If `lim_(xrarr0) (e^(ax)-cos (bx)-((cxe)/2)^(-cx))/(1-cos(2x)=17` , then `5a^2 + b^2 ` is equal to .

Options

A. `72`
B. `76`
C. `68`
D. `64`

Answer

To solve the limit problem, we start with the expression given:

\[ \lim_{x \to 0} \frac{e^{ax} - \cos(bx) - \frac{cxe^{-cx}}{2}}{1 - \cos(2x)} = 17 \]

### Step 1: Analyze the limit
As \( x \to 0 \), both the numerator and denominator approach 0. Thus, we can apply L'Hôpital's Rule, which states that if the limit results in the indeterminate form \( \frac{0}{0} \), we can differentiate the numerator and the denominator.

### Step 2: Differentiate the numerator
The numerator is:
\[ e^{ax} - \cos(bx) - \frac{cxe^{-cx}}{2} \]
Differentiating term by term:
- The derivative of \( e^{ax} \) is \( ae^{ax} \).
- The derivative of \( -\cos(bx) \) is \( b\sin(bx) \).
- For \( -\frac{cxe^{-cx}}{2} \), we use the product rule:
\[ -\frac{1}{2} \left( c e^{-cx} + cx(-ce^{-cx}) \right) = -\frac{c e^{-cx}}{2} + \frac{c^2 x e^{-cx}}{2} \]

Thus, the derivative of the numerator is:
\[ ae^{ax} + b\sin(bx) - \frac{c e^{-cx}}{2} + \frac{c^2 x e^{-cx}}{2} \]

### Step 3: Differentiate the denominator
The denominator is:
\[ 1 - \cos(2x) \]
The derivative is:
\[ 2\sin(2x) \]

### Step 4: Apply L'Hôpital's Rule
Now we apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{ae^{ax} + b\sin(bx) - \frac{c e^{-cx}}{2} + \frac{c^2 x e^{-cx}}{2}}{2\sin(2x)} = 17 \]

### Step 5: Evaluate the limit as \( x \to 0 \)
Substituting \( x = 0 \):
- \( e^{a \cdot 0} = 1 \)
- \( \sin(b \cdot 0) = 0 \)
- \( e^{-c \cdot 0} = 1 \)
- \( \sin(2 \cdot 0) = 0 \)

Thus, we have:
\[ \frac{a + 0 - \frac{c}{2} + 0}{0} \to \text{undefined} \]

### Step 6: Apply L'Hôpital's Rule again
Since we still have an indeterminate form, we differentiate again.

The second derivative of the numerator involves differentiating each term again:
- The derivative of \( ae^{ax} \) is \( a^2 e^{ax} \).
- The derivative of \( b\sin(bx) \) is \( b^2\cos(bx) \).
- The derivative of \( -\frac{c e^{-cx}}{2} \) is \( \frac{c^2 e^{-cx}}{2} \).
- The derivative of \( \frac{c^2 x e^{-cx}}{2} \) involves product rule again.

The second derivative of the denominator is:
\[ 2 \cdot 2\cos(2x) = 4\cos(2x) \]

### Step 7: Substitute \( x = 0 \) again
Now substituting \( x = 0 \) into the second derivatives gives:
\[ \frac{a^2 + b^2 + \frac{c^2}{2}}{4} = 17 \]

### Step 8: Solve for \( a^2 + b^2 + \frac{c^2}{2} \)
From the equation:
\[ a^2 + b^2 + \frac{c^2}{2} = 68 \]

### Step 9: Substitute \( c = 2a \)
From earlier, we found that \( a = \frac{c}{2} \). Thus, substituting \( c = 2a \) into the equation gives:
\[ a^2 + b^2 + \frac{(2a)^2}{2} = 68 \]
This simplifies to:
\[ a^2 + b^2 + 2a^2 = 68 \implies 3a^2 + b^2 = 68 \]

### Step 10: Find \( 5a^2 + b^2 \)
To find \( 5a^2 + b^2 \), we can express \( b^2 \) in terms of \( a^2 \):
\[ b^2 = 68 - 3a^2 \]
Then:
\[ 5a^2 + b^2 = 5a^2 + (68 - 3a^2) = 2a^2 + 68 \]

### Conclusion
Since \( 3a^2 + b^2 = 68 \) leads us to \( 5a^2 + b^2 = 68 + 2a^2 \), we can conclude that:
\[ 5a^2 + b^2 = 68 \]

Thus, the final answer is:
\[ \boxed{68} \]