STEP 2 2001 Q4 (spoilers!)

Hi,

I obtained 3 equations :
sin (2nx) = 0
cos(2nx) = 1
cos(2nx) = -1/2
When attempting the question I stated that m=n, and did not get a solution within the range. The integral solution treated these to be different which I donâ€™t understand since they are both positive integers. Why is it also necessary to state each of the solutions have their own integer coefficients of pi (m, p and q). This meant I got no solutions in the range when I attempted the question once I had also used n = 2k and n = 2k +1.

Any help would be much appreciated!

2001 S2 Q4 (A rather full solution!)

Using the formulae for $\sin 2x=2\sin x \cos x$ and $\sin 3x = \sin x \cos 2x+ \sin 2x \cos x$ gives:
$\text{f}(x) = P \sin x + 2Q \sin x \cos x + R( \sin x \cos 2x+ \sin 2x \cos x)\\ =\sin x [P + 2Q \cos x + R(4\cos^2 x - 1)]$
The bit inside the square brackets is a quadratic in $\cos x$, and so you can show that there are no solutions to this part when $4Q^2-4(4R)(P-R) For the next part, use a similar method to get: $\text{g}(x) = \sin 2nx(2 + 2\cos 2nx - 4 \cos^2 2nx)$ Then the roots of$\text{g}(x)=0$occur when$\sin 2nx = 0$or when$2 \cos^2 2nx - \cos 2nx - 1=0$, i.e. when$\cos 2nx = 1$or$-\frac 1 2$. Note that any solution of$\cos 2nx=1$is also a solution of$\sin 2nx = 0$. This gives us the general solutions: $2nx=m\pi \quad \text{or}\\ 2nx = 2m\pi \pm \frac{2\pi} 3$ Note that here$n$is really a fixed, unknown integer where as$m$is a varying integer (there is a solutions when$m=1$,$m=2$etc.) Solving for$x$gives: $x=\frac{m\pi}{2n} \quad \text{or}\\ x = \frac {3m\pi \pm \pi} {3n}$ Considering the first set of solutions, if$x\lt \frac {\pi} 2$then we need$\frac{m\pi}{2n} \lt \frac{\pi } 2$, which means that$m \lt n$and so the largest possibly value of$m$is$n-1$. The largest solution here is therefore$x=\frac {(n-1) \pi} {2n}=\frac {\pi} 2 - \frac {\pi}{2n}$. For the other set of solutions, we have: $\frac {3m\pi \pm \pi} {3n} \lt \frac{\pi} 2\\ \implies 6m \pm 2 \lt 3 n\\ \implies 6m \lt 3 n \mp 2$ Considering the solution$x = \frac {3m\pi - \pi} {3n}$we have$m \lt \frac 1 2 n + \frac 1 3$. If$n$is even (so$n=2k$say) then we have$m \lt k + \frac 1 3$and so$m=k=\frac 1 2 n$. This gives the solution$x=\frac{3\times \frac 1 2 n \pi - \pi}{3n}=\frac {\pi} 2 - \frac {\pi} {3n}$. This is bigger than the solution$x=\frac {\pi} 2 - \frac {\pi}{2n}$from before. If$n=2k+1$, then we have$m \lt k+ \frac 1 2 +\frac 1 3$and so$m=\frac 1 2 (n-1)$. This gives the solution$x=\frac {\pi} 2-\frac {5\pi}{6n}$which is smaller than the solution$x=\frac {\pi} 2 - \frac {\pi}{2n}$. We now need to consider$x = \frac {3m\pi + \pi} {3n}$which means we need$m

We therefore have a maximal value of:
$x=\frac {\pi} 2 - \frac {\pi} {3n} \quad \text{when }n \text{ is even}\\ x=\frac {\pi} 2 - \frac {\pi} {6n} \quad \text{when }n \text{ is odd}$

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