1.
43x–1 = 1.5625 × 10–2
(3x – 1)log10 4 = log10 1.5625 – 2
(M1)
⇒ 3x – 1 =
(A1)
⇒ 3x – 1 = –3
(A1)
⇒x=–
(A1) (C4)
[4]
2.
METHOD 1
(A1)
(A1)
(A1)
(A1)(A1)
(A1) (C6)
METHOD 2
(A1)(A1)(A1)
(A1)
(A2)
[6]
IB Questionbank Maths SL
1
3.
9x–1 =
32x–2 = 3–2x
2x – 2 = –2x
(M1) (A1)
(A1)
x=
(A1) (C4)
[4]
4.
(a)
log5 x2 = 2 log5 x
= 2y
(b)
log5
= –log5 x
= –y
(c)
(M1)
(A1) (C2)
log25 x =
=
(M1)
(A1) (C2)
(M1)
(A1) (C2)
[6]
5.
(a)
(b)
log2 5 =
(M1)
=
(A1) (C2)
loga 20 = loga 4 + loga 5 or loga 2 + loga 10
= 2 loga 2 + loga 5
= 2x + y
(M1)
(A1) (C2)
[4]
IB Questionbank Maths SL
2
6.
log10
= 2log10
(M1)
2log10
= 2(log10P – log10(QR3))
(M1)
= 2(1og10P – log10Q – 3log10R)
= 2(x – y – 3z)
= 2x – 2y – 6z or 2(x – y – 3z)
(M1)
(A1)
[4]
7.
(a)
Initial mass ⇒ t = 0
mass = 4
(A1)
(A1) (C2)
(b)
1.5 = 4e–0.2t (or 0.375 = e–0.2t)
ln 0.375 = –0.2t
t = 4.90 hours
(M2)
(M1)
(A1) (C4)
[6]
8.
METHOD 1
log9 81 + log9
⇒
+ log9 3 = 2 – 1 +
= log9 x
(M1)
(A1)
⇒x=
⇒ x = 27
(M1)
(A1) (C4)
METHOD 2
log 81 + log9
= log9 27
⇒ x = 27
+ log9 3 = log9
(M2)
(A1)
(A1) (C4)
[4]
IB Questionbank Maths SL
3
9.
log27 (x(x – 0.4)) = l
x2 – 0.4x = 27
x = 5.4 or x = –5
x = 5.4
(M1)(A1)
(M1)
(G2)
(A1) (C6)
Note: Award (C5) for giving both roots.
[6]
10.
Required term is
(3x)5(–2)3
Therefore the coefficient of x5 is 56 × 243 × –8
= –108864
(A1)(A1)(A1)
(A1) (C4)
[4]
11.
(a)
10
(A2) (C2)
(b)
(3x2)3
[for correct exponents]
33 x6
(M1)(A1)
(A1)
constant = 2268
(A1) (C4)
[6]
12.
(5a + b)7 = ...+
=
(5a)3(b)4 + ...
× 53 × (a3b4) = 35 × 53 × a3b4
So the coefficient is 4375
(M1)
(M1)(A1)
(A1) (C4)
[4]
IB Questionbank Maths SL
4
13.
METHOD 1
Using binomial expansion
(M1)
(A1)
= 27 + 27
+ 63 + 7
(A2)
(so p = 90 , q = 34 )
(A1)(A1)(C3)(C3)
METHOD 2
For multiplying
(M1)
(A1)
= 27 + 9
( = 27 + 27
+ 18
+ 42 + 21 + 7
+ 63 + 7
)
(so p = 90 , q = 34 )
(A2)
(A1)(A1)(C3)(C3)
[6]
14.
The constant term will be the term independent of the variable x.
(R1)
(M1)
(A1)
= –672
(A1)
[4]
15.
(a)
(b)
evidence of expanding
e.g. (x – 2)4 = x4 + 4x3(–2) + 6x2(–2)2 + 4x(–2)3 + (–2)4
M1
(x – 2)4 = x4 – 8x3 + 24x2 – 32x + 16
A2
N2
(A1)(A1)
A1
N3
finding coefficients, 3 × 24 (= 72),4 × (–8)(= –32)
term is 40x3
[6]
IB Questionbank Maths SL
5