2013 – NST First Year Maths

2013 Paper 1 Question 11

12th Oct 20201st Nov 2020nstmathsupervisorLeave a comment

(a)

$\left( \begin{matrix} 1 & 2 & 3 \\ 1 & -2 & 1 \\ 2 & 1 & -1 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) = \left( \begin{matrix} 14 \\ 0 \\ 1 \end{matrix} \right)$

(i)

In Gaussian elimination, reduce the matrix to upper triangular form, then back-substitute.

Eliminate:

$\begin{aligned} R_2 &\to R_2 - R_1 \\ R_3 &\to R_3 - 2R_1\end{aligned}$

$\left( \begin{matrix} 1 & 2 & 3 \\ 0& -4 & -2 \\ 0 & -3 & -7 \end{matrix} \right| \left. \begin{matrix} 14 \\ -14 \\ -27 \end{matrix} \right)$

$R_3 \to R_3 - (-3)/(-4)R_2$

$\left( \begin{matrix} 1 & 2 & 3 \\ 0& -4 & -2 \\ 0 & 0 & -11/2 \end{matrix} \right| \left. \begin{matrix} 14 \\ -14 \\ -33/2 \end{matrix} \right)$

Back-substitute:

$R_3 \Rightarrow z = 3$

$R_2 \Rightarrow \begin{aligned} -4y -2(3) &= -14 \\ y &= 2 \end{aligned}$

$R_1 \Rightarrow \begin{aligned} x +2(2) +3(3) &= 14 \\ x &= 1 \end{aligned}$

The solution is

$\left( \begin{matrix} x \\ y \\ z \end{matrix} \right) = \left( \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right)$

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

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

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

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