Let `f(x)=p[x]+qe^(-[x])+r|x|^(2)`, where p,q and r are real constants, If f(x) is differential at x=0. Then,

Options

A. `q=0,r=0, p in R`
B. `p=0,r=0, q in R`
C. `p=0,q=0, r in R`
D. none of these

Explanation

To determine the conditions under which the function \( f(x) = p[x] + q e^{-[x]} + r |x|^2 \) is differentiable at \( x = 0 \), we will analyze each term in the function.

### Step 1: Analyze the greatest integer function \([x]\)

The greatest integer function \([x]\) is defined as the largest integer less than or equal to \( x \). At \( x = 0 \), we have:
\[ [x] = 0 \quad \text{for } x = 0 \]
However, as \( x \) approaches \( 0 \) from the left (\( x \to 0^- \)), \([x] = -1\), and from the right (\( x \to 0^+ \)), \([x] = 0\).

### Step 2: Evaluate the left-hand limit and right-hand limit of \( f(x) \)

1. **Left-hand limit** as \( x \to 0^- \):
\[ f(0^-) = p[-1] + q e^{-[0]} + r |0|^2 = -p + q \cdot 1 + 0 = -p + q \]

2. **Right-hand limit** as \( x \to 0^+ \):
\[ f(0^+) = p[0] + q e^{-[0]} + r |0|^2 = 0 + q \cdot 1 + 0 = q \]

### Step 3: Set the left-hand limit equal to the right-hand limit for continuity

For \( f(x) \) to be differentiable at \( x = 0 \), it must first be continuous at that point. Therefore, we set the left-hand limit equal to the right-hand limit:
\[ -p + q = q \]
This simplifies to:
\[ -p = 0 \implies p = 0 \]

### Step 4: Substitute \( p = 0 \) into the function

Now, substituting \( p = 0 \) into the function:
\[ f(x) = 0 + q e^{-[x]} + r |x|^2 = q e^{-[x]} + r |x|^2 \]

### Step 5: Check differentiability at \( x = 0 \)

Now we need to check if the function \( f(x) = q e^{-[x]} + r |x|^2 \) is differentiable at \( x = 0 \).

1. **Left-hand derivative**:
\[ f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{q e^{-h} + r h^2 - (q + 0)}{h} = \lim_{h \to 0^-} \frac{q (e^{-h} - 1) + r h^2}{h} \]
As \( h \to 0^- \), \( e^{-h} \to 1 \), so:
\[ f'(0^-) = \lim_{h \to 0^-} \frac{q (0) + r h^2}{h} = \lim_{h \to 0^-} r h = 0 \]

2. **Right-hand derivative**:
\[ f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{q e^{-h} + r h^2 - (q + 0)}{h} = \lim_{h \to 0^+} \frac{q (e^{-h} - 1) + r h^2}{h} \]
As \( h \to 0^+ \), \( e^{-h} \to 1 \), so:
\[ f'(0^+) = \lim_{h \to 0^+} \frac{q (0) + r h^2}{h} = \lim_{h \to 0^+} r h = 0 \]

### Conclusion

Since both the left-hand and right-hand derivatives at \( x = 0 \) are equal to \( 0 \), the function is differentiable at \( x = 0 \).

Thus, the conditions for differentiability at \( x = 0 \) are:
\[ p = 0, \quad q \text{ is any real number}, \quad r \text{ is any real number} \]