2011 – NST First Year Maths

2011 Paper 1 Question 14

8th Oct 20201st Nov 2020nstmathsupervisorLeave a comment

$\sqrt3 \frac{dy}{dx} + y = 4 \sin x \qquad(*)$

(a)

$f(x)=\frac{1}{2}a_0 + \sum\limits_{r=1}^\infty a_r \cos \left(\frac{2\pi rx}{L}\right) + \sum\limits_{r=1}^\infty b_r \sin \left(\frac{2\pi rx}{L}\right)$

The period $L=2\pi$ so

$f(x)=\frac{1}{2}a_0 + \sum\limits_{r=1}^\infty a_r \cos \left( r x\right) + \sum\limits_{r=1}^\infty b_r \sin \left(r x \right)$

(b)

Assuming such an expansion exists for $y$

$\frac{dy}{dx} = \sum\limits_{r=1}^\infty a_r (-)\sin \left( r x\right)r + b_r \cos \left(r x \right) r$

Substitute in (*)

$\frac{1}{2}a_0 + \sum\limits_{r=1}^\infty \left( \sqrt3 r b_r + a_r \right) \cos{(rx)} + \left( b_r - \sqrt3 r a_r \right) \sin{(rx)} = 4 \sin{x}$

This yields the constraints on the coefficients, by comparing constants, coefficients of $\cos$ and $\sin$

$\begin{aligned}a_0 &= 0 \\ \sqrt3 r b_r + a_r &= 0 \\ b_r - \sqrt3 r a_r &= 4\delta_{1,r} \end{aligned}$

Here, $\delta_{1,r}$ is the Kronecker delta.

2011 Paper 1 Question 11

7th Oct 20201st Nov 2020nstmathsupervisorLeave a comment

(a)

Given it is for 4 marks, I suspect they desire the remainder term including.

$f(x) = f(a) + f^{(1)}(a)(x-a) + f^{(2)}(a)\frac{(x-a)^2}{2!}+f^{(3)}(\xi)\frac{(x-a)^3}{3!}$

where $a < \xi < x$ .

(b)

For $x=0$ , the Taylor series becomes

$f(x) = f(0) + f^{(1)}(0)x + f^{(2)}(0)\frac{x^2}{2!}+f^{(3)}(0)\frac{x^3}{3!}+\cdots$

(i)

$f(x) = \frac{1}{(x^2+9)^{1/2}}$

$f(0) = 1/3$

Note $f(x)$ is an even function so we expect $f^{(1)}(0) =f^{(3)}(0) =0$ but we still need to differentiate to work out $f^{(2)}(0)$ .

$\begin{aligned}f^{(1)}(x) &= (-1/2)(x^2+9)^{-3/2}\,2x \\ &= -\frac{x}{(x^2+9)^{3/2}} \\ &= -x\,f^3 \end{aligned}$

As expected $f^{(1)}(0)=0$

Notice also the very useful device here of expressing the derivative in terms of the original function. It can save a lot of effort sometimes.

$\begin{aligned} -f^{(2)}(x) &= x (3)f^2 \, f^{(1)}(x) + f^3 \\ \Rightarrow -f^{(2)}(0) &=\left( f(0) \right)^3 \\ &= 1/27 \end{aligned}$

Putting it all together we obtain

${f(x) = 1/3 - 1/54 \, x^2 }$

2011 Paper 1 Question 17

15th Sep 20201st Nov 2020nstmathsupervisorLeave a comment

(b)

$z^6+z^3+1=0 \quad (1)$

Let $z^3=w$

$\therefore \; w^2+w+1=0$

This is a quadratic equation

$\Rightarrow w = \frac{-1 \pm \sqrt{-3}}{2} = -\frac{1}{2} \pm i \frac{\sqrt{3}}{2}$

Recognise $1/2$ and $\sqrt3/2$ as standard trigonometric ratios. Consider the $+$ root

$w_+ = e^{i 2\pi/3} = z^3$

Compute the cube roots

$z=\exp{\left[ i \left( \frac{2}{9}\pi+\frac{2k}{3}\pi \right) \right] }, \quad k = 0,1,2$

Or in full, placing angles in the range $(-\pi,\pi]$

$z=\exp{ (i \theta)}, \quad \theta \in {\frac{2}{9}\pi, \frac{8}{9}\pi, -\frac{4}{9}\pi}$

As equation $(1)$ has real coefficients, the roots occur in complex-conjugate pairs. The full solution set is therefore

$z=\exp{ (i \theta)}, \quad \theta \in {\pm \frac{2}{9}\pi, \pm \frac{8}{9}\pi, \pm \frac{4}{9}\pi}$