Question

The domain of the function \[f(x) = lo{g_{(2x - 1)}}(x - 1)\] is
A. \[(1,\infty )\]
B. \[(\dfrac{1}{2},\infty )\]
C. \[(0,\infty )\]
D.None of these

Answer 1

Hint: In this question we have to find the domain of the function. We will use the definition of logarithm function to find the solution of the problem. i.e. in the logarithm function , \[{\log _a}x\] we must have \[x > 0\] , \[a > 0\] and \[a \ne 0\] for the function to be defined . Hence we will check these conditions for the given function \[f(x) = lo{g_{(2x - 1)}}(x - 1)\] to get the domain of the given function.

Complete step-by-step answer:
This problem is based on the logarithm function. Logarithm function is the inverse of exponential function. It is denoted as \[{\text{f(x)}} = {\text{lo}}{{\text{g}}_a}x\] where \[a\] is called the base of the function.
The logarithm function must satisfy the following condition:
 \[x > 0\]
 \[a > 0\] and
 \[a \ne 0\]
Consider the given question,
We are given the function \[f(x) = lo{g_{(2x - 1)}}(x - 1)\]
We have to find the domain of this given function. Therefore we check the condition for the logarithm function. i.e.
 \[(x - 1) > 0\]
On solving, the inequality we get
 \[x > 1\]
 \[(2x - 1) > 0\]
On solving, we have
 \[
   \Rightarrow 2x > 1 \\
   \Rightarrow x > \dfrac{1}{2} \;
\]
 \[(2x - 1) \ne 1\]
On solving , we get
 \[
   \Rightarrow 2x \ne 1 \\
   \Rightarrow x \ne \dfrac{1}{2} \;
 \]
Hence we have \[x > 1\] , \[x > \dfrac{1}{2}\] and \[x \ne 1\] .
Therefore, the common interval satisfying the above condition is \[(1,\infty )\] .
Hence the domain of the function \[f(x) = lo{g_{(2x - 1)}}(x - 1)\] is \[(1,\infty )\]
Hence Option \[A \] is correct.
So, the correct answer is “Option A”.

Note: While finding the common interval (i.e. domain) , we have to choose one which is satisfied by all . For example: consider the condition , \[x > 1\] satisfy the interval \[(1,\infty )\] and \[x > \dfrac{1}{2}\] satisfy the interval \[(\dfrac{1}{2},\infty )\] . But when we consider both the conditions simultaneously, we must choose the interval wisely. Hence we have to choose one that contains both (i.e. interval \[(1,\infty )\] ).
Here the bracket ( ) is called an open interval. For example, \[(a,b)\] means \[a\] and \[b\] is not included in the interval but all numbers in-between them .
We always put open brackets at infinity ( \[\infty \] ).