Question

If \[y = {3^{x - 1}} + {3^{ - x - 1}}\] (x real), then the least value of y is
\[\left( 1 \right){\text{ }}2\]
\[\left( 2 \right){\text{ 6}}\]
\[\left( 3 \right){\text{ }}\dfrac{2}{3}\]
\[\left( 4 \right)\] None of these

Answer 1

Hint: To make the given equation of y simple, first you have to change the negative exponents to the positive reciprocals. Then we observe that both the terms in the equation are positive and real. So we will apply here the AM-GM inequality to find the least value of y. The formula we will use for this is given as \[\dfrac{{a + b}}{2} \geqslant \sqrt {ab} \] . This is easy to solve just a calculation and nothing.

Complete step-by-step answer:
It is given that \[y = {3^{x - 1}} + {3^{ - x - 1}}\] where x is a real number and we have to find the least value of y.
First you should know what are negative exponents before proceeding. A negative exponent means how many times to divide by the number. For example:- \[{{\text{6}}^{ - 1}}\] is \[\dfrac{1}{6}\] .So, negative exponents can be expressed as the positive reciprocal of the base multiplied by itself x times. The larger the negative exponent, the smaller the number it represents because negative exponents represent repeated division.
As it is given to us that \[y = {3^{x - 1}} + {3^{ - x - 1}}\] where the terms in the power are negative so we can write the above equation as
\[y = \dfrac{{{3^x}}}{3} + \dfrac{1}{{{3^x}.3}}\] --------- (i)
Here we will use AM-GM inequality, or inequality of arithmetic and geometric means which states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list. i.e., \[\dfrac{{a + b}}{2} \geqslant \sqrt {ab} \]
We are using this inequality here because the terms \[\dfrac{{{3^x}}}{3}\] and \[\dfrac{1}{{{3^x}.3}}\] are non-negative terms.
So, according to the inequality formula our a term is \[\dfrac{{{3^x}}}{3}\] and b term is \[\dfrac{1}{{{3^x}.3}}\] .So now by putting these values in the formula we get
\[\dfrac{{a + b}}{2} \geqslant \sqrt {ab} \]
\[ \Rightarrow \dfrac{{\dfrac{{{3^x}}}{3} + \dfrac{1}{{{3^x}.3}}}}{2} \geqslant \sqrt {\dfrac{{{3^x}}}{3}.\dfrac{1}{{{3^x}.3}}} \]
Take L.C.M. in the numerator at the left hand side of the equation. Whereas in the right hand side the terms \[{3^x}\] will cancel out each other .
 \[ \Rightarrow \dfrac{{\dfrac{{{3^x}}}{3} + \dfrac{1}{{{3^x}.3}}}}{2} \geqslant \sqrt {\dfrac{1}{3}.\dfrac{1}{3}} \]
On further solving we get
\[ \Rightarrow \dfrac{{\dfrac{{{3^x}}}{3} + \dfrac{1}{{{3^x}.3}}}}{2} \geqslant \sqrt {\dfrac{1}{9}} \]
We can also write the root term as the power of \[\dfrac{1}{2}\] . Therefore,
\[ \Rightarrow \dfrac{{\dfrac{{{3^x}}}{3} + \dfrac{1}{{{3^x}.3}}}}{2} \geqslant {\left( {\dfrac{1}{9}} \right)^{\dfrac{1}{2}}}\]
Or we can write it as
\[ \Rightarrow \dfrac{{\dfrac{{{3^x}}}{3} + \dfrac{1}{{{3^x}.3}}}}{2} \geqslant \dfrac{1}{{{9^{\dfrac{1}{2}}}}}\]
As we know that \[9\] is the perfect square of number \[3\] .Therefore the above expression will be
\[ \Rightarrow \dfrac{{\dfrac{{{3^x}}}{3} + \dfrac{1}{{{3^x}.3}}}}{2} \geqslant \dfrac{1}{3}\]
Now when we shift \[\dfrac{1}{2}\] to the right hand side it becomes \[2\] .So,
\[ \Rightarrow \dfrac{{{3^x}}}{3} + \dfrac{1}{{{3^x}.3}} \geqslant \dfrac{2}{3}\]
 Now from equation (i) we can write the above equation as
 \[y \geqslant \dfrac{2}{3}\] or we can also write it as \[\dfrac{2}{3} \leqslant y\] . This means that y is greater than equal to \[\dfrac{2}{3}\] . Therefore the least value of y is \[\dfrac{2}{3}\] .
Hence, the correct option is \[\left( 3 \right){\text{ }}\dfrac{2}{3}\]
So, the correct answer is “Option 3”.

Note: Remember that AM-GM inequality is used for non-negative real numbers. And keep in mind its formula too. Don’t forget to use this formula. With the help of this you are able to find the least value of y. Take care of the inequality sign. Here the inequality sign was not changed because \[2\] is a positive number but if there was \[ - 2\] instead of \[2\] then there will be a change in the inequality sign.