Standard +0.3 This is a straightforward differentiation problem requiring knowledge of standard hyperbolic function derivatives (d/dx(cosh x) = sinh x, d/dx(sinh x) = cosh x) and solving a simple equation. While hyperbolic functions are a Further Maths topic, the question itself involves routine application of rules with no conceptual difficulty—just differentiate, set equal to zero, and solve using the definition of hyperbolic functions to find ln form. Slightly above average difficulty only because it's Further Maths content.
The curve \(C\) has equation
$$y = 9 \cosh x + 3 \sinh x + 7x$$
Use differentiation to find the exact \(x\) coordinate of the stationary point of \(C\), giving your answer as a natural logarithm.
[6]
Question 1:
1 | y(cid:32)9coshx(cid:14)3sinhx(cid:14)7x
dy
(cid:32)9sinhx(cid:14)3coshx(cid:14)7
dx | Correct derivative | B1
(cid:11) ex (cid:16) e(cid:16)x(cid:12) (cid:11) ex+e(cid:16)x(cid:12)
9 (cid:14)3 (cid:14)7(cid:32)0
2 2 | Replaces sinhx and coshx by
the correct exponential
forms | M1
Note that the first 2 marks can score the other way round:
(cid:11) ex+e(cid:16)x(cid:12) (cid:11) ex (cid:16)e(cid:16)x(cid:12)
M1: y (cid:32)9 (cid:14)3 (cid:14)7x
2 2
(cid:11) ex (cid:16)e(cid:16)x(cid:12) (cid:11) ex+e(cid:16)x(cid:12)
dy
B1: (cid:32)9 (cid:14)3 (cid:14)7
dx 2 2
12e2x (cid:14)14ex (cid:16)6(cid:32)0 oe | M1: Obtains a quadratic in
ex | M1 A1
A1: Correct quadratic
(cid:11) 3ex (cid:16)1 (cid:12)(cid:11) 2ex (cid:14)3 (cid:12) (cid:32)0(cid:159)ex (cid:32)... | Solves their quadratic as far
as ex = ... | M1
(cid:167)1(cid:183)
x(cid:32)ln
(cid:168) (cid:184)
(cid:169)3(cid:185) | 3
cso (Allow –ln3) ex (cid:32)(cid:16)
2
need not be seen. Extra
answers, award A0 | A1
Alternative
dy
(cid:32)9sinhx(cid:14)3coshx(cid:14)7
dx | Correct derivative | B1
9sinhx(cid:32)(cid:16)3coshx(cid:16)7(cid:159)81sinh2 x(cid:32)9cosh2 x(cid:14)42coshx(cid:14)49
72cosh2 x(cid:16)42coshx(cid:16)130(cid:32)0 | Squares and attempts
quadratic in coshx | M1
5
(cid:11)3coshx(cid:16)5(cid:12)(cid:11)12coshx(cid:14)13(cid:12)(cid:32)0(cid:159)coshx(cid:32)
3 | M1: Solves quadratic | M1 A1
A1: Correct value
(cid:167) 2 (cid:183)
5 (cid:167)5(cid:183)
x(cid:32)ln(cid:168) (cid:114) (cid:16)1(cid:184)
(cid:168) (cid:184)
(cid:168)3 (cid:169)3(cid:185) (cid:184)
(cid:169) (cid:185) | Use of ln form of arcosh | M1
(cid:167)1(cid:183)
x(cid:32)ln
(cid:168) (cid:184)
(cid:169)3(cid:185) | cso (Allow – ln3) | A1
NB: Ignore any attempts to find the y coordinate
(6 marks)
Question | Scheme | Marks
The curve $C$ has equation
$$y = 9 \cosh x + 3 \sinh x + 7x$$
Use differentiation to find the exact $x$ coordinate of the stationary point of $C$, giving your answer as a natural logarithm.
[6]
\hfill \mbox{\textit{Edexcel F3 2018 Q1 [6]}}