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: Rewrite the Limit
We recognize that as \( x \to 0 \), \( \csc x \) approaches infinity, and \( 1 + px + qx^2 \) approaches 1. This gives us the indeterminate form \( 1^\infty \). We can use the logarithmic limit property:

\[ \lim_{x \to 0} (1 + u)^{v} = e^{\lim_{x \to 0} v \cdot \ln(1 + u)} \]

where \( u = px + qx^2 \) and \( v = \csc x \).

### Step 2: Calculate \( \ln(1 + u) \)
Using the Taylor expansion for \( \ln(1 + u) \) around \( u = 0 \):

\[ \ln(1 + u) \approx u - \frac{u^2}{2} + O(u^3) \]

Substituting \( u = px + qx^2 \):

\[ \ln(1 + px + qx^2) \approx px + qx^2 \]

### Step 3: Substitute into the Limit
Now we substitute back into our limit expression:

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

### Step 4: Simplify the Expression
Recall that \( \csc x = \frac{1}{\sin x} \). As \( x \to 0 \), \( \sin x \approx x \), thus \( \csc x \approx \frac{1}{x} \):

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

### Step 5: Set the Limit Equal to 5
We know from the problem statement that this limit equals 5:

\[ p = 5 \]

### Step 6: Determine the Value of \( q \)
Since the limit expression \( qx^2 \) becomes negligible as \( x \to 0 \), \( q \) can be any real number. Thus, \( q \in \mathbb{R} \).

### Conclusion
The values we found are:

\[ p = 5 \quad \text{and} \quad q \in \mathbb{R} \]