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If three positive real numbers a,b,c are in AP such that abc=4, then the minimum value of b isA. `2^(1//3)`B. `2^(2//3)`C. `2^(1//2)`D. `2^(3//2)` |
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Answer» Correct Answer - B Since a,b,c are in A.P., therefore, b-a=d and c-b=d, where d is the common difference of the A.P. `therefore a=b-d and c=b+d` Now, abc=4 `rArr(b-d)b(b+d)=4` `rArrb(b^(2)-d^(2))=4` But, `b(b^(2)-d^(2))ltbxxb^(2)` `rArrb(b^(2)-d^(2))ltb^(3)` `rArr4ltb^(3)` `rArrb^(3)gt4` `rArrbgt2^(2//3)` Hence, the minimum value of b is `2^(2//3)`. |
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