If `{:(lim),(xrarr0):}(1+px+qx^2)^("cosec"x)=e^5` ,then

Options

A. `p=5,qinR`
B. `p=5,qgtR`
C. `p=5,qinR`
D. `q=5,p=0`

Answer

To solve the problem, we need to evaluate the limit:

\[ \lim_{x \to 0} (1 + px + qx^2)^{\csc x} = e^5 \]

### Step 1: Identify the form of the limit
As \( x \to 0 \), we have:
- \( \csc x \to \infty \)
- \( 1 + px + qx^2 \to 1 \)

This gives us the indeterminate form \( 1^\infty \).

### Step 2: Rewrite the limit using the exponential function
We can rewrite the limit in the form suitable for applying the exponential limit property:

\[ \lim_{x \to 0} (1 + px + qx^2)^{\csc x} = e^{\lim_{x \to 0} \csc x \cdot \ln(1 + px + qx^2)} \]

### Step 3: Simplify the logarithm
Using the Taylor expansion for \( \ln(1 + u) \) where \( u \) is small, we have:

\[ \ln(1 + px + qx^2) \approx px + qx^2 \quad \text{as } x \to 0 \]

### Step 4: Substitute back into the limit
Now substituting this back into our limit, we get:

\[ \lim_{x \to 0} \csc x \cdot \ln(1 + px + qx^2) \approx \lim_{x \to 0} \csc x \cdot (px + qx^2) \]

### Step 5: Express \( \csc x \)
Recall that \( \csc x = \frac{1}{\sin x} \). Near \( x = 0 \), we can use the approximation \( \sin x \approx x \):

\[ \csc x \approx \frac{1}{x} \]

Thus, we can rewrite our limit as:

\[ \lim_{x \to 0} \frac{px + qx^2}{x} = \lim_{x \to 0} (p + qx) = p \]

### Step 6: Set the limit equal to \( e^5 \)
From the original equation, we have:

\[ e^{p} = e^5 \]

This implies:

\[ p = 5 \]

### Step 7: Determine the value of \( q \)
Since the term involving \( q \) vanishes as \( x \to 0 \) (because it is multiplied by \( x \)), \( q \) can take any real value. Thus, we conclude:

\[ q \in \mathbb{R} \]

### Final Answer
The values are:
- \( p = 5 \)
- \( q \) can be any real number.