If `lim_(xrarr0)(1+px+qx^(2))^("cosec "x)=2048`, then the value of `(p)/(11)` is equal to (take `ln2=0.69`)

Answer

To solve the limit problem given in the question, we will follow these steps:

### Step 1: Rewrite the Limit
We start with the limit:
\[ \lim_{x \to 0} (1 + px + qx^2)^{\csc x} \]
We know that \(\csc x = \frac{1}{\sin x}\), so we can rewrite the limit as:
\[ \lim_{x \to 0} (1 + px + qx^2)^{\frac{1}{\sin x}} \]

### Step 2: Analyze the Form of the Limit
As \(x\) approaches 0, \(1 + px + qx^2\) approaches 1. Therefore, we have the form \(1^{\infty}\), which is indeterminate. We can use the property:
\[ \lim_{x \to a} (1 + f(x))^{g(x)} = e^{\lim_{x \to a} f(x)g(x)} \]
if \(f(x) \to 0\) and \(g(x) \to \infty\).

### Step 3: Identify \(f(x)\) and \(g(x)\)
Let:
\[ f(x) = px + qx^2 \]
and
\[ g(x) = \csc x = \frac{1}{\sin x} \]
As \(x \to 0\), \(f(x) \to 0\) and \(g(x) \to \infty\).

### Step 4: Calculate the Limit of \(f(x)g(x)\)
We need to find:
\[ \lim_{x \to 0} (px + qx^2) \cdot \frac{1}{\sin x} \]
Using L'Hospital's Rule for the limit of the form \(0 \cdot \infty\), we rewrite it as:
\[ \lim_{x \to 0} \frac{px + qx^2}{\sin x} \]
This is a \(0/0\) form, so we apply L'Hospital's Rule:
\[ \text{Differentiate the numerator: } p + 2qx \]
\[ \text{Differentiate the denominator: } \cos x \]
Thus, we have:
\[ \lim_{x \to 0} \frac{p + 2qx}{\cos x} \]
As \(x \to 0\), \(\cos x \to 1\), so:
\[ \lim_{x \to 0} \frac{p + 2qx}{\cos x} = p \]

### Step 5: Substitute Back into the Limit
Now we substitute back:
\[ \lim_{x \to 0} (1 + px + qx^2)^{\csc x} = e^{p} \]
We know this limit equals 2048, so:
\[ e^{p} = 2048 \]

### Step 6: Express 2048 in Terms of \(e\)
We can express 2048 as a power of 2:
\[ 2048 = 2^{11} \]
Taking the natural logarithm on both sides:
\[ p = \ln(2048) = \ln(2^{11}) = 11 \ln(2) \]

### Step 7: Find \(\frac{p}{11}\)
Now we need to find:
\[ \frac{p}{11} = \frac{11 \ln(2)}{11} = \ln(2) \]
Given that \(\ln(2) \approx 0.69\), we conclude:
\[ \frac{p}{11} = 0.69 \]

### Final Answer
Therefore, the value of \(\frac{p}{11}\) is:
\[ \boxed{0.69} \]