CAT 2002 — QA Question 3

x + y + z = 5                    …(i)
xy + yz + zx = 3               …(ii)

From equations (i) and (ii), we get,
(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
∴ x² + y² + z² = 19         …(iii)

When x² is maximum, y² + z² is minimum,

∵ y² + z² ≥ 0 or least value of y² + z² = 0

∴ y = z = 0, This is not possible as it does not satisfy equation (ii)

∴ y ≠ 0 and z ≠ 0

∴ y² + z² will be minimum when y = z [Using AM ≥ GM rule]

Substituting z = y in (i) and (ii), we get,

x = 5 − 2y                       …(iii)

xy + y² + xy = 3              …(iv)

Solving (iii) and (iv) for x and y, we get,

y = 3 or 1/3

∴ If y = 3; x = −1 and z = 3

∴ If y = 1/3; x = 13/3 and z = 1/3

∴ The maximum value of x is 13/3.

Hence, option (c).