Question 17:
17 | Recalls and uses
a n d 1 2�𝑒𝑒 𝜃𝜃 −𝑒𝑒 −𝜃𝜃 �
s in h 𝜃𝜃 =
PI 1 2�𝑒𝑒 𝜃𝜃 +𝑒𝑒 −𝜃𝜃 �
cosh𝜃𝜃 = | 1.2 | B1 | 1 𝑠𝑠 −𝑠𝑠 1 𝑠𝑠 −𝑠𝑠 1 𝑠𝑠 −𝑠𝑠
�𝑒𝑒 −𝑒𝑒 �×� �𝑒𝑒 −𝑒𝑒 �+ �𝑒𝑒 +𝑒𝑒 ��= 1
2 2 2
1 𝑠𝑠 −𝑠𝑠 1 𝑠𝑠 1 𝑠𝑠
�𝑒𝑒 −𝑒𝑒 � ×� 𝑒𝑒 + 𝑒𝑒 �= 1
2 2 2
2𝑠𝑠 0
𝑒𝑒 −𝑒𝑒 = 2
2𝑠𝑠
𝑒𝑒 = 3
2𝜃𝜃 = ln3
1
Forms equation and rearranges to obtain exactly one exponential term | 3.1a | M1
Takes logarithms of an equation of the form where
2𝑠𝑠 | 1.1a | M1
𝑒𝑒 = 𝑘𝑘 𝑘𝑘 > 0
Obtains correct answer in required form | 1.1b | A1
ALT
17 | Use of
PI 2 2
𝑐𝑐𝑐𝑐𝑠𝑠ℎ 𝜃𝜃−𝑠𝑠𝑖𝑖𝑛𝑛ℎ 𝜃𝜃 = 1 | 3.1a | M1 | 2 2 2
𝑠𝑠𝑖𝑖𝑛𝑛ℎ 𝜃𝜃+𝑠𝑠𝑖𝑖𝑛𝑛ℎ𝜃𝜃𝑐𝑐𝑐𝑐𝑠𝑠ℎ𝜃𝜃 = 𝑐𝑐𝑐𝑐𝑠𝑠ℎ 𝜃𝜃−𝑠𝑠𝑖𝑖𝑛𝑛ℎ 𝜃𝜃
2 2 2
𝑠𝑠𝑖𝑖𝑛𝑛ℎ 𝜃𝜃 𝑠𝑠𝑖𝑖𝑛𝑛ℎ𝜃𝜃𝑐𝑐𝑐𝑐𝑠𝑠ℎ𝜃𝜃 𝑐𝑐𝑐𝑐𝑠𝑠ℎ 𝜃𝜃 𝑠𝑠𝑖𝑖𝑛𝑛ℎ 𝜃𝜃
2 + 2 = 2 − 2
𝑐𝑐𝑐𝑐𝑠𝑠ℎ 𝜃𝜃 𝑐𝑐𝑐𝑐𝑠𝑠ℎ 𝜃𝜃 𝑐𝑐𝑐𝑐𝑠𝑠ℎ 𝜃𝜃 𝑐𝑐𝑐𝑐𝑠𝑠ℎ 𝜃𝜃
2 2
𝑡𝑡 𝑎𝑎𝑛𝑛ℎ 𝜃𝜃+𝑡𝑡𝑎𝑎𝑛𝑛ℎ𝜃𝜃 = 1−𝑡𝑡𝑎𝑎𝑛𝑛ℎ 𝜃𝜃
2
2𝑡𝑡𝑎𝑎𝑛𝑛ℎ 𝜃𝜃+𝑡𝑡𝑎𝑎𝑛𝑛ℎ𝜃𝜃−1 = 0
(2𝑡𝑡𝑎𝑎𝑛𝑛ℎ𝜃𝜃− 1 )o(r𝑡𝑡 𝑎𝑎 𝑛𝑛 ℎ𝜃𝜃+1)= 0
1
𝑡𝑡𝑎𝑎𝑛𝑛ℎ𝜃𝜃 =2 𝑡𝑡𝑎𝑎𝑛𝑛ℎ𝜃𝜃 = −1
but only
1
𝑡𝑡𝑎𝑎𝑛𝑛ℎ𝜃𝜃 ≠ −1 ∴ 𝑡𝑡𝑎𝑎𝑛𝑛ℎ𝜃𝜃 = 2
1
𝜃𝜃 = artanh� �
Recalls and uses
𝑠𝑠𝑠𝑠𝑛𝑛ℎ𝑠𝑠
𝑡𝑡𝑎𝑎𝑛𝑛ℎ𝜃𝜃 = 𝑐𝑐𝑐𝑐𝑠𝑠ℎ𝑠𝑠 | 1.2 | B1
Solves a three-term quadratic in (oe)
𝑡𝑡𝑎𝑎𝑛𝑛ℎ𝜃𝜃 | 1.1a | M1
Obtains the correct answer.
ISW
Condone lack of reference to | 1.1b | A1
𝑡𝑡𝑎𝑎𝑛𝑛ℎ𝜃𝜃 ≠ −1
Total | 4 | 2
Q | Marking instructions | AO | Mark | Typical solution