Let `f: R-> R` be a continuous function defined by `f(x)""=1/(e^x+2e^(-x))` . Statement-1: `f(c)""=1/3,` for some `c in R` . Statement-2: `0""<""f(x)lt=1/(2sqrt(2)),` for all `x in R` . (1) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is false (3) Statement-1 is false, Statement-2 is true (4) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1

Options

A. statement 1 is false statement 2 is true
B. satement 1 si true , statement 2 is true statement 2 is a correct explanation for statement 1
C. statement 1 is true statement 2 is true statement 2 is not a correct explanation for statement 2
D. statement 1 is true , statement 2 is false

Explanation

To solve the problem, we need to analyze the function \( f(x) = \frac{1}{e^x + 2e^{-x}} \) and evaluate the two statements given.

### Step 1: Analyze the Function
The function is defined as:
\[ f(x) = \frac{1}{e^x + 2e^{-x}} \]
We need to determine the range of this function.

### Step 2: Find the Limits of \( f(x) \)
First, let's find the limit of \( f(x) \) as \( x \) approaches positive and negative infinity.

- As \( x \to +\infty \):
\[ f(x) = \frac{1}{e^x + 2e^{-x}} \to \frac{1}{\infty} = 0 \]

- As \( x \to -\infty \):
\[ f(x) = \frac{1}{e^x + 2e^{-x}} \to \frac{1}{0 + 2 \cdot \infty} = 0 \]

### Step 3: Find Critical Points
To find the maximum value of \( f(x) \), we can differentiate \( f(x) \) and find its critical points.

Using the quotient rule:
\[ f'(x) = \frac{(0)(e^x + 2e^{-x}) - (1)(e^x - 2e^{-x})}{(e^x + 2e^{-x})^2} \]
Setting the numerator to zero for critical points:
\[ e^x - 2e^{-x} = 0 \implies e^{2x} = 2 \implies x = \frac{1}{2} \ln(2) \]

### Step 4: Evaluate \( f(x) \) at Critical Points
Now we evaluate \( f \) at the critical point:
\[ f\left(\frac{1}{2} \ln(2)\right) = \frac{1}{e^{\frac{1}{2} \ln(2)} + 2e^{-\frac{1}{2} \ln(2)}} \]
Calculating this gives:
\[ f\left(\frac{1}{2} \ln(2)\right) = \frac{1}{\sqrt{2} + 2 \cdot \frac{1}{\sqrt{2}}} = \frac{1}{\sqrt{2} + \sqrt{2}} = \frac{1}{2\sqrt{2}} \]

### Step 5: Establish the Range of \( f(x) \)
From the limits and the maximum value found:
- \( f(x) \) approaches 0 as \( x \to \pm\infty \)
- The maximum value is \( \frac{1}{2\sqrt{2}} \)

Thus, the range of \( f(x) \) is:
\[ (0, \frac{1}{2\sqrt{2}}] \]

### Step 6: Evaluate the Statements
1. **Statement 1**: \( f(c) = \frac{1}{3} \) for some \( c \in \mathbb{R} \).
- Since \( \frac{1}{3} < \frac{1}{2\sqrt{2}} \) and \( f(x) \) is continuous, by the Intermediate Value Theorem, there exists some \( c \) such that \( f(c) = \frac{1}{3} \). Thus, Statement 1 is **true**.

2. **Statement 2**: \( 0 < f(x) \leq \frac{1}{2\sqrt{2}} \) for all \( x \in \mathbb{R} \).
- We have established that \( f(x) \) is always positive and reaches a maximum of \( \frac{1}{2\sqrt{2}} \). Thus, Statement 2 is also **true**.

### Conclusion
Both statements are true, and Statement 2 provides a correct explanation for Statement 1. Therefore, the correct option is:
**(4) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1.**