Key Concepts and Formulas

• Continuity: A function $f(x)$ is continuous at a point $c$ if $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$. For a piecewise function, we need to check continuity at the points where the definition of the function changes.
• Differentiability: A function $f(x)$ is differentiable at a point $c$ if the limit of the difference quotient exists. For a piecewise function, this means the left-hand derivative (LHD) must equal the right-hand derivative (RHD) at the point where the definition changes, and the function must also be continuous at that point.
  • Left-Hand Derivative (LHD) at $c$: $\lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h}$ or by differentiating the left piece of the function and evaluating at $c$.
  • Right-Hand Derivative (RHD) at $c$: $\lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h}$ or by differentiating the right piece of the function and evaluating at $c$.
• Inverse Trigonometric Functions: We will use the property that $\tan^{-1}(1) = \frac{\pi}{4}$ and $\tan^{-1}(-1) = -\frac{\pi}{4}$.

Step-by-Step Solution

1. Rewrite the function in a more explicit piecewise form: The given function is defined using $|x|$. We can rewrite it by considering the cases $|x| \le 1$ (which means $-1 \le x \le 1$) and $|x| > 1$ (which means $x > 1$ or $x < -1$). f\left( x \right) = \left\{ {\matrix{ {{\pi \over 4} + {{\tan }^{ - 1}}x,} & { - 1 \le x \le 1} \cr {{1 \over 2}\left( {x - 1} \right),} & {x > 1} \cr {{1 \over 2}\left( { - x - 1} \right),} & {x < - 1} \cr } } \right. Note that for $|x| > 1$, if $x > 1$, then $|x|=x$, so $\frac{1}{2}(|x|-1) = \frac{1}{2}(x-1)$. If $x < -1$, then $|x|=-x$, so $\frac{1}{2}(|x|-1) = \frac{1}{2}(-x-1)$.

2. Check for continuity at $x = 1$: We need to evaluate the left-hand limit (LHL), the right-hand limit (RHL), and the function value at $x=1$.

   • LHL: As $x$ approaches $1$ from the left ($x \to 1^-$), $x$ is in the interval $[-1, 1]$, so we use the first part of the function. $\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {{\pi \over 4} + {{\tan }^{ - 1}}x} \right) = {\pi \over 4} + {\tan ^{ - 1}}(1) = {\pi \over 4} + {\pi \over 4} = {\pi \over 2}$
   • Function Value: At $x=1$, we use the first part of the function. $f(1) = {\pi \over 4} + {\tan ^{ - 1}}(1) = {\pi \over 4} + {\pi \over 4} = {\pi \over 2}$
   • RHL: As $x$ approaches $1$ from the right ($x \to 1^+$), $x > 1$, so we use the second part of the function. $\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {{1 \over 2}\left( {x - 1} \right)} \right) = {1 \over 2}(1 - 1) = 0$ Since LHL (${\pi \over 2}$) $\ne$ RHL ($0$), the function $f(x)$ is discontinuous at $x = 1$. A function must be continuous at a point to be differentiable at that point. Therefore, $f(x)$ is not differentiable at $x=1$.

3. Check for continuity at $x = -1$: We need to evaluate the left-hand limit (LHL), the right-hand limit (RHL), and the function value at $x=-1$.

   • LHL: As $x$ approaches $-1$ from the left ($x \to -1^-$), $x < -1$, so we use the third part of the function. $\mathop {\lim }\limits_{x \to { - 1^ - }} f(x) = \mathop {\lim }\limits_{x \to { - 1^ - }} \left( {{1 \over 2}\left( { - x - 1} \right)} \right) = {1 \over 2}( - (-1) - 1) = {1 \over 2}(1 - 1) = 0$
   • Function Value: At $x=-1$, we use the first part of the function. $f(-1) = {\pi \over 4} + {\tan ^{ - 1}}(-1) = {\pi \over 4} - {\pi \over 4} = 0$
   • RHL: As $x$ approaches $-1$ from the right ($x \to -1^+$), $x$ is in the interval $[-1, 1]$, so we use the first part of the function. $\mathop {\lim }\limits_{x \to { - 1^ + }} f(x) = \mathop {\lim }\limits_{x \to { - 1^ + }} \left( {{\pi \over 4} + {{\tan }^{ - 1}}x} \right) = {\pi \over 4} + {\tan ^{ - 1}}(-1) = {\pi \over 4} - {\pi \over 4} = 0$ Since LHL ($0$) = RHL ($0$) = $f(-1)$ ($0$), the function $f(x)$ is continuous at $x = -1$.

