If `f(x) = {{:(ax^(2) + b, b ne 0, x le 1),(bx^(2) + ax + c,, x gt 1):}`
Then f[x] is continuous and differentiable at x = 1 if:

Options

A. c=0, a=2b
B. `a=b, c in R`
C. `a=b, c=0`
D. `a=b, c ne 0`

Answer

To determine the conditions under which the function \( f(x) \) is continuous and differentiable at \( x = 1 \), we need to analyze the left-hand limit and right-hand limit at that point.

Given the piecewise function:
\[ f(x) = \begin{cases} ax^2 + b & \text{if } x \leq 1 \\ bx^2 + ax + c & \text{if } x > 1 \end{cases} \]

### Step 1: Check for Continuity at \( x = 1 \)

For \( f(x) \) to be continuous at \( x = 1 \), the left-hand limit as \( x \) approaches 1 must equal the right-hand limit at \( x = 1 \), and both must equal \( f(1) \).

**Left-hand limit:**
\[ \lim_{x \to 1^-} f(x) = f(1) = a(1)^2 + b = a + b \]

**Right-hand limit:**
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (bx^2 + ax + c) = b(1)^2 + a(1) + c = b + a + c \]

**Setting the limits equal for continuity:**
\[ a + b = b + a + c \]
This simplifies to:
\[ c = 0 \]

### Step 2: Check for Differentiability at \( x = 1 \)

For \( f(x) \) to be differentiable at \( x = 1 \), the left-hand derivative must equal the right-hand derivative at that point.

**Left-hand derivative:**
\[ f'(x) = \frac{d}{dx}(ax^2 + b) = 2ax \]
Thus,
\[ \lim_{h \to 0} \frac{f(1-h) - f(1)}{-h} = \lim_{h \to 0} \frac{(a(1-h)^2 + b) - (a + b)}{-h} \]
Calculating this gives:
\[ = \lim_{h \to 0} \frac{a(1 - 2h + h^2) + b - a - b}{-h} = \lim_{h \to 0} \frac{-2ah + ah^2}{-h} = \lim_{h \to 0} (2a - ah) = 2a \]

**Right-hand derivative:**
\[ f'(x) = \frac{d}{dx}(bx^2 + ax + c) = 2bx + a \]
Thus,
\[ \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{(b(1+h)^2 + a(1+h) + c) - (a + b)}{h} \]
Substituting \( c = 0 \):
\[ = \lim_{h \to 0} \frac{(b(1 + 2h + h^2) + a(1 + h)) - (a + b)}{h} \]
This simplifies to:
\[ = \lim_{h \to 0} \frac{b + 2bh + bh^2 + a + ah - a - b}{h} = \lim_{h \to 0} \frac{(2b + a)h + bh^2}{h} = 2b + a \]

**Setting the derivatives equal for differentiability:**
\[ 2a = 2b + a \]
This simplifies to:
\[ a = 2b \]

### Conclusion

The conditions for \( f(x) \) to be continuous and differentiable at \( x = 1 \) are:
1. \( c = 0 \)
2. \( a = 2b \)