Standard +0.8 This question requires applying the quotient rule to a composite function involving e^(sin x) and cos²x, then solving the resulting transcendental equation (cos x = 2sin x tan x) for stationary points. The differentiation is moderately complex, and finding all solutions in [0, 2π] requires careful trigonometric manipulation beyond routine exercises.
5 The equation of a curve is \(y = \frac { e ^ { \sin x } } { \cos ^ { 2 } x }\) for \(0 \leqslant x \leqslant 2 \pi\).
Find \(\frac { \mathrm { dy } } { \mathrm { dx } }\) and hence find the \(x\)-coordinates of the stationary points of the curve.
Obtain correct derivative \(\dfrac{e^{\sin x}\cos^3 x - (-2e^{\sin x}\sin x\cos x)}{(\cos^2 x)^2}\) or equivalent
A1
Equate numerator to zero
DM1
Obtain equation in one unknown
DM1
E.g. \(\sin^2 x - 2\sin x - 1 = 0\)
Solve a 3 term quadratic in \(\sin x\) to find a value for \(x\)
M1
Obtain a correct solution, e.g. \(3.57°\)
A1
At least 3sf
Obtain a further correct solution, e.g. \(x = 5.86°\) and no others in the interval
A1FT
At least 3sf. FT \(3\pi - their\ 3.57\)
Alternative Method — first 3 marks:
Take logarithms of both sides and simplify
(\*M1)
\(\ln y = \sin x - 2\ln\cos x\) or equivalent
Obtain \(\dfrac{1}{y}\dfrac{dy}{dx} = \cos x + 2\dfrac{\sin x}{\cos x}\)
(A1)
Or equivalent
Equate \(\dfrac{dy}{dx}\) to zero
(DM1)
## Question 5:
| Answer | Mark | Guidance |
|--------|------|----------|
| Use correct quotient (or product) rule | \*M1 | |
| Obtain correct derivative $\dfrac{e^{\sin x}\cos^3 x - (-2e^{\sin x}\sin x\cos x)}{(\cos^2 x)^2}$ or equivalent | A1 | |
| Equate numerator to zero | DM1 | |
| Obtain equation in one unknown | DM1 | E.g. $\sin^2 x - 2\sin x - 1 = 0$ |
| Solve a 3 term quadratic in $\sin x$ to find a value for $x$ | M1 | |
| Obtain a correct solution, e.g. $3.57°$ | A1 | At least 3sf |
| Obtain a further correct solution, e.g. $x = 5.86°$ and no others in the interval | A1FT | At least 3sf. FT $3\pi - their\ 3.57$ |
| **Alternative Method — first 3 marks:** | | |
| Take logarithms of both sides and simplify | (\*M1) | $\ln y = \sin x - 2\ln\cos x$ or equivalent |
| Obtain $\dfrac{1}{y}\dfrac{dy}{dx} = \cos x + 2\dfrac{\sin x}{\cos x}$ | (A1) | Or equivalent |
| Equate $\dfrac{dy}{dx}$ to zero | (DM1) | |
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5 The equation of a curve is $y = \frac { e ^ { \sin x } } { \cos ^ { 2 } x }$ for $0 \leqslant x \leqslant 2 \pi$.\\
Find $\frac { \mathrm { dy } } { \mathrm { dx } }$ and hence find the $x$-coordinates of the stationary points of the curve.\\
\hfill \mbox{\textit{CAIE P3 2024 Q5 [7]}}