Question 12:
12 | (a) | (i) | At A, cosht−2sinht =0tanht = 1
2
1 1 + 1212
 t = ln
2 1 −
= 12 l n 3 | M1*
A1
M1dep
A1 | 3.1a
1.1
1.1
1.1 | Use of tanh𝑡 = sinh𝑡/cosh𝑡 in 𝑦 = 0 equation
Correct value of tanh𝑡
1
Use of artanh formula with their
2
Must be exact, www
Alternative method 1
𝑡 −𝑡 𝑡 −𝑡
e +e e −e
cosh𝑡−2sinh𝑡 = 0⟹ −2( ) = 0
2 2 | M1* | Exponential definitions used
3e−𝑡−e𝑡 = 0 | A1 | Collecting 𝑒𝑡 terms and 𝑒−𝑡 terms correctly
3−e2𝑡 = 0
2𝑡 = ln3 | M1dep | Logs taken to isolate a term in 𝑡 correctly
1
𝑡 = ln3
2 | A1 | A1 | Must be exact www | Must be exact www
Alternative method 2
cosh2𝑡 = 4sinh2𝑡
1+sinh2𝑡 = 4sinh2𝑡 | or | cosh2𝑡 = 4(cosh2𝑡−1) | M1* | Squaring both sides and using cosh2𝑡−sinh2𝑡 = 1
1
sinh𝑡 =
√3 | or | 2
cosh𝑡 =
√3 | 2
cosh𝑡 =
√3 | A1 | A1 | 1
Correct value of sinh𝑡 or cosh𝑡. Condone sinh𝑡 = ±
√3
2
but not cosh𝑡 = ±
√3
1 1
𝑡 = ln( +√ +1)
√3 3 | or | 2 4
𝑡 = ln( +√ −1)
√3 3 | M1dep | 1 2
Use of arsinh or arcosh formula with their or
√3 √3 | 1 2
Use of arsinh or arcosh formula with their or
√3 √3
= 12 l n 3 | = 12 l n 3 | A1 | Must be exact, www. Any sinh𝑡 < 0 must be explicitly
rejected.
[4]
12 | (a) | (ii) | 1 1
x = 2cosh ( ln3) + sinh ( ln3)
2 2
= 2.89 (3 s.f.) | M1
A1
[2] | 1.1
1.1 | Substituting their 1ln3(= 0.5493) into correct
2
expression for x. FT a decimal for their exact value.
Must be 3sf
e | −𝑡 | −e | 𝑡 = 0
12 | (b) | d y d y d x
=
d x d t d t
sinht−2cosht
=
2sinht+cosht
d y
When t = 0 , s i n h t = 0 , c o s h t = 1  = − 2
d x
B is (2, 1)
so equation of tangent is y − 1 = −2(x − 2)
 y = −2x + 5 | M1
A1
A1
B1
M1
A1
[6] | 3.1a
2.1
2.1
2.1
2.1
2.2a | d𝑦
Parametric differentiation leading to an expression for
d𝑥
with correct numerator or denominator. Implied by correct
d𝑦
expression for .
d𝑥
e𝑡+3e−𝑡
Could be in exponential form, e.g.
e−𝑡−3e𝑡
soi by correct use in equation for tangent
Use of y = mx + c or y − y = m(x − x ) with their B(2,1)
1 1
d𝑦
and their value for
d𝑥
or 2𝑥+𝑦 = 5 or 2𝑥+𝑦−5 = 0. Must be simplified.