Let `f (x)= {((1+ax )^(1//x), x lt 0),( ((x+c)^(1//3)-1)/((x+1)^(1//2) -1), x gt 0):},` is continous at `x=0,` then `3 (e ^(a)+b+c)` is equal to:

Options

A. 3
B. 6
C. 7
D. 8

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 0 \). This means that the left-hand limit (LHL) as \( x \) approaches 0 from the left must equal the right-hand limit (RHL) as \( x \) approaches 0 from the right, and both must equal \( f(0) \).

The function is defined as follows:
\[ f(x) = \begin{cases} (1 + ax)^{\frac{1}{x}}, & x < 0 \\ \frac{(x + c)^{\frac{1}{3}} - 1}{(x + 1)^{\frac{1}{2}} - 1}, & x > 0 \end{cases} \]

### Step 1: Calculate the Left-Hand Limit (LHL)
For \( x < 0 \):
\[ LHL = \lim_{x \to 0^-} (1 + ax)^{\frac{1}{x}} \]
As \( x \) approaches 0 from the left, \( ax \) approaches 0, leading to the form \( (1 + 0)^{\infty} \). We can rewrite this using the exponential limit:
\[ LHL = e^{\lim_{x \to 0^-} \frac{ax}{x}} = e^{a} \]

### Step 2: Calculate the Right-Hand Limit (RHL)
For \( x > 0 \):
\[ RHL = \lim_{x \to 0^+} \frac{(x + c)^{\frac{1}{3}} - 1}{(x + 1)^{\frac{1}{2}} - 1} \]
Substituting \( x = 0 \) gives us the indeterminate form \( \frac{0}{0} \). We can apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule
Differentiate the numerator and denominator:
- The derivative of the numerator:
\[ \frac{d}{dx}((x + c)^{\frac{1}{3}} - 1) = \frac{1}{3}(x + c)^{-\frac{2}{3}} \]
- The derivative of the denominator:
\[ \frac{d}{dx}((x + 1)^{\frac{1}{2}} - 1) = \frac{1}{2}(x + 1)^{-\frac{1}{2}} \]

Now we can rewrite the limit:
\[ RHL = \lim_{x \to 0^+} \frac{\frac{1}{3}(x + c)^{-\frac{2}{3}}}{\frac{1}{2}(x + 1)^{-\frac{1}{2}}} \]
This simplifies to:
\[ RHL = \lim_{x \to 0^+} \frac{2}{3} \cdot \frac{(x + 1)^{\frac{1}{2}}}{(x + c)^{\frac{2}{3}}} \]
Substituting \( x = 0 \):
\[ RHL = \frac{2}{3} \cdot \frac{1}{c^{\frac{2}{3}}} \]

### Step 4: Set LHL equal to RHL
For continuity at \( x = 0 \):
\[ e^{a} = \frac{2}{3c^{\frac{2}{3}}} \]

### Step 5: Determine values of \( b \) and \( c \)
To ensure the function is continuous, we also need to set \( b = c \) and find their values. From the limit, we can set \( b = c = 1 \) to avoid any further complications.

### Step 6: Substitute the values into the equation
Now we have:
- \( e^{a} = \frac{2}{3} \)
- \( b = 1 \)
- \( c = 1 \)

### Step 7: Calculate \( 3(e^{a} + b + c) \)
Substituting these values into the expression:
\[ 3(e^{a} + b + c) = 3\left(\frac{2}{3} + 1 + 1\right) = 3\left(\frac{2}{3} + 2\right) = 3\left(\frac{2 + 6}{3}\right) = 3 \cdot \frac{8}{3} = 8 \]

Thus, the final answer is:
\[ \boxed{8} \]