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To solve this problem, we need to find the equation of the line perpendicular to the tangent line of the function \( y = f(x) = 2 \sqrt{x^2 + 3} \) at \( x = 1 \).

Let's go through the steps:

### Step 1: Differentiate \( f(x) \)
To find the slope of the tangent line, we first need \( f'(x) \), the derivative of \( f(x) \).
Given:
\[
f(x) = 2 \sqrt{x^2 + 3}
\]
We can rewrite \( f(x) \) as:
\[
f(x) = 2 (x^2 + 3)^{1/2}
\]
Using the chain rule:
\[
f'(x) = 2 \cdot \frac{1}{2} (x^2 + 3)^{-1/2} \cdot 2x
\]
Simplifying, we get:
\[
f'(x) = \frac{2x}{\sqrt{x^2 + 3}}
\]

### Step 2: Find \( f'(1) \)
Now, substitute \( x = 1 \) to find the slope of the tangent line at that point:
\[
f'(1) = \frac{2 \cdot 1}{\sqrt{1^2 + 3}} = \frac{2}{\sqrt{4}} = \frac{2}{2} = 1
\]
So, the slope of the tangent line at \( x = 1 \) is 1.

### Step 3: Determine the Perpendicular Slope
The slope of a line perpendicular to a line with slope \( m \) is \( -\frac{1}{m} \). Since the slope of the tangent line is 1, the perpendicular slope is:
\[
-1
\]

### Step 4: Find the Point on the Curve at \( x = 1 \)
To find the point of tangency, calculate \( f(1) \):
\[
f(1) = 2 \sqrt{1^2 + 3} = 2 \sqrt{4} = 2 \cdot 2 = 4
\]
So, the point of tangency is \( (1, 4) \).

### Step 5: Write the Equation of the Perpendicular Line
Now, we have a point \( (1, 4) \) and a slope of \( -1 \). Using the point-slope form of the line equation \( y - y_1 = m(x - x_1) \), we get:
\[
y - 4 = -1(x - 1)
\]
Expanding this:
\[
y - 4 = -x + 1
\]
\[
y = -x + 5
\]

### Conclusion
The equation of the perpendicular line is:
\[
\boxed{y = -x + 5}
\]
Thus, the correct answer is **(a)**.

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Would you like further details or have any questions?

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Here are some related questions to explore further:

1. How would you find the tangent line equation instead of the perpendicular line?
2. What are the steps for differentiating more complex functions?
3. How does the chain rule apply in different scenarios?
4. How would this problem change if the function was modified to include higher powers of \( x \)?
5. How do you determine if two lines are parallel or perpendicular given their equations?

**Tip:** Remember, perpendicular lines have slopes that are negative reciprocals of each other.

The equation of the perpendicular to the tangent line of the function y = f(x) = 2√(x^2 + 3) at x = 1 is:

Step-by-step solution to find the perpendicular line to the tangent of y = 2√(x^2 + 3) at x = 1. This calculus problem involves finding derivatives, applying the chain rule, and using point-slope form for line equations. Suitable for Grades 11-12.

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