Key Concepts and Formulas

• Differentiability: A function $f(x)$ is differentiable at a point $x=a$ if the limit defining its derivative exists: $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ This limit must exist and be finite. For a function involving absolute values, we often need to check the left-hand derivative (LHD) and the right-hand derivative (RHD) separately.
  • LHD at $x=a$: $\lim_{h \to 0^-} \frac{f(a+h) - f(a)}{h}$
  • RHD at $x=a$: $\lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h}$ The function is differentiable at $x=a$ if and only if LHD = RHD.
• Absolute Value Properties: $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$. This means functions involving $|x|$ might have points of non-differentiability at $x=0$. Similarly, $|x-\pi|$ might have a point of non-differentiability at $x=\pi$.
• Properties of Exponential and Trigonometric Functions: $e^x$ and $\sin x$ are differentiable everywhere. Their behavior at $x=0$ is important for limits. Specifically, $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$ and $\lim_{x \to 0} \frac{\sin x}{x} = 1$.

Step-by-Step Solution

The problem asks us to find the set $S$ of all real numbers $t$ where the function $f(x) = |x - \pi| (e^{|x|} - 1) \sin|x|$ is not differentiable. We need to check the differentiability of $f(x)$ at points where the absolute value functions might cause a change in behavior, which are $x=0$ and $x=\pi$.

Step 1: Analyze the function $f(x)$ The function is given by $f(x) = |x - \pi| (e^{|x|} - 1) \sin|x|$. We can simplify the absolute values based on the sign of their arguments:

• For $|x-\pi|$:
  • If $x \ge \pi$, $|x-\pi| = x-\pi$.
  • If $x < \pi$, $|x-\pi| = -(x-\pi) = \pi - x$.
• For $|x|$:
  • If $x \ge 0$, $|x| = x$.
  • If $x < 0$, $|x| = -x$.

This suggests we should examine the differentiability at $x=0$ and $x=\pi$.

Step 2: Check differentiability at $x=0$. We need to calculate the LHD and RHD at $x=0$. First, let's find $f(0)$: $f(0) = |0 - \pi| (e^{|0|} - 1) \sin|0| = |-\pi| (e^0 - 1) \sin(0) = \pi (1 - 1) \cdot 0 = 0$.

Now, let's calculate the LHD at $x=0$: LHD = $\mathop {\lim }\limits_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \mathop {\lim }\limits_{h \to 0^-} \frac{f(h) - 0}{h}$ Since $h \to 0^-$, $h$ is negative. Therefore, $|h| = -h$ and $|h-\pi| = \pi - h$ (since $h < \pi$). LHD = $\mathop {\lim }\limits_{h \to 0^-} \frac{|h - \pi|(e^{|h|} - 1)\sin|h|}{h}$ LHD = $\mathop {\lim }\limits_{h \to 0^-} \frac{(\pi - h)(e^{-h} - 1)\sin(-h)}{h}$ Using the property $\sin(-h) = -\sin(h)$: LHD = $\mathop {\lim }\limits_{h \to 0^-} \frac{(\pi - h)(e^{-h} - 1)(-\sin h)}{h}$ LHD = $-\mathop {\lim }\limits_{h \to 0^-} (\pi - h) \cdot \mathop {\lim }\limits_{h \to 0^-} \frac{e^{-h} - 1}{h} \cdot \mathop {\lim }\limits_{h \to 0^-} \frac{\sin h}{h}$

Let's evaluate each limit:

• $\mathop {\lim }\limits_{h \to 0^-} (\pi - h) = \pi - 0 = \pi$.
• $\mathop {\lim }\limits_{h \to 0^-} \frac{e^{-h} - 1}{h}$: Let $u = -h$. As $h \to 0^-$, $u \to 0^+$. This limit becomes $\mathop {\lim }\limits_{u \to 0^+} \frac{e^u - 1}{-u} = -\mathop {\lim }\limits_{u \to 0^+} \frac{e^u - 1}{u} = -1$.
• $\mathop {\lim }\limits_{h \to 0^-} \frac{\sin h}{h} = 1$.

Substituting these values back into the LHD expression: LHD = $-(\pi) \cdot (-1) \cdot (1) = \pi$.

