Question

If \( \alpha, \beta \) are the zeroes of the polynomial :
\( f(x) = x^2 - p(x+1) - C \) such that \( (\alpha+1)(\beta+1) = 0 \) then C=

• A. 1
• B. 0
• C. -1
• D. 2

Hint:

1. Expand \(f(x)\): \(x^2 - p(x+1) - C = x^2 - px - p - C = x^2 - px - (p+C)\). 2. From Vieta's formulas for \(ax^2+bx+c'\): Sum of roots: \(\alpha+\beta = -(\text{coeff. of } x) / (\text{coeff. of } x^2) = -(-p)/1 = p\). Product of roots: \(\alpha\beta = (\text{constant term}) / (\text{coeff. of } x^2) = -(p+C)/1 = -(p+C)\). 3. Expand given condition: \((\alpha+1)(\beta+1) = \alpha\beta + \alpha + \beta + 1 = 0\). 4. Substitute: \(-(p+C) + p + 1 = 0\). 5. Simplify: \(-p - C + p + 1 = 0 \implies -C + 1 = 0 \implies C = 1\).

Answer 1

The Correct Option is A

Concept: For a quadratic polynomial \(ax^2+bx+c\), if \(\alpha\) and \(\beta\) are its zeroes, then:
Sum of zeroes: \(\alpha + \beta = -b/a\)
Product of zeroes: \(\alpha\beta = c/a\) Step 1: Rewrite the polynomial \(f(x)\) in standard quadratic form \(ax^2+bx+c'\) The given polynomial is \( f(x) = x^2 - p(x+1) - C \). Expand and rearrange: \( f(x) = x^2 - px - p - C \) \( f(x) = x^2 - px - (p+C) \) Comparing this with \(ax^2+bx+c'\): Here, \(a=1\), \(b=-p\), and \(c' = -(p+C)\). Step 2: Use Vieta's formulas for sum and product of zeroes For \(f(x) = x^2 - px - (p+C)\): Sum of zeroes: \(\alpha + \beta = -(-p)/1 = p\). Product of zeroes: \(\alpha\beta = -(p+C)/1 = -(p+C)\). Step 3: Use the given condition \((\alpha+1)(\beta+1) = 0\) Expand the given condition: \( (\alpha+1)(\beta+1) = \alpha\beta + \alpha(1) + 1(\beta) + (1)(1) = 0 \) \( \alpha\beta + \alpha + \beta + 1 = 0 \) This can be written as: \( (\alpha\beta) + (\alpha+\beta) + 1 = 0 \) Step 4: Substitute the expressions for sum and product of zeroes into this equation We have: \(\alpha\beta = -(p+C)\) \(\alpha+\beta = p\) Substitute these into \( (\alpha\beta) + (\alpha+\beta) + 1 = 0 \): \[ -(p+C) + (p) + 1 = 0 \] Simplify the equation: \[ -p - C + p + 1 = 0 \] The terms \(-p\) and \(+p\) cancel out: \[ -C + 1 = 0 \] Step 5: Solve for C \[ 1 = C \] So, \( C = 1 \).