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 can not 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.
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