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 find the number of integral values that the expression \( p = 144^{\sin^2 x} + 144^{\cos^2 x} \) can take. Let's break this down step by step.

### Step 1: Rewrite the expression
We can rewrite \( 144 \) as \( 12^2 \). Therefore, we have:
\[ p = 144^{\sin^2 x} + 144^{\cos^2 x} = (12^2)^{\sin^2 x} + (12^2)^{\cos^2 x} \]
Using the property of exponents \( (a^m)^n = a^{mn} \), we can simplify this to:
\[ p = 12^{2\sin^2 x} + 12^{2\cos^2 x} \]

### Step 2: Use trigonometric identities
We know that \( \sin^2 x + \cos^2 x = 1 \). Let \( t = \cos^2 x \). Then \( \sin^2 x = 1 - t \). Thus, we can express \( p \) as:
\[ p = 12^{2(1-t)} + 12^{2t} = 12^{2 - 2t} + 12^{2t} \]

### Step 3: Simplify further
Now, we can rewrite \( p \):
\[ p = \frac{12^2}{12^{2t}} + 12^{2t} = \frac{144}{12^{2t}} + 12^{2t} \]
Let \( u = 12^{2t} \). Then, we can express \( p \) as:
\[ p = \frac{144}{u} + u \]

### Step 4: Find the range of \( u \)
Since \( t = \cos^2 x \) varies from \( 0 \) to \( 1 \), \( u = 12^{2t} \) will vary from:
- When \( t = 0 \), \( u = 12^{0} = 1 \)
- When \( t = 1 \), \( u = 12^{2} = 144 \)

Thus, \( u \) varies from \( 1 \) to \( 144 \).

### Step 5: Analyze the function \( p(u) \)
Now we need to analyze the function \( p(u) = \frac{144}{u} + u \) for \( u \) in the interval \( [1, 144] \).

### Step 6: Find the derivative
To find the minimum and maximum values of \( p(u) \), we take the derivative:
\[ p'(u) = -\frac{144}{u^2} + 1 \]
Setting \( p'(u) = 0 \):
\[ -\frac{144}{u^2} + 1 = 0 \implies 1 = \frac{144}{u^2} \implies u^2 = 144 \implies u = 12 \]
(Note: We only consider \( u = 12 \) since \( u \) must be positive.)

### Step 7: Evaluate \( p(u) \) at critical points
Now we evaluate \( p(u) \) at the endpoints and the critical point:
- At \( u = 1 \):
\[ p(1) = \frac{144}{1} + 1 = 145 \]
- At \( u = 12 \):
\[ p(12) = \frac{144}{12} + 12 = 12 + 12 = 24 \]
- At \( u = 144 \):
\[ p(144) = \frac{144}{144} + 144 = 1 + 144 = 145 \]

### Step 8: Determine the range of \( p \)
From the evaluations:
- Minimum value of \( p \) is \( 24 \) (at \( u = 12 \))
- Maximum value of \( p \) is \( 145 \) (at \( u = 1 \) and \( u = 144 \))

### Step 9: Count the integral values
The integral values of \( p \) range from \( 24 \) to \( 145 \). The number of integral values is given by:
\[ 145 - 24 + 1 = 122 \]

### Final Answer
Thus, the number of integral values that \( p \) can take is \( \boxed{122} \).

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