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Question: Which of the following has no domain? \[\left( \text{a} \right)f\left( x \right)={{\log }_{x-1}}\l...

Which of the following has no domain?
(a)f(x)=log⁡x−1(2−[x]−[x]2)\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.
(c)h(x)=ln⁡ln⁡(cos⁡x)\left( \text{c} \right)h\left( x \right)=\ln \ln \left( \cos x \right)
(d)f(x)=1sec⁡−1(sgn(e−x))\left( \text{d} \right)f\left( x \right)=\dfrac{1}{{{\sec }^{-1}}\left( \text{sgn} \left( {{e}^{-x}} \right) \right)}

Explanation

Solution

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∈(0,∞),x\in \left( 0,\infty \right), the domain of {{\cos }^{-1}}\left\\{ x \right\\} is x∈[−1,1].x\in \left[ -1,1 \right]. The domain of log⁡a7{{\log }_{a}}7 is a∈(0,1)∪(1,∞),a\in \left( 0,1 \right)\cup \left( 1,\infty \right), the domain of sec⁡−1x{{\sec }^{-1}}x is x∈(−∞,−1]∪[1,∞).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(x)=log⁡x−1(2−[x]−[x]2)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≠1x-1\ne 1
⇒x≠2.....(i)\Rightarrow x\ne 2.....\left( i \right)
Also, x – 1 > 0
⇒x>1....(ii)\Rightarrow x>1....\left( ii \right)
Now, the argument part of the logarithmic function should be positive. Thus,
2−[x]−[x]2>02-\left[ x \right]-{{\left[ x \right]}^{2}}>0
⇒[x]2+[x]−2<0\Rightarrow {{\left[ x \right]}^{2}}+\left[ x \right]-2<0
⇒[x]2+2[x]−[x]−2<0\Rightarrow {{\left[ x \right]}^{2}}+2\left[ x \right]-\left[ x \right]-2<0
⇒[x]([x]+2)−1([x]+2)<0\Rightarrow \left[ x \right]\left( \left[ x \right]+2 \right)-1\left( \left[ x \right]+2 \right)<0
⇒([x]−1)([x]+2)<0\Rightarrow \left( \left[ x \right]-1 \right)\left( \left[ x \right]+2 \right)<0
⇒[x]∈(−2,1)\Rightarrow \left[ x \right]\in \left( -2,1 \right)
Now, [x] > – 2. So,
x≥−1......(iii)x\ge -1......\left( iii \right)
Similarly, [x] < 1. So,
x<2.....(iv)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∈(1,2)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∈[−1,1].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(x)=ln⁡ln⁡(cos⁡x)h\left( x \right)=\ln \ln \left( \cos x \right)
Now, we know that the domain of ln x is x∈(0,∞).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∈(0,∞)t\in \left( 0,\infty \right)
⇒0\[⇒0<ln⁡cos⁡x<∞\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∈(0,∞).\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(x)=1sec⁡−1(sgn(e−x))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{{e}^{-x}} is positive for any value of x, so sgn(e−x)=1.sgn \left( {{e}^{-x}} \right)=1. Thus,
f(x)=1sec⁡−1(1)f\left( x \right)=\dfrac{1}{{{\sec }^{-1}}\left( 1 \right)}
Now,
sec⁡−11=0{{\sec }^{-1}}1=0
So,
f(x)=10f\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∈(0,∞)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
=(2−1−)=\left( 2-{{1}^{-}} \right)
=(1+)=\left( {{1}^{+}} \right)
But the maximum value of the domain of cos⁡−1(x){{\cos }^{-1}}\left( x \right) is 1. And we have 2 – {x} > 1. So, there is no domain.