Let `g (x ) = x ^(C )e ^(Cx) and f (x) = int _(0)^(x) te ^(2r) (1+3t ^(2))^(1//2) dt. If L = lim _(x to oo) (f'(x ))/(g '(x))` is non-zero finite number then :
The valur of C is:

Options

A. 7
B. `3/2`
C. `2`
D. 3

Answer

To solve the problem step-by-step, we will find the value of \( C \) such that the limit \( L = \lim_{x \to \infty} \frac{f'(x)}{g'(x)} \) is a non-zero finite number.

### Step 1: Define the functions
We have:
- \( g(x) = x^C e^{Cx} \)
- \( f(x) = \int_0^x t e^{2t} (1 + 3t^2)^{1/2} dt \)

### Step 2: Find \( g'(x) \)
Using the product rule:
\[ g'(x) = \frac{d}{dx}(x^C) \cdot e^{Cx} + x^C \cdot \frac{d}{dx}(e^{Cx}) \]
Calculating the derivatives:
\[ g'(x) = C x^{C-1} e^{Cx} + x^C \cdot C e^{Cx} = C e^{Cx} (x^{C-1} + x^C) = C e^{Cx} x^{C-1} (1 + x) \]

### Step 3: Find \( f'(x) \)
Using the Leibniz rule for differentiation under the integral sign:
\[ f'(x) = x e^{2x} (1 + 3x^2)^{1/2} \]

### Step 4: Set up the limit
Now we need to evaluate:
\[ L = \lim_{x \to \infty} \frac{f'(x)}{g'(x)} = \lim_{x \to \infty} \frac{x e^{2x} (1 + 3x^2)^{1/2}}{C e^{Cx} x^{C-1} (1 + x)} \]

### Step 5: Simplify the limit
We can simplify the expression:
\[ L = \lim_{x \to \infty} \frac{x e^{2x} (1 + 3x^2)^{1/2}}{C e^{Cx} x^{C-1} (1 + x)} = \lim_{x \to \infty} \frac{(1 + 3x^2)^{1/2}}{C x^{C-2} (1 + x)} e^{(2 - C)x} \]

### Step 6: Analyze the limit
For \( L \) to be a non-zero finite number as \( x \to \infty \):
1. The exponential term \( e^{(2 - C)x} \) must not diverge to infinity or vanish to zero.
2. This implies \( 2 - C = 0 \) or \( C = 2 \).

### Conclusion
Thus, the value of \( C \) is:
\[ \boxed{2} \]