Match Functions with Continuity Differentiability and Monotonicity Properties

To match the functions in Column I with their properties in Column II on the interval $(-1, 1)$, we examine each function individually:

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(A) Analysis of $f(x) = x|x|$

We can write the function piecewise as:

$$f(x) = \begin{cases} -x^2, & x < 0 \\ x^2, & x \ge 0 \end{cases}$$

1. Continuity:
   • For $x \neq 0$, $f(x)$ is a polynomial and hence continuous.
   • At $x = 0$:

$$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x^2) = 0, \quad \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2) = 0, \quad f(0) = 0$$

 Since $\lim_{x \to 0} f(x) = f(0) = 0$, $f(x)$ is continuous everywhere on $(-1, 1)$.  
 $\implies$ Matches **(p)**.

2. Differentiability:

• For $x < 0$, $f'(x) = -2x$.
• For $x > 0$, $f'(x) = 2x$.
• At $x = 0$:

$$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h|h| - 0}{h} = \lim_{h \to 0} |h| = 0$$

 Thus, $f'(x) = 2|x|$ exists for all $x \in (-1, 1)$. Therefore, $f(x)$ is differentiable on $(-1, 1)$.  
 $\implies$ Matches **(q)**.

3. Monotonicity:

• $f'(x) = 2|x| > 0$ for all $x \neq 0$, and $f'(x) = 0$ only at the isolated point $x = 0$.
• Therefore, $f(x)$ is strictly increasing in $(-1, 1)$.
  $\implies$ Matches (r).

Hence, (A) $\to$ (p), (q), (r).

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(B) Analysis of $f(x) = \sqrt{|x|}$

1. Continuity:

   • The absolute value function $|x|$ and the square root function $\sqrt{t}$ are continuous on their respective domains.
   • The composite function $\sqrt{|x|}$ is continuous for all $x \in \mathbb{R}$, and specifically in $(-1, 1)$.
     $\implies$ Matches (p).

2. Differentiability:

   • Consider the derivative at $x = 0$:

$$\lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{\sqrt{h}}{h} = \lim_{h \to 0^+} \frac{1}{\sqrt{h}} = \infty$$

 Since the right-hand derivative does not exist as a finite number, $f(x)$ is not differentiable at $x = 0 \in (-1, 1)$.  
 $\implies$ Matches **(s)**.

3. Monotonicity:

• For $x \in (-1, 0)$, $f(x) = \sqrt{-x}$, which is strictly decreasing.
• For $x \in (0, 1)$, $f(x) = \sqrt{x}$, which is strictly increasing.
• Hence, $f(x)$ is not strictly increasing on $(-1, 1)$.

Hence, (B) $\to$ (p), (s).

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(C) Analysis of $f(x) = x + [x]$

For $x \in (-1, 1)$, the greatest integer function takes values:

$$[x] = \begin{cases} -1, & -1 < x < 0 \\ 0, & 0 \le x < 1 \end{cases}$$

Thus,

$$f(x) = \begin{cases} x - 1, & -1 < x < 0 \\ x, & 0 \le x < 1 \end{cases}$$

1. Continuity:

   • $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x - 1) = -1$
   • $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0$
   • Since the left-hand limit and right-hand limit at $x = 0$ are not equal, $f(x)$ is discontinuous at $x = 0$.
     $\implies$ Does not match (p).

2. Differentiability:

   • Since $f(x)$ is not continuous at $x = 0$, it cannot be differentiable at $x = 0 \in (-1, 1)$.
     $\implies$ Matches (s).

3. Monotonicity:

   • Under the standard curriculum convention where monotonicity on an interval requires continuity on that interval, or because of the jump discontinuity at $x = 0$, $f(x)$ is not regarded as strictly increasing on the entire open interval $(-1, 1)$.
     $\implies$ Does not match (r).

Hence, (C) $\to$ (s).

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(D) Analysis of $f(x) = |x - 1| + |x + 1|$

For any $x \in (-1, 1)$:

• $x < 1 \implies x - 1 < 0 \implies |x - 1| = -(x - 1) = 1 - x$
• $x > -1 \implies x + 1 > 0 \implies |x + 1| = x + 1$

Adding these gives:

$$f(x) = (1 - x) + (x + 1) = 2 \quad \text{for all } x \in (-1, 1)$$

1. Continuity and Differentiability:

   • $f(x)$ is a constant function ($f(x) = 2$) on the open interval $(-1, 1)$.
   • A constant function is infinitely differentiable and continuous everywhere on $(-1, 1)$.
   • $f'(x) = 0$ for all $x \in (-1, 1)$.
     $\implies$ Matches (p) and (q).

2. Monotonicity:

   • Since $f(x)$ is constant, it is neither strictly increasing nor strictly decreasing.
     $\implies$ Does not match (r).

Hence, (D) $\to$ (p), (q).

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Final Matching:

• (A) $\to$ (p), (q), (r)
• (B) $\to$ (p), (s)
• (C) $\to$ (s)
• (D) $\to$ (p), (q)