JEE MAIN - Mathematics (2020 - 8th January Morning Slot - No. 18)
Let ƒ : R $$ \to $$ R be such that for all
x $$ \in $$ R
(21+x + 21–x), ƒ(x) and (3x + 3–x) are in A.P.,
then the minimum value of ƒ(x) is
(21+x + 21–x), ƒ(x) and (3x + 3–x) are in A.P.,
then the minimum value of ƒ(x) is
2
0
3
4
Explanation
f(x) = $${{2\left( {{2^x} + {2^{ - x}}} \right) + \left( {{3^x} + {3^{ - x}}} \right)} \over 2} \ge 3$$
As we know, A.M > G.M
As we know, A.M > G.M
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