Sequences And Series Ques 7
- If $a, b, c$ are positive real numbers such that $a+b+c+d=2$, then $M=(a+b)(c+d)$ satisfies the relation
$(2000,2 \mathrm{M})$
(a) $0<M \leq 1$
(b) $1 \leq M \leq 2$
(c) $2 \leq M \leq 3$
(d) $3 \leq M \leq 4$
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Answer:
Correct Answer: 7.(a)
Solution: (a) Since, $\mathrm{AM} \geq \mathrm{GM}$, then
$ \frac{(a+b)+(c+d)}{2} \geq \sqrt{(a+b)(c+d)} \Rightarrow M \leq 1 $
Also, $ (a+b)+(c+d)>0 \quad [\because a, b, c, d>0]$
$\therefore \quad 0<M \leq 1 $
BETA
