Question 8:
8 | (a) | y =4sinhx+3coshx
dy
⇒ =4coshx+3sinhx
dx
=0 when 4coshx+3sinhx=0
ex +e−x  ex −e−x 
⇒4 +3 =0
 2   2 
1
⇒e2x =−
7
which is not possible as e2x >0 so no turning points | M1
M1
A1 | 1.1
2.1
2.4 | Diffn (Hyperbolics or exponentials)
Set = 0 and use exponential forms – can change to
exponentials before diffn.
Conclusion with justification
Alternative method | M1
M1
A1 | Differentiate
Set = 0 and use formula for tanh
Conclusion with justification
y =4sinhx+3coshx
dy
⇒ =4coshx+3sinhx
dx
4
=0 when tanhx=−
3
But tanhx <1 for all x.
dy
So there are no values of x for which =0
dx
So no turning points
[3]
M1
M1
A1
Differentiate
Set = 0 and use formula for tanh
Conclusion with justification
(b) | y =4sinhx+3coshx=5
ex −e−x  ex+e−x 
⇒4 +3 =5
 2   2 
⇒7ex −e−x =10⇒7e2x −10ex −1=0
10± 100+28 5+ 32 5− 32
ex = = or
14 7 7
5− 32
But ex >0 so cannot =
7
5+ 32
So the only root is ex =
7
5+4 2 
⇒ x=ln 
 
 7  | M1
M1
A1
A1
A1 | 3.1a
3.1a
1.1
2.3
1.1 | Use of exponentials
equation of the form ae2x +bex +c=0(for non-zero a, b
and c)
Two roots for ex
One rejected plus reason
Ignore inclusion of 2nd root
Alternative method (see appendix for full working) | M1
M1
A1
A1
A1 | Use Pythagoras
Quadratic in cosh (or sinh)
Two roots
One rejected plus reason
4sinhx+3coshx=5⇒4sinhx=5−3coshx
∴16sinh2x=16(cosh2x−1)=25−30coshx+9cosh2x
7cosh2x+30coshx−41=0
−15+16 2
coshx≥1⇒coshx=
7
 
−15+16 2 −15+16 2+4 43−30 2
⇒x=cosh−1 =±ln 
7  7 
 
But the negative root does not work in the original
equation since LHS would be negative while RHS
would be positive (but equal when squared).
 
−15+16 2+4 43−30 2
∴x=ln 
 7 
 
[5]
M1
M1
A1
A1
A1
Use Pythagoras
Quadratic in cosh (or sinh)
Two roots
One rejected plus reason