Let `x^2-2(a - 1)x + a - 1 = 0 (a in R)` be a quadratic equation, then find the value of 'a' for which
(a) Both the roots are positive
(b)Both the roots are negative
(c) Both the roots are opposite in sign.
(d) Both the roots are greater than 1.
(e)Both the roots are smaller than 1.
(f) One root is small than 1 and the other root is greater than 1.

Answer

To solve the quadratic equation \(x^2 - 2(a - 1)x + (a - 1) = 0\) for different conditions of the roots, we will analyze the coefficients and the discriminant of the quadratic equation.

### Step 1: Identify the coefficients
The given quadratic equation can be rewritten as:
\[ x^2 - 2(a - 1)x + (a - 1) = 0 \]
Here, the coefficients are:
- \(A = 1\)
- \(B = -2(a - 1)\)
- \(C = a - 1\)

### Step 2: Calculate the discriminant
The discriminant \(D\) of a quadratic equation \(Ax^2 + Bx + C = 0\) is given by:
\[ D = B^2 - 4AC \]
Substituting the values of \(A\), \(B\), and \(C\):
\[ D = [-2(a - 1)]^2 - 4(1)(a - 1) \]
\[ D = 4(a - 1)^2 - 4(a - 1) \]
\[ D = 4[(a - 1)^2 - (a - 1)] \]
\[ D = 4[(a - 1)(a - 2)] \]

### Step 3: Analyze conditions for the roots

#### (a) Both roots are positive
For both roots to be positive:
1. The sum of the roots \( \alpha + \beta = 2(a - 1) > 0 \) implies \( a - 1 > 0 \) or \( a > 1 \).
2. The product of the roots \( \alpha \beta = a - 1 > 0 \) implies \( a - 1 > 0 \) or \( a > 1 \).
3. The discriminant must be non-negative: \( D \geq 0 \) gives \( (a - 1)(a - 2) \geq 0 \).

From \( (a - 1)(a - 2) \geq 0 \), we find \( a \leq 1 \) or \( a \geq 2 \). Combining this with \( a > 1 \), we conclude:
\[ \text{Both roots are positive for } a \geq 2. \]

#### (b) Both roots are negative
For both roots to be negative:
1. The sum of the roots \( 2(a - 1) < 0 \) implies \( a - 1 < 0 \) or \( a < 1 \).
2. The product of the roots \( a - 1 > 0 \) implies \( a > 1 \).
This results in a contradiction, thus:
\[ \text{No values of } a \text{ make both roots negative.} \]

#### (c) Both roots are opposite in sign
For roots to be opposite in sign:
1. The product of the roots \( a - 1 < 0 \) implies \( a < 1 \).
2. The discriminant must be non-negative: \( D \geq 0 \) gives \( (a - 1)(a - 2) \geq 0 \).

From \( (a - 1)(a - 2) \geq 0 \), we find \( a \leq 1 \) or \( a \geq 2 \). Combining with \( a < 1 \), we conclude:
\[ \text{Both roots are opposite in sign for } a < 1. \]

#### (d) Both roots are greater than 1
For both roots to be greater than 1:
1. The sum of the roots \( 2(a - 1) > 2 \) implies \( a - 1 > 1 \) or \( a > 2 \).
2. The product of the roots \( a - 1 > 1 \) implies \( a > 2 \).
3. The discriminant must be non-negative: \( D \geq 0 \) gives \( (a - 1)(a - 2) \geq 0 \).

Thus, we conclude:
\[ \text{Both roots are greater than 1 for } a > 2. \]

#### (e) Both roots are smaller than 1
For both roots to be smaller than 1:
1. The sum of the roots \( 2(a - 1) < 2 \) implies \( a - 1 < 1 \) or \( a < 2 \).
2. The product of the roots \( a - 1 < 1 \) implies \( a < 2 \).
3. The discriminant must be non-negative: \( D \geq 0 \) gives \( (a - 1)(a - 2) \geq 0 \).

Thus, we conclude:
\[ \text{Both roots are smaller than 1 for } a < 1 \text{ or } a \geq 2. \]

#### (f) One root is smaller than 1 and the other root is greater than 1
For one root to be smaller than 1 and the other greater than 1:
1. The discriminant must be positive: \( D > 0 \) gives \( (a - 1)(a - 2) > 0 \) which implies \( a < 1 \) or \( a > 2 \).
2. The sum of the roots \( 2(a - 1) > 0 \) implies \( a > 1 \).

Thus, we conclude:
\[ \text{One root is smaller than 1 and the other greater than 1 for } a > 2. \]

### Summary of Results:
- (a) \( a \geq 2 \)
- (b) No values of \( a \)
- (c) \( a < 1 \)
- (d) \( a > 2 \)
- (e) \( a < 1 \) or \( a \geq 2 \)
- (f) \( a > 2 \)