NST First Year Maths

2019 Paper 2 Question 15

8th Oct 20201st Nov 2020nstmathsupervisorLeave a comment

(b)

Using dot notation for the time derivatives

$\begin{aligned} \dot{u} &= -3u + v \quad(1) \\ \dot{v} &= -5u + v \quad(2) \end{aligned}$

(i)

We need to eliminate $v$ and it’s derivatives. Differentiate (1)

$\ddot{u} = -3 \dot{u} + \dot{v}$

Use (2) to eliminate $\dot{v}$

$\ddot{u} = -3 \dot{u} + -5u + v$

From (1)

$v = \dot{u} + 3u$

and using this to eliminate $v$

$\ddot{u} = -3 \dot{u} + -5u + \dot{u} + 3u$

Hence

$\ddot{u} + 2\dot{u} + 2u = 0$

2016 Paper 1 Question 11

8th Oct 20201st Nov 2020nstmathsupervisorLeave a comment

(a)

When $\lambda = 1$

$|z-i| = |z^*-i|$

Geometrically this means $z$ is as far from $i$ as $z^*$ .

Let $z = x +i\,y$

$\begin{aligned} |x+iy-i| &= |x-iy-i| \\ \Rightarrow x^2 + (y-1)^2 &= x^2 + (y+1)^2 \\ \Rightarrow y^2-2y+1 &= y^2 +2y +1 \\ \Rightarrow y &= 0 \end{aligned}$

$y=0$ is the equation of the straight line in the complex plane. This is consistent with the geometric interpretation.

(b)

When $\lambda = 0$ it must be that $z=i$ . This is another straight, horizontal, line.

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

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

2014 Paper 1 Question 12

6th Oct 20201st Nov 2020nstmathsupervisorLeave a comment

(b)

$\tanh z = -i$

$\begin{aligned} \Rightarrow \frac{\sinh z}{\cosh z} &= -i \\ \Rightarrow \frac{e^z-e^{-z}}{e^z+e^{-z}} &= -i \\ \Rightarrow e^{2z}-1 &= -i \, ( e^{2z}+1 ) \\ \Rightarrow (1+i)e^{2z} &= 1 - i \\ \Rightarrow e^{2z} &= \frac{1-i}{1+i} \\ \Rightarrow e^{2z} &= \frac{1}{2}(1-2i-1) \\ \Rightarrow e^{2z} &= -i \\ \Rightarrow e^{2z} &= e^{-i\left( \frac{\pi}{2} +2k\pi \right) }, \quad k \in \mathbb{Z} \\ \Rightarrow 2z &= -i\left( \frac{\pi}{2} +2k\pi \right) \\ \Rightarrow z &= -i\left( \frac{\pi}{4} +k\pi \right), \quad k \in \mathbb{Z} \\ \end{aligned}$

2014 Paper 1 Question 11

5th Oct 20201st Nov 2020nstmathsupervisorLeave a comment

(a)

$|A| = \left| \begin{matrix} 1 & -4 & 7 \\ -4 & 4 & -4 \\ 7 & -4 & 1 \end{matrix} \right|$

Clearly, you can just expand this as is, but it can be worth simplifying with some elementary row/column operations.

$R_2 \to R_2 + R_3$

$|A| = \left| \begin{matrix} 1 & -4 & 7 \\ 3 & 0 & -3 \\ 7 & -4 & 1 \end{matrix} \right|$

$C_3 \to C_3 + C_1$

$|A| = \left| \begin{matrix} 1 & -4 & 8 \\ 3 & 0 & 0 \\ 7 & -4 & 8 \end{matrix} \right|$

$C_3 \to C_3 + 2C_2$

$|A| = \left| \begin{matrix} 1 & -4 & 0 \\ 3 & 0 & 0 \\ 7 & -4 & 0 \end{matrix} \right|$

With a column of zeros in the determinant

$|A| = 0$ .

For the trace

$\text{Tr}(A) = 1 + 4 + 1 = 6$

The determinant is equal to the product of the eigenvalues, so (at least) one of the eigenvalues equals zero.

2013 Paper 2 Question 13

3rd Oct 20201st Nov 2020nstmathsupervisorLeave a comment

(a)

Let $y = x/2$

$dy = dx/2$

Therefore,

$I = 2\,\int_0^{\infty} e^{-y^2}\,dy$

Now, given

$\int_{-\infty}^{\infty}e^{-x^2}dx = \sqrt \pi$

we have, due to the even integrand

$\int_{0}^{\infty}e^{-x^2}dx = \frac{1}{2}\sqrt \pi$

Thus,

$I = 2 \left( \frac{1}{2} \sqrt \pi \right) = \sqrt \pi$

(b)

(i)

