Warlock007: As per your request, I am explaining the solution to this question. The solution is exactly as given by walker above. I will just elaborate on 'why'
Is \(\sqrt{(x-3)^2}=3-x\) ?
1)\(x\ne3\)
2) \(-x*|x| >0\)
When we say square root, we mean just the positive square root. Another way to think about it is:
\(\sqrt{x^2} = |x|\) i.e. square root is only the positive root.
Therefore, \(\sqrt{(x-3)^2} = |x - 3|\)
Now the question is, is |x - 3| = 3 - x?
Now the definition of modulus can be used here.
|x| = x when x >= 0
and |x| = -x when x < 0
i.e. |5| = 5 since 5 > 0
and |-5| = -(-5) = 5 since -5 < 0 (Works for x = 0 too)
Therefore, |x - 3| = -(x - 3) = (3 - x) if (x - 3) <= 0
Let's go back to our question: Is |x - 3| = 3 - x?
|x - 3| = 3 - x only when (x - 3) <= 0 i.e. when x <= 3
Stmnt 1: \(x\ne3\)
No idea whether x is less than 3 so not sufficient.
Stmnt 2: -x*|x| >0
Now, mods are always greater than or equal to 0 (i.e. they can never be negative)
So |x| has to be positive.
Then -x must be positive too to make -x*|x| >0
This means x must be negative (only then will -x be positive)
If x is negative, it is certainly less than 3. Hence, this stmnt alone is sufficient.
Answer (B)
You could also do this question by plugging in numbers and checking (say try x = -1, 0, 1, 3, 4) but plugging in for such questions makes me worry that I have forgotten to check for some specific condition and hence I avoid doing it.
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Karishma
Owner of Angles and Arguments at https://anglesandarguments.com/
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