This question tests a student's ability to raise a positive base to an exponent of -1.

Find the gcd $d$ of two positive integers $a$ and $b$ also find integers $x,y$ such that $ax+by=d$, using the extended Euclidean algorithm.

Expand $(ax+b)(cx+d)$.

rebelmaths

Converting to standard form in both positive and negative powers

Given a number evaluate simple power, negative power, to one half.

Given a number evaluate simple power, negative power, to one half.

Find the $x$ and $y$ components of a force which is applied at an angle to a particle. Resolve using $F \cos \theta$. The force is applied in the negative $x$ direction but the positive $y$. 

Find the $x$ and $y$ components of a force which is applied at an angle to a particle. Resolve using $F \cos \theta$. The force acts in the positive $x$ and positive $y$ direction.

Converting to standard form in both positive and negative powers

$I$ compact interval. $\displaystyle g: I\rightarrow I, g(x)=\frac{x^2}{(x-c)^{a/b}}$. Are there stationary points and local maxima, minima? Has $g$ a global max, global min? 

Customised for the Numbas demo exam

Factorise $x^2+cx+d$ into 2 distinct linear factors and then find $\displaystyle \int \frac{ax+b}{x^2+cx+d}\;dx,\;a \neq 0$ using partial fractions or otherwise.

Video in Show steps.

Converting to standard form in both positive and negative powers

$I$ compact interval. $\displaystyle g: I\rightarrow I, g(x)=\frac{x^2}{(x-c)^{a/b}}$. Are there stationary points and local maxima, minima? Has $g$ a global max, global min? 

$I$ compact interval. $\displaystyle g: I\rightarrow I, g(x)=\frac{x^2}{(x-c)^{a/b}}$. Are there stationary points and local maxima, minima? Has $g$ a global max, global min? 

Find $\displaystyle\int \frac{ax+b}{(x+c)(x+d)}\;dx,\;a\neq 0,\;c \neq d $.

Find $\displaystyle\int \frac{a}{(x+b)(x+c)}\;dx,\;b \neq c $.

Factorise $x^2+cx+d$ into 2 distinct linear factors and then find $\displaystyle \int \frac{ax+b}{x^2+cx+d}\;dx,\;a \neq 0$ using partial fractions or otherwise.

Video in Show steps.