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 as \( x \) approaches 0 from the negative side must equal the right-hand limit as \( x \) approaches 0 from the positive side, and both must equal \( f(0) \).

Given:
\[ f(x) = \begin{cases} (1 + ax)^{1/x}, & x < 0 \\ \frac{(x + c)^{1/3} - 1}{(x + 1)^{1/2} - 1}, & x > 0 \end{cases} \]

### Step 1: Calculate the left-hand limit as \( x \to 0^- \)

For \( x < 0 \):
\[ f(x) = (1 + ax)^{1/x} \]

We need to find:
\[ \lim_{x \to 0^-} (1 + ax)^{1/x} \]

Using the standard limit:
\[ \lim_{x \to 0} (1 + u)^{1/u} = e \quad \text{where } u = ax \text{ as } x \to 0 \]

Thus, we have:
\[ \lim_{x \to 0^-} (1 + ax)^{1/x} = e^{\lim_{x \to 0} \frac{ax}{x}} = e^a \]

### Step 2: Calculate the right-hand limit as \( x \to 0^+ \)

For \( x > 0 \):
\[ f(x) = \frac{(x + c)^{1/3} - 1}{(x + 1)^{1/2} - 1} \]

We need to find:
\[ \lim_{x \to 0^+} \frac{(x + c)^{1/3} - 1}{(x + 1)^{1/2} - 1} \]

Substituting \( x = 0 \):
\[ = \frac{(0 + c)^{1/3} - 1}{(0 + 1)^{1/2} - 1} = \frac{c^{1/3} - 1}{1 - 1} \text{ (which is } \frac{0}{0} \text{ form)} \]

### Step 3: Apply L'Hôpital's Rule

Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule:
\[ \text{Differentiate the numerator and denominator:} \]

Numerator:
\[ \frac{d}{dx}[(x + c)^{1/3}] = \frac{1}{3}(x + c)^{-2/3} \]

Denominator:
\[ \frac{d}{dx}[(x + 1)^{1/2}] = \frac{1}{2}(x + 1)^{-1/2} \]

Thus,
\[ \lim_{x \to 0^+} \frac{(x + c)^{1/3} - 1}{(x + 1)^{1/2} - 1} = \lim_{x \to 0} \frac{\frac{1}{3}(x + c)^{-2/3}}{\frac{1}{2}(x + 1)^{-1/2}} = \lim_{x \to 0} \frac{2}{3} \cdot \frac{(x + 1)^{1/2}}{(x + c)^{2/3}} \]

Substituting \( x = 0 \):
\[ = \frac{2}{3} \cdot \frac{1}{c^{2/3}} \]

### Step 4: Set the limits equal for continuity

For \( f(x) \) to be continuous at \( x = 0 \):
\[ e^a = \frac{2}{3c^{2/3}} \]

### Step 5: Solve for \( c \)

To ensure the limit exists, we need \( c \) to be such that the numerator approaches 0 when \( x \to 0 \):
\[ c = 1 \]

Substituting \( c = 1 \):
\[ e^a = \frac{2}{3 \cdot 1^{2/3}} \Rightarrow e^a = \frac{2}{3} \]

### Step 6: Final Calculation

Now, we need to find \( 3(e^a + b + c) \):
\[ 3\left(\frac{2}{3} + b + 1\right) = 3\left(\frac{2}{3} + b + 1\right) = 3\left(\frac{2}{3} + b + \frac{3}{3}\right) = 3\left(\frac{5}{3} + b\right) = 5 + 3b \]

### Conclusion

Thus, the final answer is:
\[ 3(e^a + b + c) = 5 + 3b \]