`f(x)={{:((3//x^2)sin2x^2,"if" x lt 0),((x^2+2x+c)/(1-3x^2), "if" x ge 0 "):}` then in order that f be continuous at x=0, the value of c is

Options

A. 2
B. 4
C. 6
D. 8

Answer

To determine the value of \( c \) such that the function \( f(x) \) is continuous at \( x = 0 \), we need to ensure that the left-hand limit and the right-hand limit at \( x = 0 \) are equal to \( f(0) \).

### Step 1: Define the function
The function is defined as:
\[ f(x) = \begin{cases} \frac{3}{x^2} \sin(2x^2) & \text{if } x < 0 \\ \frac{x^2 + 2x + c}{1 - 3x^2} & \text{if } x \geq 0 \end{cases} \]

### Step 2: Find the left-hand limit as \( x \) approaches 0
We calculate the left-hand limit:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{3}{x^2} \sin(2x^2) \]
Using the substitution \( h = -x \) (as \( x \to 0^- \), \( h \to 0^+ \)):
\[ = \lim_{h \to 0^+} \frac{3}{h^2} \sin(2h^2) \]
Using the limit property \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \):
\[ = \lim_{h \to 0^+} 3 \cdot \frac{\sin(2h^2)}{2h^2} \cdot \frac{2h^2}{h^2} = 6 \]
Thus, the left-hand limit is:
\[ f(0^-) = 6 \]

### Step 3: Find the right-hand limit as \( x \) approaches 0
Now we calculate the right-hand limit:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{x^2 + 2x + c}{1 - 3x^2} \]
Substituting \( x = 0 \):
\[ = \frac{0^2 + 2(0) + c}{1 - 3(0^2)} = \frac{c}{1} = c \]
Thus, the right-hand limit is:
\[ f(0^+) = c \]

### Step 4: Set the limits equal for continuity
For the function to be continuous at \( x = 0 \), we need:
\[ f(0^-) = f(0^+) \]
This gives us:
\[ 6 = c \]

### Conclusion
Thus, the value of \( c \) that makes the function continuous at \( x = 0 \) is:
\[ \boxed{6} \]