Question

The value of $a$ for which the sum of the squares of the roots of the equation $x^2 - (a - 2)x - (a - 1) = 0$ assumes the least value is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The Correct Option is B

We are given the quadratic equation: \[ x^2 - (a - 2)x - (a - 1) = 0 \] Let the roots of the equation be \( r_1 \) and \( r_2 \). We need to find the value of \( a \) for which the sum of the squares of the roots is minimized.

Step 1: Sum and product of the roots
From Vieta's formulas for a quadratic equation \( ax^2 + bx + c = 0 \), we know that: \[ r_1 + r_2 = -\frac{b}{a}, \quad r_1 r_2 = \frac{c}{a} \] For the equation \( x^2 - (a - 2)x - (a - 1) = 0 \), comparing with \( x^2 + bx + c = 0 \), we get: \[ r_1 + r_2 = a - 2, \quad r_1 r_2 = -(a - 1) \] 

Step 2: Expression for the sum of the squares of the roots
The sum of the squares of the roots is given by: \[ r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2 \] Substituting the values from Vieta's formulas: \[ r_1^2 + r_2^2 = (a - 2)^2 - 2(-(a - 1)) = (a - 2)^2 + 2(a - 1) \] Expanding this expression: \[ r_1^2 + r_2^2 = (a^2 - 4a + 4) + 2a - 2 = a^2 - 2a + 2 \] 

Step 3: Minimizing the sum of squares of the roots
The function to minimize is: \[ f(a) = a^2 - 2a + 2 \] To find the value of \( a \) that minimizes this function, we take the derivative of \( f(a) \): \[ f'(a) = 2a - 2 \] Setting \( f'(a) = 0 \) to find the critical points: \[ 2a - 2 = 0 \quad \Rightarrow \quad a = 1 \]

 Step 4: Verifying the minimum
To confirm that \( a = 1 \) gives a minimum, we check the second derivative: \[ f''(a) = 2 \] Since \( f''(a) > 0 \), the function is concave up, and \( a = 1 \) corresponds to a minimum.

Answer:

\[ \boxed{1} \]