Paper 1 – NST First Year Maths

2020 Paper 1 Question 6

25th Jan 2022nstmathsupervisorLeave a comment

$I = \int x \sqrt{3-2\,x}\, dx$

You can do this by substitution.

Let $u = 3 -2x$

$\Rightarrow dx = -\frac{1}{2} du$ and $x = \frac{1}{2} \left( 3-u \right )$

$\begin{aligned} \therefore I &= \int \frac{(3-u)}{2} \, u^{1/2} \, \left(-\frac{1}{2} \right) \,du \\ &= -\frac{1}{4} \int \left( 3u^{1/2} - u^{3/2} \right) \, du \\ &= -\frac{1}{4} \left( 2u^{3/2}-\frac{2}{5} u^{5/2} \right) + C \\ &= -\frac{1}{2} u^{3/2} \left( \frac{5-u}{5} \right) + C \\ &= -\frac{1}{2} (3-2x)^{3/2} \left( \frac{2+2x}{5} \right) + C \\ &= -\frac{1}{5} (3-2x)^{3/2} \left( 1+x \right) + C \end{aligned}$

This is probably the most compact form. Other forms are possible, for example by writing $(3-2x)^{3/2} = (3-2x)^{1/2}(3-2x)$ .

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

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

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.

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.

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

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