` f(x)=(1/sqrt(b-a))(((sqrt((b-a)/a))sin2x)/sqrt(1+((sqrt(b-a)/a))sinx)^2)(sqrt(a+btan^2x)` at `x=3pi/4`

Options

A. `sqrt(2)`
B. `-sqrt(2)`
C. `1`
D. non existent

Answer

To solve the given function \( f(x) = \frac{1}{\sqrt{b-a}} \left( \frac{\sqrt{\frac{b-a}{a}} \sin 2x}{\sqrt{1 + \left( \sqrt{\frac{b-a}{a}} \sin x \right)^2}} \right) \sqrt{a + b \tan^2 x} \) at \( x = \frac{3\pi}{4} \), we will follow these steps:

### Step 1: Substitute \( x = \frac{3\pi}{4} \)
First, we need to evaluate \( \sin 2x \) and \( \tan^2 x \) at \( x = \frac{3\pi}{4} \).

\[ \sin 2x = \sin\left(2 \cdot \frac{3\pi}{4}\right) = \sin\left(\frac{3\pi}{2}\right) = -1 \]

\[ \tan x = \tan\left(\frac{3\pi}{4}\right) = -1 \quad \Rightarrow \quad \tan^2 x = 1 \]

### Step 2: Substitute values into the function
Now substitute these values into the function \( f(x) \):

\[ f\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{b-a}} \left( \frac{\sqrt{\frac{b-a}{a}} \cdot (-1)}{\sqrt{1 + \left( \sqrt{\frac{b-a}{a}} \cdot (-1) \right)^2}} \right) \sqrt{a + b \cdot 1} \]

### Step 3: Simplify the expression
Now simplify the expression:

1. The numerator becomes:
\[ -\sqrt{\frac{b-a}{a}} \]

2. The denominator becomes:
\[ \sqrt{1 + \left( \sqrt{\frac{b-a}{a}} \right)^2} = \sqrt{1 + \frac{b-a}{a}} = \sqrt{\frac{a + b}{a}} = \frac{\sqrt{a + b}}{\sqrt{a}} \]

Thus, the function simplifies to:

\[ f\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{b-a}} \cdot \frac{-\sqrt{\frac{b-a}{a}}}{\frac{\sqrt{a + b}}{\sqrt{a}}} \cdot \sqrt{a + b} \]

### Step 4: Final simplification
Now, we can simplify further:

\[ f\left(\frac{3\pi}{4}\right) = \frac{-\sqrt{b-a}}{\sqrt{a} \sqrt{b-a}} \cdot \sqrt{a + b} = \frac{-\sqrt{a + b}}{\sqrt{a}} \]

### Final Result
Thus, the final result for \( f\left(\frac{3\pi}{4}\right) \) is:

\[ f\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{a + b}}{\sqrt{a}} \]

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