at 7:54 AM
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IB Math – Standard Level Year 1 – Quadratics Practice: MarkScheme
10.
11.
(a)
(i)
(ii)
(b)
a(l + l)2 + 2 = 0
a = –0.5
Alei - Desert Academy
h=–1
k=2
2
(A2)(C2)
(A1)(C1)
(M1)(A1)
(A1)(C3)
[6]
2
(a)
2x – 8x + 5 = 2(x – 4x + 4) + 5 – 8
= 2(x – 2)2 – 3
=> a = 2, p = 2, q = –3
(M1)
(A1)(A1)(A1)
(C4)
(b)
Minimum value of 2(x – 2)2 = 0 (or minimum value occurs when x = 2) (Ml)
 Minimum value of f (x) = –3
(A1)(C2)
OR
Minimum value occurs at (2, –3)
(M1)(A1)
(C2)
[6]
12.
Method 1
b2 – 4ac = 9 – 4k
9 – 4k > 0
2.25 > k
crosses the x-axis if k = 1 or k = 2
probability =
(M1)
(M1)
(A1)
(A1)(A1)
2
7
Method 2
(A1)(C6)
y
3
2
1
x
(M2)(M1)
Note: Award (M2) for one (relevant) curve;
(M1) for a second one.
k = 1 or k = 2
probability =
13.
(a)
(b)
(c)
(G1)(G1)
2
7
(A1)(C6)
Since the vertex is at (3, 1)
h=3
(A1)
k=1
(A1) 2
2
(5, 9) is on the graph  9 = a(5 – 3) + 1
(M1)
= 4a + 1
(A1)
=>9–1=4a=8
(A1)
=>a=2
(AG) 3
Note: Award (M1)(A1)(A0) for using a reverse proof, ie
substituting for a, h, k and showing that (5, 9) is on the graph.
2
y = 2(x – 3) + 1
(M1)
2
= 2x – 12x + 19
(AG) 1
C:\Users\Bob\Documents\Dropbox\Desert\SL\1Algebra&Functions\TestsQuizzesPractice\SLA1QuadraticsPractice.docx on 11/30/14 at 7:54 AM
[6]
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IB Math – Standard Level Year 1 – Quadratics Practice: MarkScheme
(d)
Graph has equation y = 2x2 – 12x + 19
(i)
dy
= 4x – 12
dx
(ii)
(iii)
14.
Alei - Desert Academy
At point (5, 9), gradient = 4(5) – 12 = 8
Equation: y – 9 = 8(x – 5)
8x – y – 31 = 0
OR
9 = 8(5) + c
c = –31
y = 8x – 31
(A1)
(A1)
(M1)(A1)
(M1)
(A1) 4
One solution  discriminant = 0
(M2)
32 – 4k = 0
(A2)
9 = 4k
9 1

.25

k = 2 ,2
(A2)(C6)
4 4

Note: If candidates correctly solve an incorrect equation, award
M2 A0 A2(ft), if they have the first line or equivalent, otherwise
award no marks.
[10]
[6]
15.
(a)
a = 3, b = 4
f (x) = (x – 3)2 + 4
(A1)
A1 (C2)
(b)
y = (x – 3)2 + 4
METHOD 1
x = (y – 3)2 + 4
x – 4 = (y – 3)2
x4 = y – 3
(M1)
y = x4 + 3
METHOD 2
y – 4 = (x – 3)2
y4 = x – 3
(M1)
(A1) 3
(M1)
(M1)
y4 + 3 = x
x4 + 3
y=
–1
 f (x) =
(c)
x4
x4 + 3
(A1) 3
(A1)(C1)
[6]
C:\Users\Bob\Documents\Dropbox\Desert\SL\1Algebra&Functions\TestsQuizzesPractice\SLA1QuadraticsPractice.docx on 11/30/14 at 7:54 AM
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