Question 6:
--- 6(a) ---
6(a) | Recalls exponential definition of
tanh | 1.2 | B1 | Let
−1
Then 𝑦𝑦 = tanh 𝑥𝑥
𝑥𝑥 = tanh𝑦𝑦
𝑦𝑦 −𝑦𝑦
e −e
𝑥𝑥 = 𝑦𝑦 −𝑦𝑦
𝑦𝑦 −e𝑦𝑦 +e𝑦𝑦 −𝑦𝑦
𝑥𝑥(e +e ) = e −e
𝑦𝑦 − 𝑦𝑦
e (𝑥𝑥−1)+e (𝑥𝑥+1)= 0
2𝑦𝑦
e (𝑥𝑥−1)+(𝑥𝑥+1) = 0
2𝑦𝑦 1+𝑥𝑥
e =
1−𝑥𝑥
1+𝑥𝑥
2𝑦𝑦 = ln� a�s required.
1−𝑥𝑥
−1 1 1+𝑥𝑥
Multiplies by their denominator
and multiplies by (or ) | 3.1a | M1
𝑦𝑦 𝑥𝑥
e e
Obtains
or 2𝑦𝑦 1+𝑥𝑥
e =1−𝑥𝑥
Obtains
OE 2𝑥𝑥 1+𝑦𝑦 | 1.1b | A1
e = 1−𝑦𝑦
Completes a rigorous argument
to show the required result | 2.1 | R1
Total | 4 | tanh 𝑥𝑥 = 2ln�1−𝑥𝑥�
Q | Marking instructions | AO | Marks | Typical solution
--- 6(b) ---
6(b) | Uses appropriate hyperbolic
identity (condone sign errors) to
obtain an equation in one
hyperbolic function
or
substitutes exponential form,
condone error in sum/difference
of exponential terms | 1.1a | M1 | 2
20(1−tanh 𝑥𝑥)−11tanh𝑥𝑥 = 1 6
2
0 = 20tanh 𝑥𝑥+11tanh𝑥𝑥−4
(5tanh𝑥𝑥+4)(4tanh𝑥𝑥−1)= 0
or
4 1
tanh𝑥𝑥 = −5 4
−1 4 1 1
𝑥𝑥 = tanh �−5�= 2ln�9�
or
−1 1 1 5
𝑥𝑥 = tanh �4�=2ln�3�
Solves their quadratic equation
to obtain two solutions in tanh x
or
quadratic in to obtain two
solutions 2𝑥𝑥
or e
quartic in to obtain at least
two solution𝑥𝑥s | 1.1a | M1
e
Obtains and OE
4 1
or
tanh𝑥𝑥 = −5 4
Obtains and OE
or 2𝑥𝑥 5 1
e = 3 9
Obtains and OE | 1.1b | A1
𝑥𝑥 √15 1
e = 3 3
Obtains correct solutions
(any correct exact log form)
ISW | 1.1b | A1
Total | 4
Question total | 8
Q | Marking instructions | AO | Marks | Typical solution