Question

Let \[f:R \to R\] be such that for all \[x \in R\], \[\left( {{2^{1 + x}} + {2^{1 - x}}} \right),f\left( x \right)\] and \[\left( {{3^x} + {3^{ - x}}} \right)\] are in the arithmetic progression. Then what is the minimum value of \[f\left( x \right)\] ?
A. 0
B. 4
C. 3
D. 2

Answer 1

Hint: Since the given three terms are in the arithmetic progression. So, apply the condition of the terms in the arithmetic progression and calculate the value of \[f\left( x \right)\]. Then apply the inequality condition of the arithmetic and geometric mean for the first and third in the arithmetic progression. After that, add both inequalities and simplify them to get the required answer.

Formula used:
If \[a,b,c\] are in the arithmetic progression, then \[2b = a + c\].
The geometric mean of two numbers \[a\], and \[b\] is: \[G.M. = \sqrt {ab} \]
The inequality between the arithmetic and geometric mean is: \[A.M. \ge G.M.\]

Complete step by step solution:
Given:
The terms \[\left( {{2^{1 + x}} + {2^{1 - x}}} \right),f\left( x \right)\] and \[\left( {{3^x} + {3^{ - x}}} \right)\] are in the arithmetic progression where \[x \in R\].
Let’s apply the condition for the terms in A.P.
We get,
\[2f\left( x \right) = \left( {{2^{1 + x}} + {2^{1 - x}}} \right) + \left( {{3^x} + {3^{ - x}}} \right)\]
Divide both sides by 2.
\[f\left( x \right) = \dfrac{{{2^{1 + x}} + {2^{1 - x}} + {3^x} + {3^{ - x}}}}{2}\]
\[ \Rightarrow f\left( x \right) = \dfrac{{{2^{1 + x}} + {2^{1 - x}}}}{2} + \dfrac{{{3^x} + {3^{ - x}}}}{2}\] \[.....\left( 1 \right)\]

Now apply the \[A.M. \ge G.M.\] inequality for the first term.
\[\dfrac{{{2^{1 + x}} + {2^{1 - x}}}}{2} \ge \sqrt {{2^{1 + x}} \cdot {2^{1 - x}}} \]
Simplify the inequality.
\[\dfrac{{{2^{1 + x}} + {2^{1 - x}}}}{2} \ge \sqrt {2 \cdot 2} \]
\[ \Rightarrow \dfrac{{{2^{1 + x}} + {2^{1 - x}}}}{2} \ge 2\] \[.....\left( 2 \right)\]

Now apply the \[A.M. \ge G.M.\] inequality for the third term.
\[\dfrac{{{3^x} + {3^{ - x}}}}{2} \ge \sqrt {{3^x} \cdot {3^{ - x}}} \]
Simplify the inequality.
\[\dfrac{{{3^x} + {3^{ - x}}}}{2} \ge \sqrt 1 \]
\[ \Rightarrow \dfrac{{{3^x} + {3^{ - x}}}}{2} \ge 1\] \[.....\left( 3 \right)\]

Now add the equations \[\left( 2 \right)\] and \[\left( 3 \right)\].
\[\dfrac{{{2^{1 + x}} + {2^{1 - x}}}}{2} + \dfrac{{{3^x} + {3^{ - x}}}}{2} \ge 2 + 1\]
From the equation \[\left( 1 \right)\], we get
\[f\left( x \right) \ge 2 + 1\]
\[ \Rightarrow f\left( x \right) \ge 3\]
Thus, the minimum value of the given function \[f\left( x \right)\] is 3.
Hence the correct option is C.

Note: Here the arithmetic mean is denoted as A.M. and the geometric mean is denoted as G.M.
Students often get confused about the concept of the geometric mean.
The geometric mean of \[n\] observations is the \[{n^{th}}\] root of the product of all \[n\] observations.
The formula of the geometric mean of \[n\] observations \[{a_1},{a_2},{a_3},....,{a_n}\] is,
\[G.M. = \sqrt[n]{{{a_1} \times {a_2} \times {a_3}.... \times {a_n}}}\] or \[G.M. = {\left( {{a_1} \times {a_2} \times {a_3}.... \times {a_n}} \right)^{\dfrac{1}{n}}}\]