(A) Given $a + b + c + d = 2$. Let $x = a + b$ and $y = c + d$. Then $x + y = 2$ and $M = xy$.By the Arithmetic Mean-Geometric Mean inequality $(AM \g

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(A) Given $a + b + c + d = 2$. Let $x = a + b$ and $y = c + d$. Then $x + y = 2$ and $M = xy$.By the Arithmetic Mean-Geometric Mean inequality $(AM \ge GM)$,we have $\frac{x + y}{2} \ge \sqrt{xy}$.Substituting the values,$\frac{2}{2} \ge \sqrt{M}$,which implies $1 \ge \sqrt{M}$,or $M \le 1$.Since $a, b, c, d$ are positive real numbers,$x > 0$ and $y > 0$,so $M = xy > 0$.Thus,the relation is $0 < M \le 1$.

Class 11 Mathematics 2 Complex Numbers and Quadratic Equations: Quadratic Equations Solution of quadratic equations and Nature of roots

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If $a, b, c, d$ are positive real numbers such that $a + b + c + d = 2$,then $M = (a + b)(c + d)$ satisfies the relation:

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A
$0 < M \le 1$
B
$1 \le M \le 2$
C
$2 \le M \le 3$
D
$3 \le M \le 4$

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2 Complex Numbers and Quadratic Equations: Quadratic Equations

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Solution of quadratic equations and Nature of roots

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If $a, b, c, d$ are positive real numbers such that $a + b + c + d = 2$,then $M = (a + b)(c + d)$ satisfies the relation:

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