Statement:1 The number of real solution of the equation `sinx = 4^(x) + 4^(-x)` is zero. And Statement-2: `|sinx| lt 1 AA x in R`.

Options

A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for statement-1
B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct expanation for statement-1
C. Statement-1 is True, Statement-2 is false
D. Statement-1 is False, Statement-2 is true

Answer

To solve the problem, we will analyze the two statements given and determine their validity step by step.

### Step 1: Analyze Statement 1
The first statement claims that the equation \( \sin x = 4^x + 4^{-x} \) has zero real solutions.

**Solution:**
1. **Understanding the Functions:**
- The function \( \sin x \) oscillates between -1 and 1 for all \( x \in \mathbb{R} \).
- The function \( 4^x + 4^{-x} \) can be rewritten using the property of exponents:
\[ 4^x + 4^{-x} = e^{x \ln 4} + e^{-x \ln 4} \]
- This function is always positive and reaches its minimum value when \( x = 0 \):
\[ 4^0 + 4^0 = 1 + 1 = 2 \]

2. **Applying the AM-GM Inequality:**
- By the AM-GM inequality:
\[ \frac{4^x + 4^{-x}}{2} \geq \sqrt{4^x \cdot 4^{-x}} = \sqrt{1} = 1 \]
- Therefore, \( 4^x + 4^{-x} \geq 2 \).

3. **Comparing Ranges:**
- The range of \( \sin x \) is \([-1, 1]\).
- The minimum value of \( 4^x + 4^{-x} \) is 2, which is greater than the maximum value of \( \sin x \).
- Hence, \( \sin x \) can never equal \( 4^x + 4^{-x} \).

**Conclusion for Statement 1:**
The statement is true; there are no real solutions to the equation \( \sin x = 4^x + 4^{-x} \).

### Step 2: Analyze Statement 2
The second statement claims that \( |\sin x| < 1 \) for all \( x \in \mathbb{R} \).

**Solution:**
1. **Understanding the Range of Sine Function:**
- The sine function oscillates between -1 and 1.
- Therefore, \( |\sin x| \) will always be less than or equal to 1.

2. **Verifying the Statement:**
- However, \( |\sin x| = 1 \) at specific points (e.g., \( x = \frac{\pi}{2} + n\pi \) for \( n \in \mathbb{Z} \)).
- Thus, the statement \( |\sin x| < 1 \) is false because there are values of \( x \) where \( |\sin x| = 1 \).

**Conclusion for Statement 2:**
The statement is false; \( |\sin x| \) can equal 1 for certain values of \( x \).

### Final Conclusion
- Statement 1 is true (there are no solutions).
- Statement 2 is false (the absolute value of sine can equal 1).