gautamsubrahmanyam wrote:
I understand that 1) is insuff
But for 2) -x|x| > 0 means x cant be +ve => |x| = -x so that -x (-x) = x^2> 0
If x is -ve => (x-3)^2 = X^2+9-6x = (-ve)^2+9-6(-ve) = +ve+9-(-ve) = +ve +9 + (+ve) = +ve
=> sqrt ((x-3)^2) = +X-3
=> sqrt ( (x-3) ^2 ) is not equal to 3-x
=> Option B
Am I right In my logic.Please help
Yes, the answer for this question is B.
Is \(\sqrt{(x-3)^2}=3-x\)?
Remember: \(\sqrt{x^2}=|x|\). Why?
Couple of things:
The point here is that
square root function cannot give negative result: wich means that \(\sqrt{some \ expression}\geq{0}\).
So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?
Let's consider following examples:
If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\);
If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).
So we got that:
\(\sqrt{x^2}=x\), if \(x\geq{0}\);
\(\sqrt{x^2}=-x\), if \(x<0\).
What function does exactly the same thing? The absolute value function! That is why \(\sqrt{x^2}=|x|\)
Back to the original question:So \(\sqrt{(x-3)^2}=|x-3|\) and the question becomes is: \(|x-3|=3-x\)?
When \(x>3\), then RHS (right hand side) is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.
When \(x\leq{3}\), then \(LHS=|x-3|=-x+3=3-x=RHS\), hence in this case equation holds true.
Basically question asks is \(x\leq{3}\)?
(1) \(x\neq{3}\). Clearly insufficient.
(2) \(-x|x| >0\), basically this inequality implies that \(x<0\), hence \(x<3\). Sufficient.Answer: B.
Hope it helps.
, in the highlighted portion above, how can we deduce that x will be less than 3 if x is less than 0? x can be 1,2 also, right? May be I am missing something. Can you please clarify?