If `f(x)={((1+|cosx|)^(p/(|cos x|)), , 0 lt x lt (pi)/2),(q, , x=(pi)/2),(e^([cotl(x-(pi)/2)//cot m (x- (pi)/2)]), , (pi)/2 lt x lt pi):}`
is a continuous function on `(0,pi)` then the value of p and q are respectively?

Options

A. `m/l , e^(m//l)`
B. `e(m//l), m/l`
C. `ml, e^(ml)`
D. `e^(m//l), ml`

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous on the interval \( (0, \pi) \). The function is defined piecewise, and we need to check the continuity at the critical point \( x = \frac{\pi}{2} \).

### Step 1: Analyze the left-hand limit as \( x \) approaches \( \frac{\pi}{2} \)

For \( 0 < x < \frac{\pi}{2} \):
\[ f(x) = (1 + |\cos x|)^{\frac{p}{|\cos x|}} \]
As \( x \) approaches \( \frac{\pi}{2} \), \( |\cos x| \) approaches \( 0 \). This leads to an indeterminate form of \( 1^{\infty} \).

To resolve this, we can rewrite the expression using the exponential function:
\[ f(x) = e^{\frac{p}{|\cos x|} \ln(1 + |\cos x|)} \]
Using the limit property \( \lim_{x \to 0} \frac{\ln(1+x)}{x} = 1 \), we can express:
\[ \ln(1 + |\cos x|) \sim |\cos x| \quad \text{as } x \to \frac{\pi}{2} \]
Thus, we have:
\[ f(x) \sim e^{\frac{p}{|\cos x|} |\cos x|} = e^{p} \]
So, the left-hand limit as \( x \to \frac{\pi}{2}^- \) is:
\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = e^{p} \]

### Step 2: Evaluate the function at \( x = \frac{\pi}{2} \)

For \( x = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = q \]

### Step 3: Analyze the right-hand limit as \( x \) approaches \( \frac{\pi}{2} \)

For \( \frac{\pi}{2} < x < \pi \):
\[ f(x) = e^{\frac{\cot(l(x - \frac{\pi}{2})}{\cot(m(x - \frac{\pi}{2})}} \]
As \( x \to \frac{\pi}{2}^+ \), both \( \cot(l(x - \frac{\pi}{2}) \) and \( \cot(m(x - \frac{\pi}{2}) \) approach \( \infty \), leading to another indeterminate form.

Taking the logarithm:
\[ \ln(f(x)) = \frac{\cot(l(x - \frac{\pi}{2})}{\cot(m(x - \frac{\pi}{2})} \]
Using the limit:
\[ \lim_{x \to \frac{\pi}{2}^+} \frac{\tan(m(x - \frac{\pi}{2}))}{\tan(l(x - \frac{\pi}{2}))} = \frac{m}{l} \]
Thus, we find:
\[ \lim_{x \to \frac{\pi}{2}^+} f(x) = e^{\frac{m}{l}} \]

### Step 4: Set the limits equal for continuity

For \( f(x) \) to be continuous at \( x = \frac{\pi}{2} \):
\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}^+} f(x) = f\left(\frac{\pi}{2}\right) \]
This gives us:
\[ e^{p} = q = e^{\frac{m}{l}} \]

### Step 5: Solve for \( p \) and \( q \)

From the equations:
1. \( q = e^{p} \)
2. \( q = e^{\frac{m}{l}} \)

Setting these equal:
\[ e^{p} = e^{\frac{m}{l}} \implies p = \frac{m}{l} \]
And substituting back:
\[ q = e^{\frac{m}{l}} \]

### Final Result

Thus, the values of \( p \) and \( q \) are:
\[ p = \frac{m}{l}, \quad q = e^{\frac{m}{l}} \]