Further Mathematics Questions and Answers

Given the equation of the circle $x^{2} + y^{2} - 8x - 2y + 1 = 0$.

The equation of a circle is given as $(x - a)^{2} + (y - b)^{2} = r^{2}$

Expanding, we have $x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2} \equiv x^{2} - 2ax + y^{2} - 2by = r^{2} - a^{2} - b^{2}$

Comparing the RHS of the equation above with the equation rewritten as $x^{2} + y^{2} - 8x - 2y = -1$, we have

$-2a = -8; -2b = -2 \implies a = 4, b = 1$

$\therefore r^{2} - 4^{2} - 1^{2} = -1 \implies r^{2} = -1 + 16 + 1 = 16$

$r = \sqrt{16} = 4$

Given $\sin \theta = \frac{3}{5}  \implies opp = 3, hyp = 5$

Using Pythagoras' Theorem, we have $adj = \sqrt{5^{2} - 3^{2}} = \sqrt{16}  = 4$

$\therefore \cos \theta = \frac{4}{5}, 0° < \theta < 90°$

In the quadrant where $180° - \theta$ lies is the 2nd quadrant and here, only $\sin \theta = +ve$.

$\therefore \cos (180 - \theta) = -ve = \frac{-4}{5}$

Recall, the terms of a GP is given by $T_{n} = ar^{n - 1}$

Given, $T_{1} = a = \frac{3}{4}$ and $T_{2} \times T_{3} = ar \times ar^{2} = 972$

$\implies \frac{3}{4}r \times \frac{3}{4}r^{2} = \frac{9}{16}r^{3} = 972$

$r^{3} = \frac{972 \times 16}{9} = 1728  \implies r = \sqrt[3]{1728} = 12$

Note that every 4- digit and 5- digit numbers formed is greater than 150. Therefore, $^{5}P_{5} + ^{5}P_{4} = \frac{5!}{(5 - 5)!} + \frac{5!}{(5 - 4)!} = 120 + 120 = 240$ numbers are greater than 150 already.

Among the 3- digit numbers, the numbers 123, 124, 125, 132, 134, 135, 142, 143 and 145 are removed from the ones that meet the criterion so we have:

$^{5}P_{3} - 9 = \frac{5!}{(5 - 3)!} = 60 - 9 = 51 \implies 240 + 51 = 291$ numbers are greater than 150. 

$\begin{vmatrix} k & k \\ 4 & k \end{vmatrix} + \begin{vmatrix}  2 & 3 \\ -1 & k \end{vmatrix} = 6$

$\begin{vmatrix} k & k \\ 4 & k \end{vmatrix} = (k^{2} - 4k)$

$\begin{vmatrix}  2 & 3 \\ -1 & k \end{vmatrix} = (2k + 3)$

$\therefore (k^{2} - 4k) + (2k + 3) = k^{2} - 2k + 3 = 6$

$k^{2} - 2k - 3 = 0$, factorising, we have $k + 1 = 0$ or $k - 3 = 0$

Since k > 0, k = 3.