Question

Which of the following statements are INCORRECT? (where {} denotes the fractional part of function)

$\square$ $f(x) = \{x\}\cos^2((\frac{2x-1}{2})\pi)$ is differentiable for all $x \in R$

$\square$ $f(x) = |\sin x|\cos^{-1}(\cos x)$ is differentiable $\forall x \in (0,2\pi)$

$\square$ $f(x) = ||x-2|-|x-6||-3|x|+2x+1$ is not differentiable at 3 points

$\square$ $f(x) = \{x\}|\sin \pi x|$ is not differentiable at all integers

• A. $f(x) = \{x\}\cos^2((\frac{2x-1}{2})\pi)$ is differentiable for all $x \in R$
• B. $f(x) = |\sin x|\cos^{-1}(\cos x)$ is differentiable $\forall x \in (0,2\pi)$
• C. $f(x) = ||x-2|-|x-6||-3|x|+2x+1$ is not differentiable at 3 points
• D. $f(x) = \{x\}|\sin \pi x|$ is not differentiable at all integers

Answer 1

Answer: (B), (C)

Statements 2 and 3 are incorrect.

Answer 2

Solution:

1. For $f(x)=\{x\}\cos^2\Big(\frac{2x-1}{2}\pi\Big)$
   Note that

   $$\cos^2\Big(\frac{2x-1}{2}\pi\Big)=\cos^2\Big(\pi x-\frac{\pi}{2}\Big) =\sin^2(\pi x)$$

   so

   $$f(x)=\{x\}\sin^2 (\pi x)$$

   Although the fractional part $\{x\}$ is not differentiable at integers, $\sin^2(\pi x)$ vanishes at every integer (since $\sin(\pi k)=0$); this “kills” the non-differentiability. Hence, $f$ is differentiable for all $x\in\mathbb R$.

2. For $f(x)=|\sin x|\cos^{-1}(\cos x)$
   Recall that the principal value:

   $$\cos^{-1}(\cos x)=\begin{cases} x, & x\in[0,\pi] \\ 2\pi-x, & x\in[\pi,2\pi] \end{cases}$$

   Also, $|\sin x|$ equals $\sin x$ for $x\in(0,\pi)$ and $-\sin x$ for $x\in(\pi,2\pi)$. At $x=\pi$ the left and right derivatives differ (one gets $-\pi$ and $\pi$ respectively), so $f$ is not differentiable at $x=\pi$ even though $\pi\in (0,2\pi)$.

3. For $f(x)= ||x-2|-|x-6|| - 3|x|+ 2x+1$
   The absolute value functions have potential kinks at $x=0, 2, 6$ and the nested absolute value gives another candidate at $x=4$ (since $|x-2|=|x-6|$ when $x=4$). A piecewise check shows that the function is nondifferentiable at $x=0,2,4,6$ (4 points) rather than at 3 points.

4. For $f(x)=\{x\}|\sin (\pi x)|$
   Here, while $\{x\}$ itself is non-differentiable at integers, $|\sin (\pi x)|$ equals zero at all integers. An analysis (using $x=k+h$) shows that the left and right derivatives at an integer $k$ are different. Thus, $f$ is not differentiable at integers.

Conclusion:

• Statement 1 is correct.
• Statement 2 is incorrect.
• Statement 3 is incorrect.
• Statement 4 is correct.

Therefore, the incorrect statements are 2 and 3.