Let `p = 144 ^(sin^(2)x) + 144^( cos^(2)x),` then the number of integral values of such p's can take are

Options

A. 124
B. 123
C. 144
D. 122

Answer

To solve the problem, we need to determine the number of integral values that the expression \( p = 144^{\sin^2 x} + 144^{\cos^2 x} \) can take.

### Step-by-Step Solution:

1. **Rewrite the Expression:**
We start with the expression:
\[ p = 144^{\sin^2 x} + 144^{\cos^2 x} \]
We can express \( 144 \) as \( 12^2 \):
\[ p = (12^2)^{\sin^2 x} + (12^2)^{\cos^2 x} = 12^{2\sin^2 x} + 12^{2\cos^2 x} \]

2. **Introduce a Substitution:**
Let \( t = \cos(2x) \). Then, we can express \( \sin^2 x \) and \( \cos^2 x \) in terms of \( t \):
\[ \sin^2 x = \frac{1 - t}{2}, \quad \cos^2 x = \frac{1 + t}{2} \]
Substituting these into \( p \):
\[ p = 12^{1 - t} + 12^{1 + t} \]

3. **Simplify the Expression:**
We can factor out \( 12 \):
\[ p = 12 \left( 12^{-t} + 12^{t} \right) \]
This can be rewritten using the property of exponents:
\[ p = 12 \left( \frac{1}{12^t} + 12^t \right) \]

4. **Define a New Variable:**
Let \( u = 12^t \). Then, we have:
\[ p = 12 \left( \frac{1}{u} + u \right) \]
The function \( f(u) = \frac{1}{u} + u \) is defined for \( u > 0 \).

5. **Find the Range of \( u \):**
Since \( t = \cos(2x) \) varies from -1 to 1, we find the range of \( u \):
\[ u = 12^{-1} \text{ to } 12^{1} \Rightarrow \frac{1}{12} \text{ to } 12 \]

6. **Analyze the Function \( f(u) \):**
We need to find the minimum and maximum values of \( f(u) \):
\[ f(u) = \frac{1}{u} + u \]
To find critical points, we differentiate:
\[ f'(u) = -\frac{1}{u^2} + 1 \]
Setting \( f'(u) = 0 \):
\[ 1 = \frac{1}{u^2} \Rightarrow u^2 = 1 \Rightarrow u = 1 \]

7. **Evaluate \( f(u) \) at Critical Points and Endpoints:**
- At \( u = \frac{1}{12} \):
\[ f\left(\frac{1}{12}\right) = 12 + \frac{1}{12} = 12.0833 \]
- At \( u = 1 \):
\[ f(1) = 1 + 1 = 2 \]
- At \( u = 12 \):
\[ f(12) = \frac{1}{12} + 12 = 12.0833 \]

8. **Determine the Range of \( p \):**
The minimum value of \( f(u) \) occurs at \( u = 1 \) and is \( 2 \). The maximum value occurs at \( u = \frac{1}{12} \) or \( u = 12 \) and is \( 12.0833 \).

Therefore, the range of \( p \) is:
\[ p \in [24, 24.0833] \]

9. **Count the Integral Values:**
The integral values \( p \) can take are \( 24 \) (since \( 24.0833 \) does not reach 25). Thus, the number of integral values is:
\[ \text{Number of integral values} = 1 \]

### Final Answer:
The number of integral values that \( p \) can take is **1**.