These Ten-Year-Series (TYS) worked solutions with video explanations for 2005 A Level H2 Mathematics are suggested by Mr Gan. For any comments or suggestions please contact us at gs.ude.htamnagmitobfsctd-8c02c7@troppus.
The indefinite integral $\int{\frac{\text{P}(x)}{{{x}^{3}}+1}\,\text{d}x}$, where $\text{P}\left( x \right)$ is a polynomial in $x$, is denoted by $\text{I}$.
(i)
Find $\text{I}$ when $\text{P}(x)={{x}^{2}}$.
[2]
(i) Find $\text{I}$ when $\text{P}(x)={{x}^{2}}$.
[2]
(ii)
By writing ${{x}^{3}}+1=\left( x+1 \right)\left( {{x}^{2}}+Ax+B \right)$, where $A$ and $B$ are constants, find $\text{I}$ when
(ii) By writing ${{x}^{3}}+1=\left( x+1 \right)\left( {{x}^{2}}+Ax+B \right)$, where $A$ and $B$ are constants, find $\text{I}$ when
(a) $\text{P}\left( x \right)={{x}^{2}}-x+1$,
[3]
(a) $\text{P}\left( x \right)={{x}^{2}}-x+1$,
[3]
(b) $\text{P}\left( x \right)=x+1$.
[3]
(b) $\text{P}\left( x \right)=x+1$.
[3]
(iii)
Using the results of parts (i) and (ii), or otherwise, find $\text{I}$ when $\text{P}\left( x \right)=1$.
[4]
(iii) Using the results of parts (i) and (ii), or otherwise, find $\text{I}$ when $\text{P}\left( x \right)=1$.
[4]
Share with your friends!
Verify that $z=\text{i}$ is a root of the equation
${{z}^{4}}-2{{z}^{3}}+6{{z}^{2}}-2z+5=0$.
[1]
Hence determine the other roots.
[4]
Share with your friends!
It is given that $a$, $b$, $c$ are the first three terms of a geometric progression. It is also given that $a$, $c$, $b$ are the first three terms of an arithmetic progression.
(i)
Show that ${{b}^{2}}=ac$ and $c=\frac{a+b}{2}$.
[2]
(i) Show that ${{b}^{2}}=ac$ and $c=\frac{a+b}{2}$.
[2]
(ii)
Hence show that $2{{\left( \frac{b}{a} \right)}^{2}}-\left( \frac{b}{a} \right)-1=0$.
[2]
(ii) Hence show that $2{{\left( \frac{b}{a} \right)}^{2}}-\left( \frac{b}{a} \right)-1=0$.
[2]
(iii)
Given that the sum to infinity of the geometric progression is $S$, find $S$ in terms of $a$.
[4]
(iii) Given that the sum to infinity of the geometric progression is $S$, find $S$ in terms of $a$.
[4]
Share with your friends!
How can we help?
