Question 5:
5 | (a) | RHS=cosh2 x+sinh2 x
ex+e−x  2 ex−e−x  2
=  + 
 2   2 
= 1 ( e 2 x + e − 2 x + 2 + e 2 x + e − 2 x − 2 )
4
= 1 ( 2 e 2 x + 2 e − 2 x ) = 1 ( e 2 x + e − 2 x )
4 2
= cosh 2x = LHS | M1
A1
[2] | 2.1
1.1 | Using definitions of cosh x and
sinh x.
AG. Intermediate working must
be seen.
(b) | cosh2x + sinh2x = cosh2x & cosh2x – sinh2x = 1
=> 2cosh2 x = cosh 2x + 1
=> cosh 2x = 2cosh2 x – 1 | B1
[1] | 2.2a
(c) | 10cosh2 x – 5 = 16cosh x + 21
=> 10c2 – 16c – 26 = 0 => 5c2 – 8c – 13 = 0
c = –1 rejected since cosh x ≥ 1 (or ≥ 0 oe)
or c = 13/5
 2 
1 3 1 3  1 3 
c o s h − 1 = ln + − 1 = ln 5
5 5 5
x=ln5 | M1
A1
M1
A1 | 1.1
2.3
1.1
2.2a | Using the identity from (b) to
reduce equation to 3 term
quadratic
Both solutions found…
… and –1 rejected explicitly with
valid reason. E.g “–1 is outside
the range of cosh x”
Use of formula for cosh–1 (or by
solving quadratic in ex).
x = ln5 or ln(1/5). Must have
both solutions. | Could be BC
Alternative method:
5 ( e 2 x + e − 2 x ) = 8 ( e x + e − x ) + 2 1
2
5 e 4 x − 1 6 e 3 x − 4 2 e 2 x − 1 6 e x + 5 = 0 | Alternative method: | M1
5 ( e 2 x + e − 2 x ) = 8 ( e x + e − x ) + 2 1
2
5 e 4 x − 1 6 e 3 x − 4 2 e 2 x − 1 6 e x + 5 = 0 | Using the exponential definition
of cosh to reduce the given
M1 | equation to a quartic equation in
ex.
ex =e
5e3(e−5)+9e2(e−5)+3e(e−5)−(e−5)=0
(e−5)(5e3+9e2 +3e−1)=0
(e−5)(e2(5e−1)+2e(5e−1)+(5e−1))=0
(e−5)(5e−1)(e2 +2e+1)=0
(e−5)(5e−1)(e+1)2 =0
1
e=−1, or 5
5 | M1 | Factorising or using quartic
solver.
Factorising or using quartic
solver.
e = –1 => ex = –1 which is not possible since | A1 | Negative solution must be
ex > 0 for all (real) x. | rejected and a valid reason given.
So ex = 5 or 1/5 => x = ln5 or ln(1/5) | So ex = 5 or 1/5 => x = ln5 or ln(1/5) | A1 | x = ln5. Must have both
solutions.
[4]
M1