Assertion (A) : If `alpha, beta` are the roots of `ax^(2) + bx + c = 0` then the equation whose roots are `(alpha-1)/(alpha), (beta-1)/(beta)` is `c(1-x)^(2)+ b(1-x)+a=0`
Reason (R): If `alpha, beta` are the roots of f(x) = 0 then the equation whose roots are `(alpha-1)/(alpha)` and `(beta-1)/(beta)` is `f((1)/(1-x))=0`

Options

A. Both A, R are true and R explain Assertion
B. Both A, R are true but R does't explain A
C. A is true R is false
D. A is false R is true

Answer

To solve the problem, we need to verify the assertion and reason provided in the question. We will derive the quadratic equation whose roots are \((\alpha - 1)/\alpha\) and \((\beta - 1)/\beta\) given that \(\alpha\) and \(\beta\) are the roots of the equation \(ax^2 + bx + c = 0\).

### Step 1: Identify the roots and their relationships
The roots of the quadratic equation \(ax^2 + bx + c = 0\) are \(\alpha\) and \(\beta\). By Vieta's formulas, we know:
- Sum of roots: \(\alpha + \beta = -\frac{b}{a}\)
- Product of roots: \(\alpha \beta = \frac{c}{a}\)

### Step 2: Find the new roots
We need to find the new roots:
\[ r_1 = \frac{\alpha - 1}{\alpha}, \quad r_2 = \frac{\beta - 1}{\beta} \]

### Step 3: Calculate the sum of the new roots
The sum of the new roots \(r_1 + r_2\) is:
\[ r_1 + r_2 = \frac{\alpha - 1}{\alpha} + \frac{\beta - 1}{\beta} \]
To combine these fractions, we find a common denominator:
\[ = \frac{(\alpha - 1)\beta + (\beta - 1)\alpha}{\alpha\beta} \]
Expanding the numerator:
\[ = \frac{\alpha\beta - \beta + \alpha\beta - \alpha}{\alpha\beta} = \frac{2\alpha\beta - (\alpha + \beta)}{\alpha\beta} \]
Substituting the values from Vieta's formulas:
\[ = \frac{2\left(\frac{c}{a}\right) + \frac{b}{a}}{\frac{c}{a}} = \frac{2c + b}{c} \]

### Step 4: Calculate the product of the new roots
The product of the new roots \(r_1 \cdot r_2\) is:
\[ r_1 \cdot r_2 = \frac{\alpha - 1}{\alpha} \cdot \frac{\beta - 1}{\beta} = \frac{(\alpha - 1)(\beta - 1)}{\alpha\beta} \]
Expanding the numerator:
\[ = \frac{\alpha\beta - \alpha - \beta + 1}{\alpha\beta} = \frac{\frac{c}{a} - \left(-\frac{b}{a}\right) + 1}{\frac{c}{a}} = \frac{c + b + a}{c} \]

### Step 5: Form the quadratic equation
Now that we have the sum and product of the new roots:
- Sum of roots: \(S = \frac{2c + b}{c}\)
- Product of roots: \(P = \frac{a + b + c}{c}\)

The quadratic equation can be expressed as:
\[ x^2 - Sx + P = 0 \]
Substituting \(S\) and \(P\):
\[ x^2 - \frac{2c + b}{c}x + \frac{a + b + c}{c} = 0 \]
Multiplying through by \(c\) to eliminate the fraction:
\[ cx^2 - (2c + b)x + (a + b + c) = 0 \]

### Step 6: Rearranging the equation
Rearranging the equation gives:
\[ c x^2 - (2c + b)x + (a + b + c) = 0 \]

### Step 7: Final form of the quadratic equation
This can be rearranged to match the form given in the assertion:
\[ c(1 - x)^2 + b(1 - x) + a = 0 \]
Thus, the assertion is true.

### Conclusion
Both the assertion and the reason provided are true, but the reason does not explain the assertion directly. Therefore, the answer is that both A and R are true, but R does not explain A.