Question 5:
5 | (a) | DR
e x − e − x e x + e − x
( R H S = 2 s i n h x c o s h x = ) 2  
2 2 | M1 | 2.1 | AG. Use of exponential
definition of sinh or cosh | Must be used on LHS or RHS
= 1( e2x +e0 −e0 −e −2x )
2
e2x −e −2x
= =sinh2x=LHS
2 | A1 | 2.2a | AG. Soan intermediate step
must be shown
LHS must be equated to RHS | e.g.
Accept full reverse argument
[2]
(b) | DR
15sinhx + 16 coshx – 12sinhxcoshx = 20 | B1 | 3.1a | Use of identity in (a).
12sc – 16c – 15s + 20 = 0
(3s – 4)(4c – 5) = 0 | M1 | 1.1 | Writing as = 0 and factorising | (where𝑠 = sinh𝑥
and 𝑐 = cosh𝑥)
 4   5 
x = s i n h − 1 o r x = c o s h − 1
3 4 | A1 | 1.1 | Complete solution in any form
(assume that cosh–1 is multi-
valued here)
 2 
4 4  4 
s i n h − 1 = ln + + 1 = ln 3
3 3 3 | A1 | 1.1
 2 
5 5 5
cosh −1 =ln  +   −1 =ln2
4 4 4 
  | A1 | 3.2a | Must show  explicitly (or have
both ln2 and ln½)
Alternative method: | B1 | Use of exponential definitions
of sinhx, coshx and sinh2x in
equation | Also award if starts with main
method before using exponentials
ex −e −x ex+e −x e2x −e −2x
15 +16 −6 =20
2 2 2
 1 5 e x − 1 5 − e x + 1 6 e x + − 1 6 e x − 6 e 2 x + 6 e − 2 x = 4 0
 1 5 e 3 x − 1 5 e x + 1 6 e 3 x + 1 6 e x − 6 e 4 x + 6 = 4 0 e 2 x
 6 e 4 x − 3 1 3 e x + 4 0 e 2 x x − e − 6 = 0 |  1 5 e x − 1 5 − e x + 1 6 e x + − 1 6 e x − 6 e 2 x + 6 e − 2 x = 4 0
 1 5 e 3 x − 1 5 e x + 1 6 e 3 x + 1 6 e x − 6 e 4 x + 6 = 4 0 e 2 x
 6 e 4 x − 3 1 3 e x + 4 0 e 2 x x − e − 6 = 0 | *M1 | Multiplying by e2x and
collecting like terms to write as
quartic equation in ex | Could see a substitution eg y = ex
leading to 6 y 4 − 3 1 y 3 + 4 0 y 2 − y − 6 = 0
Could use Pythagoras to derive
quartic in sinh or cosh
y = x  e 4 6 y 3 − + 3 1 y 4 0 y 2 − y − = 6 0
6  1 − 6  3 1 8  + − − 4 0 4 2 6 = 2 5 6 − 2 5 6 = 0
6 3 y − ( y − 2 ) 1 2 − + 9 y ( y 2 ) 2 y ( y − + 2 ) 3 ( y − 2 ) = 0
( − y 2 ) ( 3 − 6 y 2 + + 1 9 y 2 y 3 ) = 0 | *dep*
M1 | Using factor theorem to deduce
that ex = 2 (or 3 or ½) is a
solution and factorising. | or (6𝑦2 −𝑦−1)(𝑦2 −5𝑦+6)
seen | or (6𝑦2 −𝑦−1)(𝑦2 −5𝑦+6)
seen
627−199+23+3=171−171=0
6y3−19y2+2y+3=
6y2(y−3)−y(y−3)−(y−3)
=(y−3)(6y2−y−1)=(y−3)(2y−1)(3y+1) | dep*M
1 | Using factor theorem to find
another factor and fully
factorising
y = ex = 2, ½, 3 or –⅓. But ex> 0 | A1 | Must reject negative root | ScB1 for correct solution after
x = ln2, ln½ (or –ln2) or ln3 only | explicitly for A1 | B1M1M0M0
[5]
B1
Use of exponential definitions
of sinhx, coshx and sinh2x in
equation
Also award if starts with main
method before using exponentials
Multiplying by e2x and
collecting like terms to write as
quartic equation in ex
*dep*
M1
Using factor theorem to deduce
that ex = 2 (or 3 or ½) is a
solution and factorising.
dep*M