Question 7:
7 | (a) |  e u − e − u   e u + e − u 
( 2 s i n h u c o s h u  ) 2
2 2
eu −e−u eu +e−u 
2sinhucoshu 2  
 2  2 
 1 ( e 2 u + 1 − 1 − e − 2 u )  1 ( e 2 u − e − 2 u )  s i n h 2 u
2 2 | M1
A1
[2] | 1.2
2.1 | Use correct definitions of cosh u and sinh u in terms of
exponentials. Condone omission of 2 s i n h u c o s h u for this
mark. (Note 2 s i n h u c o s h u or LHS must be included for
subsequent A1.) M1 can be implied for stating that
2 s i n h u c o s h u  12 ( e − u − e − u ) ( e u + e − u ) (but not for the A
mark).
Complete proof with no errors. Allow LHS and RHS in place
of 2 sinh u cosh u and sinh 2u respectively. Minimally
acceptable proof is of the form
 e u − e − u   e u + e − u  e 2 u − e − 2 u
R H S  2   L H S in
2 2 2
either direction.
Do not condone sin for sinh or cos for cosh or x for u (unless
“let x = u” stated) for A1.
If candidate shows that both LHS and RHS are equal to some
third expression, then a conclusion must be clearly stated for
the A1 mark.
SC B1 If candidate works with given identity and concludes,
for example, that 1 = 1.
Alternative method (working left to right)
( s i n h 2 u  ) 1 ( e 2 u − e − 2 u )
2 | M1 | Use definition of s i n h 2 u in terms of exponentials. Condone
omission of sinh2ufor this mark. (Note sinh2uor RHS
must be included for subsequent A1).
1 1
sinh2u  (e2u −e−2u) (eu −e−u)(eu +e−u)
2 2
1 1
2 (eu −e−u) (eu+e−u)2sinhucoshu
2 2 | A1 | 1 1
Complete proof must see 2 (eu −e−u) (eu +e−u)
2 2
before stating correct RHS.
[2]
Alternative method (working left to right)
( s i n h 2 u  ) 1 ( e 2 u − e − 2 u )
2
M1
Use definition of s i n h 2 u in terms of exponentials. Condone
omission of sinh2ufor this mark. (Note sinh2uor RHS
1 1
Complete proof must see 2 (eu −e−u) (eu +e−u)
2 2
before stating correct RHS.
(b) | For reference: y=16coshx−sinh2x
d y
= 1 6 s i n h x − 2 c o s h 2 x
d x
d 2 y
= 1 6 c o s h x − 4 s i n h 2 x
d x 2
= 1 6 c o s h x − 4 ( 2 s i n h x c o s h x )
= 8 c o s h x ( 2 − s i n h x )
coshx0  x = s i n h − 1 2 or s i n h x = 2 | M1*
A1
M1dep*
A1
[4] | 1.1
1.1
3.1a
2.2a | Differentiating to get 16sinhxkcosh2x for some
k 0. May be in exponential form
dd yx = 8 ( e x − e − x ) − ( e 2 x + e − 2 x ) - if using exponential form
then the derivative must be of the form
dd yx =  8 ( e x − e − x )  k ( e 2 x + e − 2 x ) for some k  0 .
Correct second derivative, condone un-simplified.
May be in exponential form.
d 2 y2 = 8 ( e x + e − x ) − 2 ( e 2 x − e − 2 x ) (oe).
d x
Correct use of the result from part (a) in their second
derivative or their d 2 y2 = 8 ( e x + e − x ) − 2 ( e x − e − x ) ( e x + e − x )
d x
using difference of two squares (oe e.g. setting equal to zero
and converting to a quartic in e x which if correct is
e 4 x − 4 e 3 x − 4 e x − 1 = 0 ). If converting to exponentials at
any point, then correct exponential expression(s) for sinh
and cosh must be used.
d 2 y
Or in exponential form = 2 ( e x + e − x ) ( 4 − e x + e − x )
d x 2
Must state coshx0 or coshx 1or show that there is no
solution of the equation c o s h x = 0 (oe e.g. ‘cosh cannot be
0’, ‘cosh is at least 1’, ‘cosh has a minimum at (0, 1)’, ‘
c o s h − 1 only valid for x 1’ but not just ‘cosh x = 0 has no
solutions’) - allow stating that the only solution satisfies the
equation s i n h x = 2 . If using exponentials then must show
clearly that there is only one solution of the correct equation
(which for reference is ln(2+ 5)).
