Zarrolou wrote:
Official ExplanationIs \(|p|^2<|p|\) ?
Splitting the equation into two scenarios \(p>0\) and \(p<0\)
I)\(p>0\)
\(p^2-p>0\)
\(0<p<1\)
II)\(p<{}0\)
\((-p)^2+p<0\)
\(-1<p<0\)
We can rewrite the question as: is \(-1<p<1\) and \(p\neq{0}\)?
1.\(p^2\leq{1}\)
\(-1\leq{}p\leq{}1\)
We cannot say if given this interval p will have one of those values: \(-1<p<1\) and \(p\neq{0}\)
Not Sufficient
2.\(p^2-1\neq{0}\)
\(p\neq{}+,-1\)
Clearly not Sufficient
1+2. Using both 1 and 2 we can conclude that \(-1<p<1\) but p could equal 0. Not Sufficient
OA:
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Hey, Good Q ..... I'm bit confused ... Correct me if I'm wrong ....
The Question asks....
Is |p|^2<|p| ? , one must always keep in mind that Is |p| is always positive, no matter the sign of P ..Okay & second thing that p^2 will always remain +ve. Now, |p|^2<|p| will only be possible if |p| must be a proper fraction okay. So, we have to find if |p|
is a proper fraction or not. okay.
now statement 1 says .......... p^2 =<1 ...... this means that P can be equal to +1, -1 or p can be equal to any proper fraction as P^2 must always be less than 1 only if P is a proper fraction. so from this we have three values of P... 1, -1 & any proper fraction... however this is insufficient because of multiple possibilities.
Now statement 2 says ....... p^2-1 is not equal to Zero. that means that p is not equal to +,-1 , therefore it is clearly insufficient. .....
Now, 1+2 ........ as statement 1 says that p can be 1 or -1 or any proper fraction & statement 2 says that P is not equal to +,-1 , therefore we can conclude that p = Proper fraction & if P is a Proper fraction whether +ve Proper fraction or -ve Proper fraction, ..... |p|^2<|p| will always remain true. Hence , C.
Please correct me if I'm wrong. Pls......