Let f (x) be a continous function in `[-1,1]` such that
`f (x)= [{:((ln (ax ^(2)+ bx +c))/(x ^(2)),,, -1 le x lt 0),(1 ,,, x =0),((sin (e ^(x ^(2))-1))/(x ^(2)),,, 0 lt x le 1):}` Then which of the following is/are corrent

Options

A. `a+b+c=0`
B. `b=a+c`
C. `c=1+b`
D. `b ^(2) +c ^(2) =1`

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 0 \). The function is defined piecewise as follows:

1. For \( -1 \leq x < 0 \):
\[ f(x) = \frac{\ln(ax^2 + bx + c)}{x^2} \]

2. For \( x = 0 \):
\[ f(0) = 1 \]

3. For \( 0 < x \leq 1 \):
\[ f(x) = \frac{\sin(e^{x^2} - 1)}{x^2} \]

### Step 1: Find the left-hand limit as \( x \) approaches 0
We need to find the limit of \( f(x) \) as \( x \) approaches 0 from the left:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\ln(ax^2 + bx + c)}{x^2} \]

### Step 2: Evaluate the limit
As \( x \to 0 \):
\[ ax^2 + bx + c \to c \quad \text{(since \( ax^2 \) and \( bx \) both approach 0)} \]
Thus,
\[ \ln(ax^2 + bx + c) \to \ln(c) \]
So we have:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\ln(c)}{x^2} \]

This limit tends to \( -\infty \) unless \( c = 1 \) (since \( \ln(1) = 0 \)). Therefore, we set:
\[ c = 1 \]

### Step 3: Find the right-hand limit as \( x \) approaches 0
Now we find the limit of \( f(x) \) as \( x \) approaches 0 from the right:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sin(e^{x^2} - 1)}{x^2} \]

Using the fact that \( e^{x^2} - 1 \approx x^2 \) as \( x \to 0 \), we can apply L'Hôpital's Rule:
\[ \lim_{x \to 0^+} \frac{\sin(e^{x^2} - 1)}{x^2} = \lim_{x \to 0^+} \frac{\sin(x^2)}{x^2} = 1 \]

### Step 4: Set the limits equal for continuity
For the function to be continuous at \( x = 0 \):
\[ \lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x) \]
This gives us:
\[ 0 = 1 \]
This is not possible unless we ensure the numerator also approaches 0. Thus, we need to differentiate \( \ln(ax^2 + bx + c) \) and apply L'Hôpital's Rule.

### Step 5: Differentiate and apply L'Hôpital's Rule
Using L'Hôpital's Rule:
\[ \lim_{x \to 0^-} \frac{\ln(ax^2 + bx + c)}{x^2} = \lim_{x \to 0^-} \frac{\frac{2ax + b}{ax^2 + bx + c}}{2x} \]
Setting \( c = 1 \), we have:
\[ \lim_{x \to 0^-} \frac{2a(0) + b}{2(0)(a(0)^2 + b(0) + 1)} = \frac{b}{0} \]
This implies \( b = 0 \).

### Step 6: Relate \( a \) and \( c \)
Since we have \( c = 1 \) and \( b = 0 \), we can also find \( a \):
\[ \lim_{x \to 0^-} \frac{\ln(ax^2 + 0 + 1)}{x^2} = 0 \Rightarrow a = 1 \]

### Summary of Results
We have found:
- \( a = 1 \)
- \( b = 0 \)
- \( c = 1 \)

### Step 7: Check the options
1. \( a + b + c = 1 + 0 + 1 = 2 \) (Incorrect)
2. \( b = a + c \Rightarrow 0 = 1 + 1 \) (Incorrect)
3. \( c = a + b \Rightarrow 1 = 1 + 0 \) (Correct)
4. \( b^2 + c^2 = 0^2 + 1^2 = 1 \) (Correct)

### Final Answer
The correct options are:
- Option C: \( c = a + b \)
- Option D: \( b^2 + c^2 = 1 \)