Question

Let $S = \{ t \in R:f(x) = \left| {x - \pi } \right|.({e^{\left| x \right|}} - 1)\sin \left| x \right|{\text{ is not differentiable at t\} }}{\text{.}}$ Then the set $S$ is equal to:
A. $\{ \pi \} $
B. $\{ 0,\pi \} $
C. $\phi $(empty set)
D. $\{ 0\} $

Answer 1

Hint: We have to check the differentiability for $0,\pi $. We can do this by differentiating the given function and equating it to zero. First, we will check for zero and then for $\pi $. This will give you your required answer.

Complete step-by-step solution:
Here in this problem, we are given a set which is defined as a set of all real numbers ‘t’ where the function $f(x) = \left| {x - \pi } \right|.({e^{\left| x \right|}} - 1)\sin \left| x \right|$ is not differentiable. And with this information, we need to find the set S.
As we can see in the options that the doubtful points are$0,\pi $.
Hence, we need to check the differentiability of these points.
How will we check the differentiability of these points?
First, we will break the function $f(x)$ at the points$0,\pi $.
Like for $x < 0,\sin \left| x \right| = - \sin x$
And for $x > \pi ,\left| {x - \pi } \right| = x - \pi $
$f(x)$can be written as:
$f\left( x \right) = \left\{ \begin{gathered}
  (x - \pi )({e^{ - x}} - 1)\sin x{\text{ }}:x < 0 \\
  0{\text{ }}:x = 0 \\
\end{gathered} \right.$
And this can be further simplified as:
$ \Rightarrow f\left( x \right) = \left\{ \begin{gathered}
   - (x - \pi )({e^x} - 1)\sin x:0 < x < \pi \\
  0{\text{ }}:x = \pi \\
  (x - \pi )({e^x} - 1)\sin x{\text{ }}:x > \pi \\
\end{gathered} \right.$
Now we have to check the differentiability in the correspondence of $0,\pi $which means that for checking at every point in the vicinity of $0,\pi $.
$ \Rightarrow f'(0 - h) = ({e^{ - x}} - 1)\sin x + (x - \pi )({e^{ - x}} - 1)\cos x - (x - \pi ){e^{ - x}}\sin x$
Let us first of all check for $x < 0$
Differentiating $f(x)$ such that $x < 0$
Where $h$is very small.
As $h$is very small,
$ \Rightarrow f'(0 - h) = f'(0) \Rightarrow f'(0 - h) = 0$ , which means we will substitute $x = 0$
For$x = 0$, $f(x) = 0 \Rightarrow f'(x) = 0$
For $0 < x < \pi $
$ \Rightarrow f'(0 + h) = f'(\pi - h) = - [(x - \pi )({e^x} - 1)\cos x + (x - \pi ){e^x}.\sin x + ({e^x} - 1)\sin x]$
Substituting $x = 0$ or $x = \pi $
$ \Rightarrow f'(0 + h) = f'(\pi - h) = 0$
For $x = \pi $, $f(x) = 0 \Rightarrow f'(x) = 0$
For $x > \pi $
$ \Rightarrow f'(\pi + h) = (x - \pi )({e^x} - 1)\cos x + (x - \pi ){e^x}.\sin x + ({e^x} - 1)\sin x$
Substituting $x = \pi $
$ \Rightarrow f'(\pi + h) = 0$
Also,
$ \Rightarrow f'(0 - h) = f'(0) = f'(0 + h) = 0$
Therefore $f(x)$ is differentiable at$x = 0$.
$ \Rightarrow f'(\pi - h) = f'(\pi ) = f'(\pi + h) = 0$
Therefore $f(x)$ is differentiable at $x = \pi $
$ \Rightarrow S = \{ \phi \} $

Option C is the correct answer.

Note: In calculus, a differentiable function is a continuous function whose derivative exists at all points on its domain. That is, the graph of a differentiable function must have a (non-vertical) tangent line at each point in its domain, be relatively "smooth" (but not necessarily mathematically smooth), and cannot contain any breaks, corners, or cusps.