Question 5:
5 | (a) | 5coshx+3sinhx=4
ex +e−x  ex −e−x 
⇒5 +3 =4
 2   2 
⇒4ex +e−x =4
( )
( )2
⇒4e2x −4ex +1=0 ⇒ 2ex −1 =0
1
⇒ex =
2
⇒ x=−ln2 oe
Alternatively:
5coshx+3sinhx≡Rcosh(x+α) M1
where R= 25−9 =4,
 3
1+
tanhα= 3 ⇒α=tanh−1 3 = 1 ln   5  = 1 ln4=ln2 M1 A1
5 5 2 1− 3  2
 5
⇒4cosh(x+α)=4⇒cosh(x+α)=0
⇒x=−α=−ln2 A1 | M1
M1
A1
A1
[4] | 3.1a
3.1a
1.1
1.1 | Use of exponentials
Multiply by ex | Alternatively make
cosh the subject,
square and use
Pythagoras to give
quadratic in cosh
Alternatively use
compound angle
formula
5 | (b) | DR
1
∫( 5coshx+3sinhx ) dx=[ 5sinhx+3coshx ]1
−1
−1
 e1−e−1 e1+e−1  e−1−e1 e−1+e1
=5 +3 −5 +3 
 2 2   2 2 
= ( 4e1−e−1 ) − ( 4e−1−e1 )
5
=5e−
e
Alternatively:
5coshx+3sinhx
=5 ex +e−x   +3 ex −e−x   = 1( 8ex +2e−x ) M1
 2   2  2
⇒ ∫ 1 1( 8ex +2e−x ) dx=4ex −e−x 1 M1
 
2 −1
−1
= ( 4e−e−1 ) − ( 4e−1−e ) =5e− 5 A1
e | M1
M1
A1
[3] | 1.1
1.1
2.1 | Attempt at integral (i.e. one
function changed)
Convert sinhx and coshx to
exponential form in their
integrated function and use
limits correctly | Alternatively:
M1 convert (including
possibly using result
from (a))
M1 integrate and use
limits correctly