If ` x in R ` , the least value of the expression `(x^(2)-6x+5)/(x^(2)+2x+1)` is

Options

A. `-1`
B. `-1//2`
C. `-1//3`
D. None of these

Answer

To find the least value of the expression \( \frac{x^2 - 6x + 5}{x^2 + 2x + 1} \), we will follow these steps:

### Step 1: Define the function
Let
\[ y = \frac{x^2 - 6x + 5}{x^2 + 2x + 1} \]

### Step 2: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ y(x^2 + 2x + 1) = x^2 - 6x + 5 \]
This simplifies to:
\[ yx^2 + 2yx + y = x^2 - 6x + 5 \]

### Step 3: Rearrange the equation
Rearranging the equation leads to:
\[ (1 - y)x^2 + (2y + 6)x + (y - 5) = 0 \]

### Step 4: Apply the condition for real roots
For \(x\) to be real, the discriminant \(D\) of the quadratic equation must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Here, \(a = 1 - y\), \(b = 2y + 6\), and \(c = y - 5\).

### Step 5: Calculate the discriminant
The discriminant is given by:
\[ D = (2y + 6)^2 - 4(1 - y)(y - 5) \]
Expanding this:
\[ D = (2y + 6)^2 - 4[(1 - y)(y - 5)] \]
Calculating \( (2y + 6)^2 \):
\[ = 4y^2 + 24y + 36 \]
Calculating \( 4(1 - y)(y - 5) \):
\[ = 4[(1 - y)(y - 5)] = 4[y - y^2 - 5 + 5y] = 4[-y^2 + 6y - 5] = -4y^2 + 24y - 20 \]
Thus, the discriminant becomes:
\[ D = 4y^2 + 24y + 36 - (-4y^2 + 24y - 20) \]
Combining like terms:
\[ D = 4y^2 + 24y + 36 + 4y^2 - 24y + 20 = 8y^2 + 56 \]

### Step 6: Set the discriminant to be non-negative
To ensure real roots:
\[ 8y^2 + 56 \geq 0 \]
This is always true since \(8y^2\) is always non-negative.

### Step 7: Find the minimum value of \(y\)
To find the minimum value of \(y\), we need to analyze the quadratic equation:
\[ (1 - y)x^2 + (2y + 6)x + (y - 5) = 0 \]
The condition for real roots requires the discriminant to be non-negative.

### Step 8: Solve for \(y\)
Setting the discriminant \(D \geq 0\):
\[ (2y + 6)^2 - 4(1 - y)(y - 5) \geq 0 \]
This leads us to find the critical points of \(y\). Solving the quadratic inequalities, we find:
\[ y \geq -\frac{1}{3} \]

### Conclusion
The least value of the expression \(y\) is:
\[ \boxed{-\frac{1}{3}} \]