Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2010  OctNov  (P29709/21)  Q#1
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Question
Solve the inequality .
Solution
SOLVING INEQUALITY: PIECEWISE
Let, . We can write it as;
We have to consider both moduli separately and it leads to following cases;



When 







If then above four intervals translate to following with their corresponding inequality;
When 
When 
When 



If then above four intervals translate to following with their corresponding inequality;
When 
When 
When 



We have the inequality;
In standard form it can be written as;
We have to consider both moduli separately and it leads to following cases;







Since then above four intervals translate to;



We can see that given inequality takes following forms for these intervals.


For 
For 
For 






No 

No 

Hence, the only solution for the given inequality is;
SOLVING INEQUALITY: ALGEBRAICALLY
Let, .
Since given equation/inequality is of the form or or , we can solve this inequality by taking square of both sides;
We are given inequality;
Therefore, we can solve it algebraically;
SOLVING INEQUALITY: GRAPHICALLY
We are given inequality;
To solve the inequality graphically, we need to sketch both sides of inequality;
Let’s sketch both equations onebyone.
First we have to sketch;
Let, .
It can be written as;
We have to draw two separate graphs;
When ; 
When ; 


Therefore;

Therefore;

It is evident that and are reflection of each other in xaxis. So we can draw line of by first drawing and then reflecting in xaxis that part of the line which is below xaxis.
It can be written as;
We have to draw two separate graphs;
When ; 
When ; 


Therefore;

Therefore;

It is evident that and are reflection of each other in xaxis. So we can sketch by first drawing and then reflecting in xaxis that part of the line which is below xaxis.
Therefore, first we sketch the line .
To sketch a line we only need x and y intercepts of the line.
The point at which curve (or line) intercepts xaxis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
Therefore, we substitute in given equation of the line.
Hence, coordinates of xintercept of the line with are .
The point at which curve (or line) intercepts yaxis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
Therefore, we substitute in given equation of the line.
Hence, coordinates of yintercept of the line with are .
Hence, we get following graph for ;
We can reflect in xaxis that part of the line which is below xaxis to make it graph of , as shown below.
Next we have to solve;
Let, .
It can be written as;
We have to draw two separate graphs;
When ; 
When ; 


Therefore;

Therefore;

It is evident that and are reflection of each other in xaxis. So we can draw line of by first drawing and then reflecting in xaxis that part of the line which is below xaxis.
We have to draw two separate graphs;
When ; 
When ; 


Therefore;

Therefore;

It is evident that and are reflection of each other in xaxis. So we can sketch by first drawing and then reflecting in xaxis that part of the line which is below xaxis.
Therefore, next we sketch the line .
To sketch a line we only need x and y intercepts of the line.
The point at which curve (or line) intercepts xaxis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
Therefore, we substitute in given equation of the line.
Hence, coordinates of xintercept of the line with are .
The point at which curve (or line) intercepts yaxis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
Therefore, we substitute in given equation of the line.
Hence, coordinates of yintercept of the line with are .
Hence, we get following graph for ;
We can reflect in xaxis that part of the line which is below xaxis to make it graph of , as shown below.
When we sketch the two graphs on the same axes and we get following.
We are looking for the solution of .
It is evident from the graphs that is above (greater) than for all values of;
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