91Ó°ÊÓ

Before finding the derivative, let's simplify the composite function \(\sin^{-1}(\mathrm{f}|\mathrm{x}|)\). We already have \(\mathrm{f}(\mathrm{x}) = |1-\mathrm{x}|\). So our composite function becomes: \(\sin^{-1}(\mathrm{f}|\mathrm{x}|) = \sin^{-1}(|1-x||x|)\)

Now, we need to find the derivative of \(\sin^{-1}(|1-x||x|)\). Note that \(\sin^{-1}(u)\) is differentiable for all \(u\) such that \(-1\leq u \leq1\). Thus, we need to examine where the derivative of \(u = |1-x||x|\) doesn't exist. The prime candidates for non-differentiability are the points where the absolute value function changes from positive to negative or vice versa.

We have two absolute value functions: \(|1-x|\) and \(|x|\). Let's determine the points where their sign changes: 1. \(|1-x|\) changes sign at \(x=1\). When \(x<1\), we have \(|1-x| = 1-x\) and when \(x>1\), we have \(|1-x| = x-1\). 2. \(|x|\) changes sign at \(x=0\). When \(x<0\), we have \(|x| = -x\) and when \(x>0\), we have \(|x| = x\). Considering both of these changes, we should examine the points \(x = 0, 1, -1\).

Since we know the points when the function may not be differentiable and the corresponding expressions for each piece, we should examine the limits of the derivative as \(x\) approaches these points. 1. At \(x=-1\), we have \(|1-x||x| = 2|x|\), which is differentiable. 2. At \(x=0\), we have \(|1-x||x|=x|1-x|\). Here, when \(x\) is positive, it is equal to \(x(1-x)\), and when \(x\) is negative, it is equal to \(x(x+1)\). Both of these expressions are differentiable, but have different derivatives at \(x=0\). So, \(x=0\) is a point of non-differentiability. 3. At \(x=1\), we also have \(|1-x||x|=x|1-x|\). Therefore, \(x=1\) is also a point of non-differentiability, as it is in the same situation as \(x=0\) So, our set of points where \(\sin^{-1}(\mathrm{f}|\mathrm{x}|)\) is non-differentiable is \(\{0,1\}\). Therefore, the correct answer is: (A) \(\{0,1\}\)