if f(x) `=2e^(x) -ae^(-x) +(2a +1) x-3` monotonically increases for `AA x in R` then the minimum value of 'a' is

Options

A. 2
B. 1
C. 0
D. `-1`

Explanation

To solve the problem, we need to determine the minimum value of 'a' such that the function \( f(x) = 2e^x - ae^{-x} + (2a + 1)x - 3 \) is monotonically increasing for all \( x \in \mathbb{R} \).

### Step 1: Find the derivative of \( f(x) \)
To check for monotonicity, we first need to find the derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(2e^x) - \frac{d}{dx}(ae^{-x}) + \frac{d}{dx}((2a + 1)x) - \frac{d}{dx}(3) \]

Calculating each term gives:

\[ f'(x) = 2e^x + ae^{-x} + (2a + 1) \]

### Step 2: Set the derivative to be greater than or equal to zero
For \( f(x) \) to be monotonically increasing, we need \( f'(x) \geq 0 \) for all \( x \).

\[ f'(x) = 2e^x + ae^{-x} + (2a + 1) \geq 0 \]

### Step 3: Rewrite the derivative in terms of \( y = e^x \)
Since \( e^x > 0 \) for all \( x \), we can substitute \( y = e^x \) (where \( y > 0 \)):

\[ f'(x) = 2y + \frac{a}{y} + (2a + 1) \geq 0 \]

### Step 4: Multiply through by \( y \) (since \( y > 0 \))
Multiplying the entire inequality by \( y \):

\[ 2y^2 + a + (2a + 1)y \geq 0 \]

Rearranging gives:

\[ 2y^2 + (2a + 1)y + a \geq 0 \]

### Step 5: Analyze the quadratic inequality
This is a quadratic in \( y \). For this quadratic to be non-negative for all \( y > 0 \), the discriminant must be less than or equal to zero.

The discriminant \( D \) of the quadratic \( 2y^2 + (2a + 1)y + a \) is given by:

\[ D = (2a + 1)^2 - 4 \cdot 2 \cdot a \]

Setting the discriminant less than or equal to zero:

\[ (2a + 1)^2 - 8a \leq 0 \]

### Step 6: Solve the discriminant inequality
Expanding the left-hand side:

\[ 4a^2 + 4a + 1 - 8a \leq 0 \]

This simplifies to:

\[ 4a^2 - 4a + 1 \leq 0 \]

### Step 7: Factor the quadratic
The quadratic can be factored or solved using the quadratic formula:

\[ 4a^2 - 4a + 1 = 0 \]

Using the quadratic formula \( a = \frac{-b \pm \sqrt{D}}{2a} \):

\[ a = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4} = \frac{4 \pm \sqrt{0}}{8} = \frac{4}{8} = \frac{1}{2} \]

### Step 8: Determine the minimum value of 'a'
Since the quadratic \( 4a^2 - 4a + 1 \) opens upwards (the coefficient of \( a^2 \) is positive), it is non-negative for \( a = \frac{1}{2} \). Therefore, the minimum value of \( a \) such that \( f(x) \) is monotonically increasing is:

\[ \text{Minimum value of } a = \frac{1}{2} \]

### Final Answer
The minimum value of \( a \) is \( \frac{1}{2} \).

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