The data is sufficient if we get a unique value for the expression \(\frac{-|x| - 1}{x - 1}\)
A quick info about this expression. If x is negative then the expression will have a unique value, which is 1.
If x is 0, it will have a unique value, which is 1.
If x is positive it will have multiple values. For instance, if x = 2, the value of the expression is -3. If x = 3, the value of the expression -2.

So, at some level if we can establish that x is not positive, we will have a unique value. Else we will not have a unique value.

Statement 1: \(\sqrt{x^6} > x^3\)
\(\sqrt{x^6}\) is non negative for real x and will be |x^3|.
For instance \(\sqrt{2^6}\) = 8 and \(\sqrt{{(-2)^6}\) will also be 8.
But 2^3 will be 8 and (-2)^3 will be -8.
So, if we know that \(\sqrt{x^6} > x^3\), we can infer that x is negative.

If x is negative, the value of \(\frac{-|x| - 1}{x - 1}\) will be 1. Take any negative x and check it. Negative integer, negative non integer. It will work for all values.

So, statement 1 ALONE is sufficient.

Statement 2:\(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1\)
We can rewrite the expression as |x| + 2x + 4|x| + 8x = -16
5|x| + 10x = -16.
The sum of the expression is -16, which is negative.
5|x| cannot be negative. So, 10x has to be negative if the sum is -16 => x has to be negative.

If x is negative, the expression \(\frac{-|x| - 1}{x - 1}\) has a unique value which is 1.

Statement 2 ALONE is sufficient.

Each statement is INDEPENDENTLY sufficient.
Choice D.
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Hi, You missed to multiply 8 by x. This will change the shape of the equation to be

5|x| +10x =-16

What is the value of \(\frac{-|x| - 1}{x - 1}\)
if x<0, = \(\frac{x-1}{x-1}\) = 1
if x>0, \(\frac{-x-1}{x-1}\), the value depends on value of x

Statement 1) \(\sqrt{x^6} > x^3\)
it is clear that x is negative, because if x is non negative, \(\sqrt{x^6} = x^3\)
So, x is negative and \(\frac{x-1}{x-1}\) = 1
SUFFICIENT

Statement 2) \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1\)
Since RHS is negative, We can straight away say that x is NEGATIVE, as all terms in LHS are in addition, we cant have the additions of positive terms = negative. (There is no need to solve the equation)
So x is negative and and \(\frac{x-1}{x-1}\) = 1
SUFFICIENT

Answer D


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Statement 1) \(\sqrt{x^6} > x^3\) implies x is negative because square root is always positive so, square root of x^6 is positive and x^3 is negative. Only in the case of x being negative, this statement holds true. So, if x is negative. -|x|=x as |x| will be -x and -(-x)=x. The expression reduces to \(\frac{x - 1}{x - 1}\)=1. Sufficient.

Statement 2) \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1\). Simplifying, we get \(\frac{5*|x|}{16} + \frac{5*x}{8}= -1\) or \(\frac{5*|x|}{16} + \frac{10*x}{16}= -1\). Further simplyfying, \(5*|x| + 10*x= -16\). Now, |x| will always be positive, so, x has to be negative to equate to -16 on RHS. Also, multiple of x is higher than of multiple of |X|, so, x has to be negative in order to total to -16. If x is positive, the expression won't hold true. Following the logic in Statement 1, if x is negative, the expression in the problem reduces to \(\frac{x - 1}{x - 1}\)=1. Sufficient.

Hence, D is the correct answer.

\(x≥0:\frac{-|x| - 1}{x - 1}=\frac{-(x)-1}{x-1}=\frac{-x-1}{x-1}=\frac{-(x+1)}{(x-1)}\)
\(x<0:\frac{-|x| - 1}{x - 1}=\frac{-(-x)-1}{x-1}=\frac{x-1}{x-1}=1\)


(1) \(\sqrt{x^6} > x^3\) sufic.

\(\sqrt{x^6}>x^3…(\sqrt{x^2}=|x|)…\sqrt{x^6}=|x^{6/2}|=|x^3|…|x^3|>x^3…x<0\)


(2) \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1\) sufic.

\(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1…|x|+2x+4|x|+8x=-16…|5x|+10x=-16\)
\(x≥0:|5x|+10x=-16…(5x)+10x=-16…15x=-16…x=-16/15=invalid…(x≥0)\)
\(x<0:|5x|+10x=-16…(-5x)+10x=-16…5x=-16…x=-16/15=valid…(x<0)\)

Answer (D)