For three real numbers x, y and z, x + y + z = 5, and xy + yz + xz = 3. What is the largest value which x can take?
Explanation:
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).
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