Question

If \(\mathrm{f}(\mathrm{x})=|1-\mathrm{x}|\), then the points where $\sin ^{-1}(\mathrm{f}|\mathrm{x}|)$ is non- differentiable are (A) \(\\{0,1\\}\) (B) \(\\{0,-1\\}\) (C) \(\\{0,1,-1\\}\) (D) none of these

Short answer

Answer: (A) \(\{0,1\}\)

Step-by-step solution

Simplify the given composite function

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: \(\sin^{-1}(\mathrm{f}|\mathrm{x}|) = \sin^{-1}(|1-x||x|)\)

Analyze the derivative

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.

Determine the change points of the absolute value function

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 \(x = 0, 1, -1\).

Check the differentiability

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) \(\{0,1\}\)