Consider the following function :
1. `f(x)={{:((1)/(x),if,xne0),(0,if,x=0):}`
2. `f(x)={{:(2x+5,"if "xgt0),(x^(2)+2x+5,"if "xle0):}`
Which of the above function is/are derivable at x = 0?

Options

A. Only 1
B. Only 2
C. Both 1 and 2
D. Neither 1 nor 2

Answer

To determine which of the given functions is derivable at \( x = 0 \), we will analyze each function step by step.

### Function 1:
\[ f(x) = \begin{cases} \frac{1}{x} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

**Step 1: Check continuity at \( x = 0 \)**
To check if \( f(x) \) is continuous at \( x = 0 \), we need to find the limit as \( x \) approaches 0:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{1}{x} \]
As \( x \) approaches 0, \( \frac{1}{x} \) approaches \( \infty \) or \( -\infty \) depending on the direction of approach. Thus, the limit does not exist.

**Conclusion for Function 1:**
Since the limit does not exist, \( f(x) \) is not continuous at \( x = 0 \), and therefore, it is not derivable at \( x = 0 \).

### Function 2:
\[ f(x) = \begin{cases} 2x + 5 & \text{if } x > 0 \\ x^2 + 2x + 5 & \text{if } x \leq 0 \end{cases} \]

**Step 2: Check continuity at \( x = 0 \)**
First, we find \( f(0) \):
\[ f(0) = 0^2 + 2(0) + 5 = 5 \]
Next, we find the limit as \( x \) approaches 0:
\[ \lim_{x \to 0^+} f(x) = 2(0) + 5 = 5 \]
\[ \lim_{x \to 0^-} f(x) = 0^2 + 2(0) + 5 = 5 \]
Since both one-sided limits equal \( f(0) \), we conclude that \( f(x) \) is continuous at \( x = 0 \).

**Step 3: Check differentiability at \( x = 0 \)**
To check if \( f(x) \) is differentiable at \( x = 0 \), we calculate the left-hand derivative (LHD) and right-hand derivative (RHD):

- **Right-hand derivative (RHD)**:
\[ f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{(2h + 5) - 5}{h} = \lim_{h \to 0^+} \frac{2h}{h} = 2 \]

- **Left-hand derivative (LHD)**:
\[ f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{(h^2 + 2h + 5) - 5}{h} = \lim_{h \to 0^-} \frac{h^2 + 2h}{h} = \lim_{h \to 0^-} (h + 2) = 2 \]

Since both the RHD and LHD at \( x = 0 \) are equal:
\[ f'(0^+) = f'(0^-) = 2 \]

**Conclusion for Function 2:**
Since \( f(x) \) is continuous and the derivatives from both sides are equal, \( f(x) \) is derivable at \( x = 0 \).

### Final Conclusion:
- **Function 1** is not derivable at \( x = 0 \).
- **Function 2** is derivable at \( x = 0 \).