Bunuel wrote:
prasannajeet wrote:
Bunuel wrote:
\(x^{11}>x^3\) to hold true either \(-1<x<0\) or \(x>1\). Only -1/2 is in either of the range.
Answer: C.
Hi Bununel
Could you explain how to proceed as i stuck up after few steps???
X^11-X^3>0
=>X^3(X^8-1)>0
=>Now X has 2 solution
such as X^3>0 then X>0...1
Now X^8-1>0
=>X^8>=1
=>X>1.....2
We got 2 solution in this case I think D satisfies
Help...........
Rgds
Prasannajeet
\(x^{11}>x^3\) --> \(x^{3}(x^8-1)>0\). So, far correct.
Next, for \(x^{3}(x^8-1)\) to be positive \(x^3\)and \(x^8-1\) must have the same sign, so both of them must be positive or both of them must be negative.
\(x^3>0\) and \(x^8-1>0\):
\(x^3>0\) gives
\(x>0\);
\(x^8-1>0\) --> \(x^8>1\) -->
\(x<-1\) or \(x>1\).
Both to hold true \(x\) must be greater than 1. So, \(x^3>0\) and \(x^8-1>0\) when \(x>1\).
\(x^3<0\) and \(x^8-1<0\):
\(x^3<0\) gives
\(x<0\);
\(x^8-1<0\) --> \(x^8<1\) --> \(-1<x<1\).
Both to hold true \(x\) must be from -1 to
1. So, \(x^3<0\) and \(x^8-1<0\) when \(-1<x<1\).
Thus we have that \(x^{11}>x^3\) when \(x>1\) or \(-1<x<1\).
/quote]
Hi
Bunuel,
Hope you're well brother.
For the first green part we do not consider the first red part. But, why do we consider second red part (1) after knowing the second green part (x<0)
Thank you...