2014 – NST First Year Maths

2014 Paper 1 Question 12

6th Oct 20201st Nov 2020nstmathsupervisorLeave a comment

(b)

$\tanh z = -i$

So

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

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.

2014 Paper 1 Question 1

5th Sep 20191st Nov 2020nstmathsupervisorLeave a comment

(a)

Apply the chain rule

$\frac{d}{dx}(x^2+4)^{-1} = -(x^2+4)^{-2}(2x)=-\frac{2x}{(x^2+4)^{2}}$

(b)

Apply the chain rule

$\frac{d}{dx}e^{\sin(x)} = e^{\sin(x)}\frac{d}{dx}\sin(x) = \cos(x)e^{\sin(x)}\$