The key is to rewrite the integrand and use integration by parts,

$I_n = \int_{-\infty}^{\infty} x^{n-1}\,x\,e^{-x^2}dx$

Now let $u =x^{n-1}$ and $dv/dx=x\,e^{-x^2}$

$du = (n-1)x^{n-2}\,dx$

and

$v = -\frac{1}{2}e^{-x^2}$

$I_n = \left[ -\frac{1}{2}x^{n-1}e^{-x^2} \right]_{-\infty}^{\infty} + \frac{1}{2} \int_{-\infty}^{\infty}(n-1) x^{n-2}e^{-x^2}\,dx$

The exponential dominates the polynomial, so the first term is zero, and the final integral belongs to the family of given integrals, so

$I_n = \frac{n-1}{2} \, I_{n-2}$

2016 Paper 1 Question 12

30th Sep 20201st Nov 2020nstmathsupervisorLeave a comment

(a) With

$u = x^2-y^2, \; v = 2xy$

We can think of $f$ as

$f=f(u,v)$

Then

$\begin{aligned} \frac{\partial f}{\partial x} &= \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x} \\ &= \frac{\partial f}{\partial u}\, 2x + \frac{\partial f}{\partial v}\, 2y \end{aligned}$

and

$\begin{aligned} \frac{\partial f}{\partial y} = \frac{\partial f}{\partial u}(-2y) + \frac{\partial f}{\partial v}(2x) \end{aligned}$

Now substitute in to the given equation

$\begin{aligned} y\frac{\partial f}{\partial x} + x\frac{\partial f}{\partial y} &= y 2x \frac{\partial f}{\partial u} + \frac{\partial f}{\partial v} 2y^2 + (-2xy)\frac{\partial f}{\partial u} + 2x^2\frac{\partial f}{\partial v} \\ &= 2 (x^2+y^2) \frac{\partial f}{\partial v} \\ &= 0 \end{aligned}$

So for general $x,y$

$\frac{\partial f}{\partial v} = 0$

This means $f$ is independent of $v$ and is a function of $u$ alone. Thus

$f = f(u) = f(x^2-y^2)$

QED

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

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

2013 Paper 1 Question 20

7th Sep 20201st Nov 2020nstmathsupervisorLeave a comment

(a)

Centre the cuboid, with volume $V$ , with the lower face on the xy plane. Let the coordinates of the vertex in $x\ge0,y\ge0,z\ge 0$ where the cuboid intersects the hemisphere be $x, y, z$ .

$V=2x\,2y\,z$

The upper vertices are on the surface of the hemishpere so we get the constraint

$x^2+y^2+z^2=a^2 \quad(1)$

Form the objective function

$f=4xyz-\lambda \left(x^2+y^2+z^2-a^2 \right)$

Take partial derivates

$\begin{aligned} f_x &= 4yz - 2 \lambda x &= 0 \\ f_y &= 4xz - 2 \lambda y &= 0 \\ f_z &= 4xy - 2 \lambda z &= 0\\ f_{\lambda} &= -\left(x^2+y^2+z^2-a^2 \right) &= 0 \end{aligned}$

From $f_x$

$\lambda = \frac{2yz}{x}$

Substitute in $f_y$ , recall $x\ge0,y\ge0$

$f_y = 2xz -\frac{2yz}{x} y = 0 \Rightarrow x^2=y^2 \Rightarrow x=y$

Similarly from $f_z$ we get $x=z$ . Thus

$x=y=z$ .

Using the constraint $(1)$ ,

$x = \frac{a}{\sqrt3}$ .

Finally, again using $x=y=z$ , the required volume is

$V=\frac{4}{3\sqrt3}a^3$

2019 Paper 2 Question 13

4th Sep 20201st Nov 2020nstmathsupervisorLeave a comment

(a)

Parameterise the path

$\mathbf{x}(t)=(t,t,t) \quad 0 \le t \le 1$

In terms of the parameter

$\mathbf{F}(\mathbf{x}) = \begin{pmatrix} 2 \alpha t + \alpha^2 t + \alpha t \\ \alpha^2 t + \beta t \\ \alpha t + \beta t^2 \end{pmatrix}$

Also

$\frac{d}{dt}\mathbf{x}(t) = (1,1,1)$

$\begin{aligned} \int_\Gamma \mathbf{F} \cdot d\mathbf{x} &= \int_0^1 \begin{pmatrix} 2 \alpha t + \alpha^2 t + \alpha t \\ \alpha^2 t + \beta t \\ \alpha t + \beta t^2 \end{pmatrix} \cdot (1,1,1) \, dt \\ &= \int_0^1 (4\alpha t + 2 \alpha^2 t + \beta t + \beta t^2) \, dt \\ &= \left[ 2\alpha t^2 + \alpha^2 t^2 + \beta t^2/2 + \beta t^3/3 \right]_0^1 \\ &= 2\alpha + \alpha^2 + \beta /2 + \beta /3 \end{aligned}$

Thus

$\int_\Gamma \mathbf{F} \cdot d\mathbf{x} = 2\alpha + \alpha^2 + \frac{5}{6}\beta$