Alternative method for second and third marks
d y
= 1 6 s i n h x − 2 ( c o s h 2 x + s i n h 2 x )
d x | d y
= 1 6 s i n h x − 2 ( c o s h 2 x + s i n h 2 x )
d x | M1dep* | M1dep* | Use of c o s h 2 x = c o s h 2 x + s i n h 2 x in their first derivative.
dd yx = 8 ( e x − e − x ) − 12 ( e x + e − x ) 2 − 12 ( e x − e − x ) 2 | Use of c o s h 2 x = c o s h 2 x + s i n h 2 x in their first derivative.
d 2 y
= 1 6 c o s h x − 4 s i n h x c o s h x − 4 s i n h x c o s h x
d x 2 | A1 | Correct second derivative, condone un-simplified.
May use exponential form:
d2y
=8(ex+e−x)−(ex +e−x)(ex −e−x)−(ex −e−x)(ex+ e−x)
dx2
Alternative method for first three marks
y = 1 6 c o s h x − 2 s i n h x c o s h x | y = 1 6 c o s h x − 2 s i n h x c o s h x | B1 | B1 | Using the result from part (a). Or
y = 8 ( e x + e − x ) − 0 . 5 ( e x − e − x ) ( e x + e − x ) | Using the result from part (a). Or
d𝑦
= sinh𝑥(16−2sinh𝑥)−2cosh2𝑥
d𝑥 | M1 | Differentiating to get  1 6 s i n h x  k c o s h 2 x  k s i n h 2 x or
dd yx =  8 ( e x − e − x )  k ( e x + e − x ) 2  k ( e x − e − x ) 2 for some
k  0 .
d 2 y
= 1 6 c o s h x − 4 s i n h x c o s h x − 4 s i n h x c o s h x
d x 2 | A1 | Correct second derivative, condone un-simplified.
May use exponential form:
dd 2 y2 = 8 ( e x + e − x ) − ( e x + e − x ) ( e x − e − x ) − ( e x − e − x ) ( e x + e − x )
x
A1
Correct second derivative, condone un-simplified.
May use exponential form:
d2y
=8(ex+e−x)−(ex +e−x)(ex −e−x)−(ex −e−x)(ex+ e−x)
dx2
Differentiating to get  1 6 s i n h x  k c o s h 2 x  k s i n h 2 x or
dd yx =  8 ( e x − e − x )  k ( e x + e − x ) 2  k ( e x − e − x ) 2 for some
k  0 .
(c) | x=ln(2+ 5)
 1  ( )
c o s h ( l n ( 2 + 5 ) ) = 12 2 + 5 + = 5
2 + 5
or cosh(ln(2+ 5))= 1+22 (= 5)
y =12 5 | B1
B1FT
B1
[3] | 2.1
2.4
2.4 | Condone ln 2+ 5 .
Finding the value of cosh x in a form not involving logs or
exponentials following through their x of the form
l n ( a + b ) where a is non-zero and b > 0 and a + b  0 –
allow un-simplified. The correct y-coordinate implies this and
the next B mark. For reference:
 1 
cosh(ln(a+ b))= 1  a+ b+ .
2  a+ b 
Or from using c o s h 2 x − s i n h 2 x  1 with their value of
sinhxfrom part (b).
Need not be stated as a coordinate. Accept any exact one term
equivalent.