Question

If x = 1 + a², y = 1 + b², z = 1 + c² and (a + b + c)² = 0, then ab + bc + ca =

• A. [3 – (x + y + z)]/2
• B. 1 – (x + y + z)/2
• C. 1 + (x + y + z)/2
• D. 1 – (x + y + z)

Answer 1

The correct option is A [3 – (x + y + z)]/2

Sol:- a = 0, b = 1, c = -1. x = 1, y = 2, c = 2. ab + bc + ca = -1. Only option a) satisfies

Alternatively: -

Given

x = 1 + a², => a² = x - 1

y = 1 + b², => b² = y - 1

z = 1 + c² => c² = z - 1 and (a + b + c)² = 0

As we know, (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

Put the given values.

0 = (x - 1) + (y - 1) + (z - 1) + 2(ab + bc + ca)

(ab + bc + ca) = [3 – (x + y + z)]/2