Which of the following functions is not differentiable at x=1?

Options

A. `f(x)=(x^(2)-1)|(x-1)(x-2)|`
B. f(x)=sin(|x-1|)-|x-1|
C. f(x)=tan(|x-1|)+|x-1|
D. None of these

Answer

To determine which of the given functions is not differentiable at \( x = 1 \), we will analyze each function step by step.

### Step 1: Analyze the first function
Let’s denote the first function as:
\[ f_1(x) = (x^2 - 1) |x - 1| (x - 2) \]

To check differentiability at \( x = 1 \), we need to compute the left-hand derivative (LHD) and right-hand derivative (RHD) at this point.

**Left-hand derivative (LHD)**:
\[ LHD = \lim_{h \to 0^-} \frac{f_1(1 + h) - f_1(1)}{h} \]
Substituting \( x = 1 + h \):
\[ f_1(1 + h) = ((1 + h)^2 - 1) |(1 + h) - 1| ((1 + h) - 2) \]
Calculating this:
\[ = (1 + 2h + h^2 - 1) |h| (h - 1) = (2h + h^2) |h| (h - 1) \]
As \( h \to 0^- \), \( |h| = -h \):
\[ = (2h + h^2)(-h)(h - 1) = -(2h^2 + h^3)(h - 1) \]
Now, we can simplify and evaluate the limit:
\[ LHD = \lim_{h \to 0^-} \frac{-(2h^2 + h^3)(h - 1)}{h} \]
This will yield \( 0 \).

**Right-hand derivative (RHD)**:
\[ RHD = \lim_{h \to 0^+} \frac{f_1(1 + h) - f_1(1)}{h} \]
Following similar steps as above, we find that:
\[ RHD = 0 \]
Since \( LHD = RHD \), the first function is differentiable at \( x = 1 \).

### Step 2: Analyze the second function
Let’s denote the second function as:
\[ f_2(x) = \sin(|x - 1|) - |x - 1| \]

**Left-hand derivative (LHD)**:
\[ LHD = \lim_{h \to 0^-} \frac{f_2(1 + h) - f_2(1)}{h} \]
Calculating:
\[ f_2(1 + h) = \sin(-h) - (-h) = -\sin(h) + h \]
Thus,
\[ LHD = \lim_{h \to 0^-} \frac{-\sin(h) + h}{h} \]
This limit evaluates to \( 0 \).

**Right-hand derivative (RHD)**:
\[ RHD = \lim_{h \to 0^+} \frac{f_2(1 + h) - f_2(1)}{h} \]
Calculating:
\[ f_2(1 + h) = \sin(h) - h \]
Thus,
\[ RHD = \lim_{h \to 0^+} \frac{\sin(h) - h}{h} \]
This limit also evaluates to \( 0 \).

Since \( LHD = RHD \), the second function is differentiable at \( x = 1 \).

### Step 3: Analyze the third function
Let’s denote the third function as:
\[ f_3(x) = \tan(|x - 1|) + |x - 1| \]

**Left-hand derivative (LHD)**:
\[ LHD = \lim_{h \to 0^-} \frac{f_3(1 + h) - f_3(1)}{h} \]
Calculating:
\[ f_3(1 + h) = \tan(-h) + (-h) = -\tan(h) - h \]
Thus,
\[ LHD = \lim_{h \to 0^-} \frac{-\tan(h) - h}{h} \]
This limit evaluates to \( -\infty \).

**Right-hand derivative (RHD)**:
\[ RHD = \lim_{h \to 0^+} \frac{f_3(1 + h) - f_3(1)}{h} \]
Calculating:
\[ f_3(1 + h) = \tan(h) + h \]
Thus,
\[ RHD = \lim_{h \to 0^+} \frac{\tan(h) + h}{h} \]
This limit evaluates to \( +\infty \).

Since \( LHD \neq RHD \), the third function is **not differentiable at \( x = 1 \)**.

### Conclusion
The function that is not differentiable at \( x = 1 \) is:
\[ f_3(x) = \tan(|x - 1|) + |x - 1| \]