JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

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Application +4 | -1

\log _2(a+b)+\log _2(c+d) \geq 4

. Then the minimum value of the expression

a+b+c+d

View answer and explanation

Correct option: C

Option C, 8, is the correct minimum value because applying AM-GM inequality to

(a+b)

and

(c+d)

with the given logarithmic condition yields

a+b+c+d \geq 8

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