Question

Which of the following has no domain?
\[\left( \text{a} \right)f\left( x \right)={{\log }_{x-1}}\left( 2-\left[ x \right]-{{\left[ x \right]}^{2}} \right)\] where [x] denotes GIF
\[\left( \text{b} \right)g\left( x \right)={{\cos }^{-1}}\left( 2-\left\{ x \right\} \right)\] where {x} denotes fractional part function.
\[\left( \text{c} \right)h\left( x \right)=\ln \ln \left( \cos x \right)\]
\[\left( \text{d} \right)f\left( x \right)=\dfrac{1}{{{\sec }^{-1}}\left( \text{sgn} \left( {{e}^{-x}} \right) \right)}\]

Answer 1

Hint: To solve the given question, we will check each option one by one. While checking the options, we will use the fact that the domain of log x is \[x\in \left( 0,\infty \right),\] the domain of \[{{\cos }^{-1}}\left\{ x \right\}\] is \[x\in \left[ -1,1 \right].\] The domain of \[{{\log }_{a}}7\] is \[a\in \left( 0,1 \right)\cup \left( 1,\infty \right),\] the domain of \[{{\sec }^{-1}}x\] is \[x\in \left( -\infty ,-1 \right]\cup \left[ 1,\infty \right).\] We will use the domains of these functions to get the answer.

Complete step by step answer:
To solve the above question, we will check each and every option. So, we have,
Option (a): \[f\left( x \right)={{\log }_{x-1}}\left( 2-\left[ x \right]-{{\left[ x \right]}^{2}} \right)\]
The function given in the question is a logarithmic function. Now, the base of the logarithm should be positive and it should not be equal to 1. The base of the above logarithmic function is x – 1. So, we have,
\[x-1\ne 1\]
\[\Rightarrow x\ne 2.....\left( i \right)\]
Also, x – 1 > 0
\[\Rightarrow x>1....\left( ii \right)\]
Now, the argument part of the logarithmic function should be positive. Thus,
\[2-\left[ x \right]-{{\left[ x \right]}^{2}}>0\]
\[\Rightarrow {{\left[ x \right]}^{2}}+\left[ x \right]-2<0\]
\[\Rightarrow {{\left[ x \right]}^{2}}+2\left[ x \right]-\left[ x \right]-2<0\]
\[\Rightarrow \left[ x \right]\left( \left[ x \right]+2 \right)-1\left( \left[ x \right]+2 \right)<0\]
\[\Rightarrow \left( \left[ x \right]-1 \right)\left( \left[ x \right]+2 \right)<0\]
\[\Rightarrow \left[ x \right]\in \left( -2,1 \right)\]
Now, [x] > – 2. So,
\[x\ge -1......\left( iii \right)\]
Similarly, [x] < 1. So,
\[x<2.....\left( iv \right)\]
Now, to find the domain of the entire function, we will find the common values of x from (i), (ii), (iii) and (iv). Thus, the domain of the given function is,
\[x\in \left( 1,2 \right)\]

Option (b): \[g\left( x \right)={{\cos }^{-1}}\left( 2-\left\{ x \right\} \right)\]
Now, we know that the domain of the cos inverse function lies in the range \[x\in \left[ -1,1 \right].\] So, we can say that,
\[2-\left\{ x \right\}\in \left[ -1,1 \right]\]
\[-1\le 2-\left\{ x \right\}\le 1\]
\[\Rightarrow -3\le -\left\{ x \right\}\le -1\]
\[\Rightarrow 1\le \left\{ x \right\}\le 3.....\left( i \right)\]
Now, the range of {x} is:
\[0\le \left\{ x \right\}<1......\left( ii \right)\]
From (i) and (ii), we will take the common values of {x}. We can see there is nothing common so there will be no domain.

Option (c): \[h\left( x \right)=\ln \ln \left( \cos x \right)\]
Now, we know that the domain of ln x is \[x\in \left( 0,\infty \right).\] We will assume the value of ln cos x = t. Thus, we get, h(t) = ln t. Now, we have,
\[t\in \left( 0,\infty \right)\]
\[\Rightarrow 0\[\Rightarrow 0<\ln \cos x<\infty \]
Now, the maximum value of cos x = 1. So, the maximum value of ln cos x = ln 1 = 0. But in the above inequality, \[\ln \cos x\in \left( 0,\infty \right).\] We can see that there is nothing common here, so this function will not have any domain.

Option (d): \[f\left( x \right)=\dfrac{1}{{{\sec }^{-1}}\left( sgn \left( {{e}^{-x}} \right) \right)}\]
Now, the signum function will give the value 1 when the function inside it is positive, it will give – 1 when the function inside it is negative and it will give 0 if the function value is zero. Now, \[{{e}^{-x}}\] is positive for any value of x, so \[sgn \left( {{e}^{-x}} \right)=1.\] Thus,
\[f\left( x \right)=\dfrac{1}{{{\sec }^{-1}}\left( 1 \right)}\]
Now,
\[{{\sec }^{-1}}1=0\]
So,
\[f\left( x \right)=\dfrac{1}{0}\]
But f(x) can’t be any infinite function so, in this function, there is no domain.

Hence, option (b), (c) and (d) are correct.

Note: We can also solve the option (b) as shown below. We know that the domain of the fractional part function {x} is \[x\in \left( 0,\infty \right)\] and the range is: \[\left\{ x \right\}\in \left[ 0,1 \right).\]
So, the minimum value of (2 – {x}) is
\[=\left( 2-{{1}^{-}} \right)\]
\[=\left( {{1}^{+}} \right)\]
But the maximum value of the domain of \[{{\cos }^{-1}}\left( x \right)\] is 1. And we have 2 – {x} > 1. So, there is no domain.