Now, let's calculate the RHD at $x=0$: RHD = $\mathop {\lim }\limits_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \mathop {\lim }\limits_{h \to 0^+} \frac{f(h) - 0}{h}$ Since $h \to 0^+$, $h$ is positive. Therefore, $|h| = h$ and $|h-\pi| = \pi - h$ (since $h < \pi$). RHD = $\mathop {\lim }\limits_{h \to 0^+} \frac{|h - \pi|(e^{|h|} - 1)\sin|h|}{h}$ RHD = $\mathop {\lim }\limits_{h \to 0^+} \frac{(\pi - h)(e^{h} - 1)\sin h}{h}$ RHD = $\mathop {\lim }\limits_{h \to 0^+} (\pi - h) \cdot \mathop {\lim }\limits_{h \to 0^+} \frac{e^{h} - 1}{h} \cdot \mathop {\lim }\limits_{h \to 0^+} \frac{\sin h}{h}$

Let's evaluate each limit:

• $\mathop {\lim }\limits_{h \to 0^+} (\pi - h) = \pi - 0 = \pi$.
• $\mathop {\lim }\limits_{h \to 0^+} \frac{e^{h} - 1}{h} = 1$ (standard limit).
• $\mathop {\lim }\limits_{h \to 0^+} \frac{\sin h}{h} = 1$ (standard limit).

Substituting these values back into the RHD expression: RHD = $(\pi) \cdot (1) \cdot (1) = \pi$.

Since LHD = RHD = $\pi$, the function $f(x)$ is differentiable at $x=0$.

Step 3: Check differentiability at $x=\pi$. We need to calculate the LHD and RHD at $x=\pi$. First, let's find $f(\pi)$: $f(\pi) = |\pi - \pi| (e^{|\pi|} - 1) \sin|\pi| = |0| (e^{\pi} - 1) \sin(\pi) = 0 \cdot (e^{\pi} - 1) \cdot 0 = 0$.

Now, let's calculate the LHD at $x=\pi$: LHD = $\mathop {\lim }\limits_{h \to 0^-} \frac{f(\pi+h) - f(\pi)}{h} = \mathop {\lim }\limits_{h \to 0^-} \frac{f(\pi+h) - 0}{h}$ Since $h \to 0^-$, $h$ is negative. Let $h = -\delta$ where $\delta > 0$ and $\delta \to 0$. Then $\pi+h = \pi-\delta$, which is slightly less than $\pi$. For $\pi+h < \pi$: $|(\pi+h) - \pi| = |h| = -h$ (since $h$ is negative). Also, since $\pi+h$ is close to $\pi$ and $\pi > 0$, $|\pi+h| = \pi+h$. LHD = $\mathop {\lim }\limits_{h \to 0^-} \frac{|(\pi+h) - \pi|(e^{|\pi+h|} - 1)\sin|\pi+h|}{h}$ LHD = $\mathop {\lim }\limits_{h \to 0^-} \frac{|h|(e^{\pi+h} - 1)\sin(\pi+h)}{h}$ LHD = $\mathop {\lim }\limits_{h \to 0^-} \frac{-h(e^{\pi+h} - 1)\sin(\pi+h)}{h}$ (since $h < 0$, $|h| = -h$) LHD = $\mathop {\lim }\limits_{h \to 0^-} -(e^{\pi+h} - 1)\sin(\pi+h)$

Let's evaluate the limit: As $h \to 0^-$, $\pi+h \to \pi^-$. $e^{\pi+h} \to e^{\pi}$. $\sin(\pi+h) \to \sin(\pi) = 0$. So, LHD = $-(e^{\pi} - 1) \cdot 0 = 0$.

