Prove that the function `f(x) = a sqrt(x - 1) + b sqrt(2x - 1)-sqrt(2x^(2) - 3x + 1)`, where a + 2b = 2 and `a, b in R` always has a root in `(1, 5) AA b in R`

Answer

To prove that the function \( f(x) = a \sqrt{x - 1} + b \sqrt{2x - 1} - \sqrt{2x^2 - 3x + 1} \), where \( a + 2b = 2 \) and \( a, b \in \mathbb{R} \), always has a root in the interval \( (1, 5) \), we will evaluate the function at the endpoints of the interval and use the Intermediate Value Theorem.

### Step 1: Evaluate \( f(1) \)

Substituting \( x = 1 \) into the function:

\[ f(1) = a \sqrt{1 - 1} + b \sqrt{2 \cdot 1 - 1} - \sqrt{2 \cdot 1^2 - 3 \cdot 1 + 1} \]

Calculating each term:

\[ = a \cdot 0 + b \cdot 1 - \sqrt{2 - 3 + 1} \]
\[ = 0 + b - \sqrt{0} \]
\[ = b \]

So, we have:

\[ f(1) = b \]

### Step 2: Evaluate \( f(5) \)

Now substituting \( x = 5 \):

\[ f(5) = a \sqrt{5 - 1} + b \sqrt{2 \cdot 5 - 1} - \sqrt{2 \cdot 5^2 - 3 \cdot 5 + 1} \]

Calculating each term:

\[ = a \sqrt{4} + b \sqrt{10 - 1} - \sqrt{50 - 15 + 1} \]
\[ = 2a + b \sqrt{9} - \sqrt{36} \]
\[ = 2a + 3b - 6 \]

So, we have:

\[ f(5) = 2a + 3b - 6 \]

### Step 3: Use the condition \( a + 2b = 2 \)

From the condition \( a + 2b = 2 \), we can express \( a \) in terms of \( b \):

\[ a = 2 - 2b \]

### Step 4: Substitute \( a \) into \( f(5) \)

Substituting \( a \) into \( f(5) \):

\[ f(5) = 2(2 - 2b) + 3b - 6 \]
\[ = 4 - 4b + 3b - 6 \]
\[ = -b - 2 \]

### Step 5: Analyze the signs of \( f(1) \) and \( f(5) \)

Now we have:

\[ f(1) = b \]
\[ f(5) = -b - 2 \]

### Step 6: Determine the signs

1. If \( b > 0 \):
- \( f(1) = b > 0 \)
- \( f(5) = -b - 2 < 0 \)

2. If \( b < 0 \):
- \( f(1) = b < 0 \)
- \( f(5) = -b - 2 > 0 \) (since \( -b > 0 \))

### Step 7: Apply the Intermediate Value Theorem

In both cases, \( f(1) \) and \( f(5) \) have opposite signs. By the Intermediate Value Theorem, since \( f(x) \) is continuous on the interval \( [1, 5] \), there exists at least one \( c \in (1, 5) \) such that \( f(c) = 0 \).

### Conclusion

Thus, we have proved that the function \( f(x) \) always has a root in the interval \( (1, 5) \).

---