Challenging +1.2 This is a Further Maths question requiring differentiation of hyperbolic functions, solving a quadratic equation, and verifying uniqueness of a stationary point. While it involves multiple steps (differentiate, set to zero, solve, verify), the techniques are standard for FP2: using standard derivatives of cosh and sinh, applying the identity cosh²x - sinh²x = 1, and solving a resulting quadratic. The algebraic manipulation is moderately involved but follows a clear path with no novel insight required. Slightly above average difficulty due to the multi-step nature and Further Maths content, but remains a routine application of learned techniques.
The equation of a curve is \(y = \cosh^2 x - 3\sinh x\). Show that \(\left(\ln\left(\frac{3+\sqrt{13}}{2}\right), -\frac{5}{4}\right)\) is the only stationary point on the curve. [8]
y(cid:32)cosh2x(cid:16)3sinhx (cid:32)(cid:168) S (cid:184) (cid:16)3(cid:117)
(cid:168) (cid:184)
4 2
(cid:169) (cid:185)
completion to y(cid:32)(cid:16)5oe
Answer
Marks
4
*M1
dep*M1
E1
E1
E1
c
eM1
A1
E1
Answer
Marks
[8]
1.1
2.1
2.4
2.4
m
i2.2a
2.1
1.1
Answer
Marks
1.1
Allow sign errors only
Their f(cid:99)(x)(cid:32)0 and factorising
their expression
n
e
AG
Attempt to evaluate coshxand
substitute to find y
Correct substitution seen
Answer
Marks
AG
e.g. (cid:114)2coshxsinhx(cid:114)3coshx
Or other complete method to
find y
Question 8:
8 | 2coshxsinhx(cid:16)3coshx
2coshxsinhx(cid:16)3coshx(cid:32)0
(cid:159)coshx(cid:11)2sinhx(cid:16)3(cid:12)(cid:32)0
coshx(cid:32)0 has no roots, because coshx(cid:116)1
sinh is a 1-1 function so therefore just one
stationary point when (cid:11)2sinhx(cid:16)3(cid:12)(cid:32)0
x(cid:32)sinh(cid:16)13 = ln (cid:167) 3(cid:14) 1(cid:14) (cid:11)3(cid:12)2 (cid:183)
(cid:168) (cid:184)
2 2 2
(cid:169) (cid:185)
(cid:167)3(cid:14) 13(cid:183)
and completion to x(cid:32)ln(cid:168) (cid:184)
(cid:168) (cid:184)
2
(cid:169) (cid:185)
13
coshx(cid:32) 1(cid:14)sinh2x (cid:32)
4
p
2
(cid:167) 13(cid:183) 3
y(cid:32)cosh2x(cid:16)3sinhx (cid:32)(cid:168) S (cid:184) (cid:16)3(cid:117)
(cid:168) (cid:184)
4 2
(cid:169) (cid:185)
completion to y(cid:32)(cid:16)5oe
4 | *M1
dep*M1
E1
E1
E1
c
eM1
A1
E1
[8] | 1.1
2.1
2.4
2.4
m
i2.2a
2.1
1.1
1.1 | Allow sign errors only
Their f(cid:99)(x)(cid:32)0 and factorising
their expression
n
e
AG
Attempt to evaluate coshxand
substitute to find y
Correct substitution seen
AG | e.g. (cid:114)2coshxsinhx(cid:114)3coshx
Or other complete method to
find y
The equation of a curve is $y = \cosh^2 x - 3\sinh x$. Show that $\left(\ln\left(\frac{3+\sqrt{13}}{2}\right), -\frac{5}{4}\right)$ is the only stationary point on the curve. [8]
\hfill \mbox{\textit{OCR Further Pure Core 2 Q8 [8]}}