If `a.4^(tan x)+ a.4^(-tan x) -2=0` has real solutions, where `0 le x le pi, x ne pi//2`, then `a` lies in the interval

Options

A. `[-1,1]`
B. `[-1,0]`
C. `(0,1]`
D. `[0,oo]`

Answer

To solve the equation \( a \cdot 4^{\tan x} + a \cdot 4^{-\tan x} - 2 = 0 \) for real solutions in the interval \( 0 \leq x \leq \pi, x \neq \frac{\pi}{2} \), we can follow these steps:

### Step 1: Rearranging the Equation
Start by rearranging the equation:
\[ a \cdot 4^{\tan x} + a \cdot 4^{-\tan x} = 2 \]
This can be rewritten as:
\[ a \left( 4^{\tan x} + 4^{-\tan x} \right) = 2 \]

### Step 2: Introducing a New Variable
Let \( y = 4^{\tan x} \). Then, we can express \( 4^{-\tan x} \) as \( \frac{1}{y} \). The equation now becomes:
\[ a \left( y + \frac{1}{y} \right) = 2 \]

### Step 3: Simplifying the Expression
Multiply through by \( y \) (noting \( y > 0 \)):
\[ a(y^2 + 1) = 2y \]
Rearranging gives:
\[ ay^2 - 2y + a = 0 \]

### Step 4: Finding the Discriminant
For the quadratic equation \( ay^2 - 2y + a = 0 \) to have real solutions, the discriminant must be non-negative:
\[ D = b^2 - 4ac = (-2)^2 - 4a \cdot a = 4 - 4a^2 \]
Setting the discriminant greater than or equal to zero:
\[ 4 - 4a^2 \geq 0 \]

### Step 5: Solving the Inequality
This simplifies to:
\[ 1 - a^2 \geq 0 \]
\[ 1 \geq a^2 \]
Taking the square root gives:
\[ -1 \leq a \leq 1 \]

### Step 6: Considering the Domain of \( a \)
Since \( a \) is in the denominator in the original equation, we must also ensure \( a \neq 0 \). Thus, the valid interval for \( a \) is:
\[ 0 < a \leq 1 \]

### Final Answer
Therefore, \( a \) lies in the interval:
\[ (0, 1] \]