Let `f :R to [(3)/(4), oo)` be a surjective quadratic function with line of symmetry `2x -1=0 and f (1) =1`
If `g (x)=(f(x)+f(-x))/(2 ) then int (dx)/(sqrt(g (e ^(x))-2))`is equal to:

Options

A. `sec ^(-1) (e ^(-x))+C`
B. `sec ^(-1) (e ^(x))+C`
C. `sin ^(-1) (e ^(-x)) +C`
D. `sin ^(-1) (e ^(x))+C`

Answer

To solve the given problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Define the Quadratic Function
We know that \( f : \mathbb{R} \to \left[\frac{3}{4}, \infty\right) \) is a surjective quadratic function with a line of symmetry given by \( 2x - 1 = 0 \). This implies that the vertex of the quadratic function occurs at \( x = \frac{1}{2} \).

Since the function is surjective and has a minimum value of \( \frac{3}{4} \), we can express \( f(x) \) in vertex form:
\[ f(x) = a\left(x - \frac{1}{2}\right)^2 + \frac{3}{4} \]
where \( a > 0 \).

### Step 2: Use the Condition \( f(1) = 1 \)
We substitute \( x = 1 \) into the function:
\[ f(1) = a\left(1 - \frac{1}{2}\right)^2 + \frac{3}{4} = 1 \]
This simplifies to:
\[ a\left(\frac{1}{2}\right)^2 + \frac{3}{4} = 1 \]
\[ \frac{a}{4} + \frac{3}{4} = 1 \]
Subtracting \( \frac{3}{4} \) from both sides gives:
\[ \frac{a}{4} = \frac{1}{4} \]
Thus, \( a = 1 \).

### Step 3: Write the Function Explicitly
Now we can write the function explicitly:
\[ f(x) = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4} \]
Expanding this, we get:
\[ f(x) = x^2 - x + \frac{1}{4} + \frac{3}{4} = x^2 - x + 1 \]

### Step 4: Find \( f(-x) \)
Next, we calculate \( f(-x) \):
\[ f(-x) = (-x)^2 - (-x) + 1 = x^2 + x + 1 \]

### Step 5: Calculate \( g(x) \)
Now we can find \( g(x) \):
\[ g(x) = \frac{f(x) + f(-x)}{2} = \frac{(x^2 - x + 1) + (x^2 + x + 1)}{2} \]
This simplifies to:
\[ g(x) = \frac{2x^2 + 2}{2} = x^2 + 1 \]

### Step 6: Set Up the Integral
We need to evaluate the integral:
\[ \int \frac{dx}{\sqrt{g(e^x) - 2}} \]
Substituting \( g(e^x) \):
\[ g(e^x) = (e^x)^2 + 1 = e^{2x} + 1 \]
Thus, we have:
\[ g(e^x) - 2 = e^{2x} + 1 - 2 = e^{2x} - 1 \]
The integral now becomes:
\[ \int \frac{dx}{\sqrt{e^{2x} - 1}} \]

### Step 7: Substitute and Simplify
Let \( e^x = t \), then \( dx = \frac{dt}{t} \):
\[ \int \frac{dt/t}{\sqrt{t^2 - 1}} = \int \frac{dt}{t\sqrt{t^2 - 1}} \]
This integral can be solved using the formula:
\[ \int \frac{dt}{t\sqrt{t^2 - 1}} = \sec^{-1}(t) + C \]
Substituting back \( t = e^x \):
\[ \int \frac{dx}{\sqrt{g(e^x) - 2}} = \sec^{-1}(e^x) + C \]

### Final Answer
Thus, the final answer is:
\[ \sec^{-1}(e^x) + C \]