Some of these can't be done by normal A level thought processes:
1)x/(x-3)\<2
x/(x-3) - 2 \<0
x/(x-3) - 2(x-3)/(x-3) \<0
(6-x)/(x-3) \<0
Now we plot a number line, substituting some of the numbers they've given us:
------(0)----(1)----(2)----(3)----(4)----(5)----(6)--(7)--(
---
-ve -ve -ve undef +ve +ve 0 -ve -ve
From this, we can see that all values below x=3 and not including it (an undefined number cannot be part of the inequality) and everything above x=6 and including it (0 is included in the inequality) is negative or \<0
Hence, the answer is A
Note, the round brackets mean you do not include the value, and the square brackets mean you do include the value. The infinite values are in round brackets, because you cannot include "infinity", which is not an actual number.
2)-2x^2+5x<-12
-2x^2+5x+12<0
2x^2 - 5x - 12<0 ....... to make it easier for ourselves
using the quadratic formula, we get to a solution of:
x=(5+11)/4 and x=(5-11)/4
x=4 x=-3/2
HOWEVER, i have only equated x with 4 and -3/2 to show you how to get the CRITICAL VALUES not the solution: we are looking at inequalities, not definite roots to quadratic equations.
Using AS level maths, we can find that
x<-3/2 and x>4
The two values not being included as shown.
Therefore, the answer is C
3)We just have to arrange and cancel out the equation to make it easier for us to comprehend:
(x-4)/((x^2)-x-12)
(x-4)/(x-4)(x+3) ........... factorising
1/(x+3) ......... cancelling out x-4 terms
Thus we only have to consider the above form of the equation. Much easier now huh
No matter what we put into 1/(x+3) we will get a real number. Except x=-3. If we substitute x=-3, we get f(x) as being undefined (which is obviously not possible) and so we must exclude -3 from the domain.
Hence, the answer is A.
If you have anymore trouble understanding my explanations still, check out
www.wolframalpha.com . Type in an equation, and the site will immediately list everything you can possibly know or calculate from the equation given (graphs, alternate forms etc.)
Hope this helped, and hope I am not too late =P
