The number of solution(s) of `2cos^(2)(x/2)sin^(2)x=x^(2)+1/x^(2), 0 le x le pi//2`, is/ are-

Options

A. 0
B. 1
C. infinite
D. none of these

Explanation

To find the number of solutions of the equation \( 2\cos^2\left(\frac{x}{2}\right)\sin^2(x) = x^2 + \frac{1}{x^2} \) for \( 0 \leq x \leq \frac{\pi}{2} \), we will analyze both sides of the equation step by step.

### Step 1: Rewrite the left-hand side
We start with the left-hand side of the equation:
\[ y = 2\cos^2\left(\frac{x}{2}\right)\sin^2(x) \]
Using the identity \( \cos^2\left(\frac{x}{2}\right) = \frac{1 + \cos(x)}{2} \), we can rewrite this as:
\[ y = 2 \cdot \frac{1 + \cos(x)}{2} \cdot \sin^2(x) = (1 + \cos(x))\sin^2(x) \]

### Step 2: Analyze the right-hand side
Now, consider the right-hand side:
\[ x^2 + \frac{1}{x^2} \]
This expression is always greater than or equal to 2 for \( x > 0 \) (by AM-GM inequality), and it approaches infinity as \( x \) approaches 0 or infinity.

### Step 3: Find the range of the left-hand side
Next, we need to analyze the range of \( y = (1 + \cos(x))\sin^2(x) \) for \( 0 \leq x \leq \frac{\pi}{2} \):
- At \( x = 0 \): \( y = (1 + \cos(0))\sin^2(0) = 2 \cdot 0 = 0 \)
- At \( x = \frac{\pi}{2} \): \( y = (1 + \cos(\frac{\pi}{2}))\sin^2(\frac{\pi}{2}) = (1 + 0) \cdot 1 = 1 \)

Since \( \cos(x) \) decreases from 1 to 0 and \( \sin^2(x) \) increases from 0 to 1, the product \( (1 + \cos(x))\sin^2(x) \) will vary between 0 and 1.

### Step 4: Find the range of the right-hand side
Now, consider the right-hand side \( x^2 + \frac{1}{x^2} \):
- As previously mentioned, this expression is always greater than or equal to 2 for \( x > 0 \).

### Step 5: Compare the ranges
From the analysis:
- The left-hand side \( y \) ranges from 0 to 1.
- The right-hand side \( x^2 + \frac{1}{x^2} \) is always greater than or equal to 2.

### Conclusion
Since the left-hand side can never reach the values of the right-hand side within the given range \( 0 \leq x \leq \frac{\pi}{2} \), there are no solutions to the equation.

Thus, the number of solutions is:
\[ \text{Number of solutions} = 0 \]