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Exams
Joint Entrance Examination
Mathematics
Sequences and Series
Question 241
Question 241 of 519
Medium
MCQ
The minimum value of
f
(
x
)
=
a
a
x
+
a
1
−
a
x
f(x) = a^{a^x}+a^{1-a^x}
f
(
x
)
=
a
a
x
+
a
1
−
a
x
, where
a
,
x
∈
R
a,x\in\mathbb{R}
a
,
x
∈
R
and
a
>
0
a>0
a
>
0
, is equal to:
A
a
+
1
a+1
a
+
1
B
a
+
1
a
a+\dfrac1a
a
+
a
1
C
2
a
2\sqrt a
2
a
D
2
a
2a
2
a
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Solutions · 1
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Official solution
We know, \[AM \ge GM\] \[ \therefore \] \[{{{a^{a^x}} + {a \over {{a^{a^x}}}}} \over 2} \ge {\left( {{a^{a^x}}\,.\,{a \over {{a^{a^x}}}}} \right)^{1/2}} \] \[\Rightarrow {a^{a^x}} + {a^{1 - a^x}} \ge 2\sqrt a \]
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