Now, let's calculate the RHD at $x=\pi$: RHD = $\mathop {\lim }\limits_{h \to 0^+} \frac{f(\pi+h) - f(\pi)}{h} = \mathop {\lim }\limits_{h \to 0^+} \frac{f(\pi+h) - 0}{h}$ Since $h \to 0^+$, $h$ is positive. Let $h = \delta$ where $\delta > 0$ and $\delta \to 0$. Then $\pi+h = \pi+\delta$, which is slightly greater than $\pi$. For $\pi+h > \pi$: $|(\pi+h) - \pi| = |h| = h$ (since $h$ is positive). Also, since $\pi+h$ is positive, $|\pi+h| = \pi+h$. RHD = $\mathop {\lim }\limits_{h \to 0^+} \frac{|(\pi+h) - \pi|(e^{|\pi+h|} - 1)\sin|\pi+h|}{h}$ RHD = $\mathop {\lim }\limits_{h \to 0^+} \frac{|h|(e^{\pi+h} - 1)\sin(\pi+h)}{h}$ RHD = $\mathop {\lim }\limits_{h \to 0^+} \frac{h(e^{\pi+h} - 1)\sin(\pi+h)}{h}$ (since $h > 0$, $|h| = h$) RHD = $\mathop {\lim }\limits_{h \to 0^+} (e^{\pi+h} - 1)\sin(\pi+h)$

Let's evaluate the limit: As $h \to 0^+$, $\pi+h \to \pi^+$. $e^{\pi+h} \to e^{\pi}$. $\sin(\pi+h) \to \sin(\pi) = 0$. So, RHD = $(e^{\pi} - 1) \cdot 0 = 0$.

Since LHD = RHD = 0, the function $f(x)$ is differentiable at $x=\pi$.

Step 4: Consider differentiability for $x \ne 0$ and $x \ne \pi$. For $x > \pi$, $f(x) = (x-\pi)(e^x - 1)\sin x$. This is a product of differentiable functions, so it is differentiable. For $0 < x < \pi$, $f(x) = (\pi-x)(e^x - 1)\sin x$. This is a product of differentiable functions, so it is differentiable. For $x < 0$, $f(x) = (\pi-x)(e^{-x} - 1)\sin(-x) = -(\pi-x)(e^{-x} - 1)\sin x$. This is a product of differentiable functions, so it is differentiable.

We have confirmed that $f(x)$ is differentiable at $x=0$ and $x=\pi$. Since the function is composed of differentiable components (polynomials, exponential, sine) and the absolute value functions are handled correctly at their critical points, the function is differentiable for all real numbers $x$.

Step 5: Determine the set S. The set $S$ consists of all real numbers $t$ where $f(x)$ is not differentiable. Since we have shown that $f(x)$ is differentiable everywhere, there are no such points. Therefore, the set $S$ is the empty set.

Common Mistakes & Tips

• Incorrectly handling absolute values: Be very careful when splitting cases for absolute values, especially when they are arguments to other functions or part of a product. For example, $|-h|$ when $h \to 0^-$ is $-h$, not $h$.
• Forgetting to check $f(a)$: When calculating LHD and RHD, ensure you correctly use $f(a)$ in the numerator. In this case, $f(0)=0$ and $f(\pi)=0$, simplifying the calculations.
• Using standard limits incorrectly: While $\lim_{x \to 0} \frac{e^x-1}{x}=1$ and $\lim_{x \to 0} \frac{\sin x}{x}=1$ are useful, they apply when the argument goes to 0. In the LHD calculation at $x=0$, we had $\frac{e^{-h}-1}{h}$ where $-h$ went to $0^+$, so it required a substitution.
• Assuming differentiability at critical points without proof: Points where the absolute value function's argument is zero (like $x=0$ for $|x|$ and $x=\pi$ for $|x-\pi|$) are potential points of non-differentiability. Always verify using the definition of the derivative.

Summary

To find the set $S$ of points where the function $f(x) = |x - \pi| (e^{|x|} - 1) \sin|x|$ is not differentiable, we analyzed the function at the critical points arising from the absolute value expressions, namely $x=0$ and $x=\pi$. We calculated the left-hand and right-hand derivatives at these points. At $x=0$, both LHD and RHD were found to be $\pi$, indicating differentiability. At $x=\pi$, both LHD and RHD were found to be 0, also indicating differentiability. Since the function is composed of standard differentiable functions and is differentiable at the points where the absolute value arguments are zero, it is differentiable for all real numbers. Therefore, the set $S$ of points where the function is not differentiable is empty.

The final answer is $\boxed{\phi}$.