4. Check for differentiability at $x = -1$: Since the function is continuous at $x=-1$, we can proceed to check for differentiability. We need to compute the left-hand derivative (LHD) and the right-hand derivative (RHD) at $x=-1$. First, let's find the derivative of each piece of the function:

   • For $-1 < x < 1$, $f'(x) = \frac{d}{dx} \left( \frac{\pi}{4} + \tan^{-1}x \right) = \frac{1}{1+x^2}$.
   • For $x > 1$, $f'(x) = \frac{d}{dx} \left( \frac{1}{2}(x-1) \right) = \frac{1}{2}$.
   • For $x < -1$, $f'(x) = \frac{d}{dx} \left( \frac{1}{2}(-x-1) \right) = -\frac{1}{2}$.

   Now, evaluate the LHD and RHD at $x=-1$:

   • LHD at $x=-1$: This corresponds to the derivative of the function for $x < -1$ evaluated at $x=-1$. $LHD = \lim_{x \to -1^-} f'(x) = \lim_{x \to -1^-} \left(-\frac{1}{2}\right) = -\frac{1}{2}$
   • RHD at $x=-1$: This corresponds to the derivative of the function for $-1 < x < 1$ evaluated at $x=-1$. $RHD = \lim_{x \to -1^+} f'(x) = \lim_{x \to -1^+} \left(\frac{1}{1+x^2}\right) = \frac{1}{1+(-1)^2} = \frac{1}{1+1} = \frac{1}{2}$ Since LHD ($-\frac{1}{2}$) $\ne$ RHD ($\frac{1}{2}$), the function $f(x)$ is not differentiable at $x = -1$.

5. Analyze continuity and differentiability for other values of $x$:

   • For $|x| < 1$ (i.e., $-1 < x < 1$), $f(x) = \frac{\pi}{4} + \tan^{-1}x$. This is a combination of a constant and an inverse tangent function, both of which are continuous and differentiable for all real numbers. Thus, $f(x)$ is continuous and differentiable on $(-1, 1)$.
   • For $x > 1$, $f(x) = \frac{1}{2}(x-1)$. This is a linear function, which is continuous and differentiable for all real numbers. Thus, $f(x)$ is continuous and differentiable on $(1, \infty)$.
   • For $x < -1$, $f(x) = \frac{1}{2}(-x-1)$. This is a linear function, which is continuous and differentiable for all real numbers. Thus, $f(x)$ is continuous and differentiable on $(-\infty, -1)$.

6. Synthesize the results:

   • The function is discontinuous at $x=1$.
   • The function is discontinuous at $x=1$, so it is not differentiable at $x=1$.
   • The function is continuous at $x=-1$, but not differentiable at $x=-1$.
   • The function is continuous on $(-\infty, -1) \cup (-1, 1) \cup (1, \infty)$.
   • The function is differentiable on $(-\infty, -1) \cup (-1, 1) \cup (1, \infty)$.

   Combining these, the function is continuous on $\mathbb{R} - \{1\}$ (since it's continuous at $-1$ but not at $1$). The function is differentiable on $\mathbb{R} - \{-1, 1\}$ (since it's not differentiable at $-1$ and not at $1$).

Common Mistakes & Tips

• Careless handling of absolute value: Ensure that $|x|$ is correctly expanded into $x$ or $-x$ based on the interval.
• Confusing continuity and differentiability conditions: Remember that differentiability at a point implies continuity at that point, but the converse is not always true. Always check continuity first.
• Incorrect evaluation of limits at boundary points: Double-check the limits and function values at the points where the function definition changes.

Summary

We analyzed the given piecewise function by first rewriting it with explicit intervals for $x$. We then checked for continuity and differentiability at the critical points $x=1$ and $x=-1$. We found that the function is discontinuous at $x=1$, and thus not differentiable there. At $x=-1$, the function is continuous but not differentiable because the left-hand derivative and the right-hand derivative are not equal. For all other values of $x$, the individual pieces of the function are continuous and differentiable. Therefore, the function is continuous on $\mathbb{R}-\{1\}$ and differentiable on $\mathbb{R}-\{-1, 1\}$.

The final answer is \boxed{A}.