Further Mathematics Questions and Answers

$^{6}C_{4}(1)^{6-4}(-2x)^{4}$ = $15\times1\times16x^{4} = 240x^{4}$

The coefficient of $x^{4}$= 240

Taking the LCM of the right hand side of the equation, we have

$\frac{4(2x-3) - 2(x+5)}{(x+5)(2x-3)} = \frac{6x+m}{2x^{2}+7x-15}$

Comparing the numerators, we have

$4(2x-3) - 2(x+5) = 6x+m$

$8x-12-2x-10 = 6x -22 = 6x + m$

$\implies m = -22$

Let $f(x) = y$, then we have

$y = \frac{x+1}{2} \implies 2y = x+1; x = 2y-1$

Let $f^{1}(x) = x; x = 2y-1$, replacing y with x,

$f^{1}(x) = 2x - 1 \implies f^{1}(-2) = 2(-2) -1= -5$

$f : x \to \sqrt{4 -2x}$ defined on the set of real numbers, R, which has range from $(-\infty, \infty)$ but because of the root sign, it is defined from $[0, \infty)$. 

This is because the root of numbers only has real number values from 0 and upwards.

$\sqrt{4-2x} \geq 0 \implies 4-2x \geq 0$

$-2x \geq -4; x \leq 2$

The equation for a circle with centre coordinates (a, b) and radius r is

$(x-a)^{2} + (y-b)^{2} = r^{2}$

Expanding the above equation, we have

$x^{2} - 2ax +a^{2} + y^{2} - 2by + b^{2} - r^{2} = 0$ so that

$x^{2} - 2ax + y^{2} - 2by = r^{2} - a^{2} - b^{2}$

Taking the original equation given, $3x^{2} + 3y^{2} - 4x + 8y = 2$ and making the coefficients of $x^{2}$ and $y^{2}$ = 1,

$x^{2} + y^{2} - \frac{4x}{3} + \frac{8y}{3} = \frac{2}{3}$, comparing, we have

$2a = \frac{4}{3}; 2b = \frac{-8}{3}$

$\implies a = \frac{2}{3}; b = \frac{-4}{3}$