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Matching :
If $x, y, z$ are connected by the relations $x + y + z = 5$ and $x y + y z + z x = 8,$ then
Column - I Column - II
A) Minimum value of $x ....$ P) 1
B) Maximum value of $x ....$ Q) 4/3
C) When $x$ is minimum, value of $y,$ is.... R) 2
D) When $x$ is maximum, value of $y,$ is.... S) 7/3

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$A \to P; B \to S; C \to R; D \to Q$

$A \to Q; B \to P; C \to R; D \to S$

$A \to P; B \to R; C \to Q; D \to S$

$A \to P; B \to Q; C \to S; D \to R$

answer is D.

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Detailed Solution

$x y + (5 - x - y) (x + y) = 0, 1 \leq x \leq \frac{7}{3}$

Since $y$ is real ${(x - 5)}^{2} - 4 (x^{2} - 5 x + 8) \geq 0$ which implies $1 \leq x \leq \frac{7}{3}$

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