If `underset (xrarr0)"lim"(cosx+asinbx)^(1/x)=e^(2)` then the possible values of a and b are

Options

A. a =1 , b = -2
B. `a = 2sqrt2, b = sqrt2`
C. `a=2sqrt2 , b = 1/ sqrt2`
D. a= -2 , b = 1

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) such that:

\[ \lim_{x \to 0} ( \cos x + a \sin(bx) )^{\frac{1}{x}} = e^2 \]

### Step 1: Take the natural logarithm of both sides

Taking the natural logarithm of both sides gives:

\[ \lim_{x \to 0} \frac{1}{x} \ln(\cos x + a \sin(bx)) = 2 \]

### Step 2: Rewrite the limit

We can rewrite the limit as:

\[ \lim_{x \to 0} \frac{\ln(\cos x + a \sin(bx))}{x} = 2 \]

### Step 3: Analyze the expression inside the logarithm

As \( x \to 0 \):

- \( \cos x \to 1 \)
- \( \sin(bx) \to bx \)

Thus, we have:

\[ \cos x + a \sin(bx) \to 1 + 0 = 1 \]

This means we need to evaluate the limit carefully since both the numerator and denominator approach 0.

### Step 4: Apply L'Hôpital's Rule

Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{\ln(\cos x + a \sin(bx))}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}[\ln(\cos x + a \sin(bx))]}{\frac{d}{dx}[x]} \]

### Step 5: Differentiate the numerator

Using the chain rule, we differentiate the numerator:

\[ \frac{d}{dx}[\ln(\cos x + a \sin(bx))] = \frac{1}{\cos x + a \sin(bx)} \left( -\sin x + ab \cos(bx) \right) \]

### Step 6: Evaluate the limit

Now substituting \( x = 0 \):

- \( \cos(0) = 1 \)
- \( \sin(0) = 0 \)
- \( \cos(b \cdot 0) = 1 \)

Thus, we have:

\[ \lim_{x \to 0} \frac{-\sin(0) + ab \cdot 1}{1 + 0} = ab \]

So we need:

\[ ab = 2 \]

### Step 7: Find possible values of \( a \) and \( b \)

The equation \( ab = 2 \) has multiple solutions. For example:

1. \( a = 2, b = 1 \)
2. \( a = 1, b = 2 \)
3. \( a = -2, b = -1 \)
4. \( a = -1, b = -2 \)

### Conclusion

The possible values of \( a \) and \( b \) that satisfy the equation \( ab = 2 \) can be any pairs of real numbers whose product is 2.