Consider two function `y=f(x) and y=g(x)` defined as
`f(x)={{:(ax^(2)+b,,0lexle1),(bx+2b,,1ltxle3),((a-1)x+2c-3,,3ltxle4):}`
`and" "g(x)={{:(cx+d,,0lexle2),(ax+3-c,,2ltxlt3),(x^(2)+b+1,,3gexle4):}`
`lim_(xrarr2) (f(x))/(|g(x)|+1)` exists and f is differentiable at x = 1. The value of limit will be

Options

A. `-2`
B. `-1`
C. 0
D. 2

Explanation

To solve the problem, we need to analyze the two piecewise functions \( f(x) \) and \( g(x) \) and find the limit \( \lim_{x \to 2} \frac{f(x)}{|g(x)| + 1} \) given that \( f \) is differentiable at \( x = 1 \).

### Step 1: Analyze the function \( f(x) \)

The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} ax^2 + b & \text{if } 0 \leq x \leq 1 \\ bx + 2b & \text{if } 1 < x \leq 3 \\ (a-1)x + 2c - 3 & \text{if } 3 < x \leq 4 \end{cases} \]

To ensure \( f(x) \) is continuous at \( x = 1 \):
\[ f(1) = a(1)^2 + b = a + b \]
\[ \lim_{x \to 1^+} f(x) = b(1) + 2b = 3b \]
Setting these equal for continuity:
\[ a + b = 3b \implies a = 2b \]

### Step 2: Ensure differentiability at \( x = 1 \)

The derivative from the left at \( x = 1 \):
\[ f'(x) = \frac{d}{dx}(ax^2 + b) = 2ax \Rightarrow f'(1) = 2a \]

The derivative from the right at \( x = 1 \):
\[ f'(x) = \frac{d}{dx}(bx + 2b) = b \Rightarrow f'(1) = b \]

Setting these equal for differentiability:
\[ 2a = b \]
Substituting \( a = 2b \):
\[ 2(2b) = b \implies 4b = b \implies b = 0 \]
Thus, \( a = 0 \).

### Step 3: Analyze the function \( g(x) \)

The function \( g(x) \) is defined as:
\[ g(x) = \begin{cases} cx + d & \text{if } 0 \leq x \leq 2 \\ ax + 3 - c & \text{if } 2 < x < 3 \\ x^2 + b + 1 & \text{if } 3 \leq x \leq 4 \end{cases} \]

To ensure continuity at \( x = 2 \):
\[ g(2) = 2c + d \]
\[ \lim_{x \to 2^-} g(x) = 2c + d \]
\[ \lim_{x \to 2^+} g(x) = 2a + 3 - c \]
Setting these equal:
\[ 2c + d = 2a + 3 - c \implies 3c + d = 2a + 3 \]

### Step 4: Calculate the limit

Now we need to find \( \lim_{x \to 2} \frac{f(x)}{|g(x)| + 1} \):
- From previous steps, we know \( f(2) = 2b = 0 \) (since \( b = 0 \)).
- For \( g(2) \), we have \( g(2) = 2c + d \).

Thus, we need to evaluate:
\[ \lim_{x \to 2} \frac{0}{|g(2)| + 1} = \frac{0}{|2c + d| + 1} = 0 \]

### Final Answer:
The value of the limit is \( \boxed{0} \).