Standard +0.3 This is a straightforward stationary point problem requiring differentiation using the chain rule, solving a trigonometric equation in a given interval, and substituting back. The chain rule application is standard (differentiating tan(x/2) and sin(x/2)), and the equation sec²(x/2) + (3/2)cos(x/2) = 0 leads to a routine solution. Slightly above average difficulty due to the trigonometric manipulation and interval restriction, but still a standard textbook exercise.
3 The equation of a curve is \(y = \tan \frac { 1 } { 2 } x + 3 \sin \frac { 1 } { 2 } x\). The curve has a stationary point \(M\) in the interval \(\pi < x < 2 \pi\). Find the coordinates of \(M\), giving each coordinate correct to 3 significant figures.
Equate first derivative to zero and produce \(\cos^3\frac{1}{2}x = k_3\)
\*M1
Use correct process to find one value of \(x\)
DM1
Dep on \*M, allow for obtaining \(1.609\ldots\), \(92.2°\) or \(268°\)
Obtain \(x = 4.67\)
A1
Allow \(x = 4.67\) or better for A1
Obtain \(y = 1.12\)
A1
Allow \(y = 1.12\) from \(x = 4.66\) but nothing else
Total
6
## Question 3:
| Answer | Mark | Guidance |
|--------|------|----------|
| Differentiate to obtain form $k_1\sec^2\frac{1}{2}x + k_2\cos\frac{1}{2}x$ | M1 | If a factor of $0.5$ is missed, can still get 5/6, penalise at first A1 |
| Obtain $\frac{1}{2}\sec^2\frac{1}{2}x + \frac{3}{2}\cos\frac{1}{2}x$ | A1 | |
| Equate first derivative to zero and produce $\cos^3\frac{1}{2}x = k_3$ | \*M1 | |
| Use correct process to find one value of $x$ | DM1 | Dep on \*M, allow for obtaining $1.609\ldots$, $92.2°$ or $268°$ |
| Obtain $x = 4.67$ | A1 | Allow $x = 4.67$ or better for A1 |
| Obtain $y = 1.12$ | A1 | Allow $y = 1.12$ from $x = 4.66$ but nothing else |
| **Total** | **6** | |
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3 The equation of a curve is $y = \tan \frac { 1 } { 2 } x + 3 \sin \frac { 1 } { 2 } x$. The curve has a stationary point $M$ in the interval $\pi < x < 2 \pi$. Find the coordinates of $M$, giving each coordinate correct to 3 significant figures.\\
\hfill \mbox{\textit{CAIE P2 2017 Q3 [6]}}