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Question: Which of the following functions have no domain? A.\[f\left( x \right) = {\log _{x - 1}}\left( {2 ...

Which of the following functions have no domain?
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)
B.g\left( x \right) = {\cos ^{ - 1}}\left( {2 - \left\\{ x \right\\}} \right)
C.h(x)=ln⁡ln⁡(cos⁡x)h(x) = \ln \ln \left( {\cos x} \right)
D.f(x)=1sec⁡−1(sgn(e−x))f\left( x \right) = \dfrac{1}{{{{\sec }^{ - 1}}({\mathop{\rm sgn}} \left( {{e^{ - x}}} \right))}}
Where [x]\left[ x \right] denotes the greatest integer function & \left\\{ x \right\\} denotes the fractional part function.

Explanation

Solution

Here, you can proceed with [x]\left[ x \right] and \left\\{ x \right\\} where the function lies between and their domain what says and calculate what the function wants and which part of the domain lies in between according to the function.

Complete step-by-step answer:
Now, We are calculating domain for:
(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)
As, we know x−1>0x - 1 > 0
Therefore, x>1⇒x > 1 \Rightarrow equation 1
Also, 2−[x]−[x]2>02 - \left[ x \right] - {\left[ x \right]^2} > 0
Adding and subtracting on left hand side.
We get,
2−[x]−[x]2+[x]−[x]>02 - \left[ x \right] - {\left[ x \right]^2} + \left[ x \right] - \left[ x \right] > 0
So, 2−[x]−[x]2+[x]−[x]>02 - \left[ x \right] - {\left[ x \right]^2} + \left[ x \right] - \left[ x \right] > 0
2−2[x]−[x]2+[x]>02 - 2\left[ x \right] - {\left[ x \right]^2} + \left[ x \right] > 0
Now, we will do the factorisation.
2(1−[x])+[x](1−[x])>02\left( {1 - \left[ x \right]} \right) + {\left[ x \right]^{}}\left( {1 - \left[ x \right]} \right) > 0
Taking out 1−[x]1 - \left[ x \right] common:
We get, (2+[x])(1−[x])>0\left( {2 + {{\left[ x \right]}^{}}} \right)\left( {1 - \left[ x \right]} \right) > 0
Changing the sign 1−[x]1 - \left[ x \right] and also changing the signs on right hand side.
([x]+2)([x]−1)<0\left( {\left[ x \right] + {2^{}}} \right)\left( {\left[ x \right] - 1} \right) < 0
Taking separately [x]+2<0\left[ x \right] + {2^{}} < 0 and [x]−1<0\left[ x \right] - 1 < 0 .
We get, [x]<\-2\left[ x \right] < \- 2 and [x]<1\left[ x \right] < 1 .
Make one equations of both: −2<[x]<1 - 2 < \left[ x \right] < 1 (Not possible according to equation 1)
That means, 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) has no solution.
(B) g\left( x \right) = {\cos ^{ - 1}}\left( {2 - \left\\{ x \right\\}} \right)
As, we know cos⁡−1x{\cos ^{ - 1}}x only exist for [−1,1]\left[ { - 1,1} \right] .
Therefore, g\left( x \right) = {\cos ^{ - 1}}\left( {2 - \left\\{ x \right\\}} \right) must exist in the [−1,1]\left[ { - 1,1} \right] for atleast one value for x.
So, from above we can say that - 1 \le \left( {2 - \left\\{ x \right\\}} \right) \le 1 .
On, solving this further equation:
We get, - 3 \le - \left\\{ x \right\\} \le - 1
Therefore, we can say that 1 \le \left\\{ x \right\\} \le 3 \Rightarrow equation 1
From equation1 it is clear that \left\\{ x \right\\} must always be greater than or equal to 1.
But, we know that \left\\{ x \right\\} always lies between [0,1)\left[ {0,\left. 1 \right)} \right. in which 0 is included but 1 is not included.
From the above lines, it is clear that both lines don't match the format so it is not possible.
That means, g\left( x \right) = {\cos ^{ - 1}}\left( {2 - \left\\{ x \right\\}} \right) is correct.
(C) h(x)=ln⁡ln⁡(cos⁡x)h(x) = \ln \ln \left( {\cos x} \right)
We know ln⁡(cos⁡x)\ln (\cos x) is operated inside logarithm, So ln⁡(cos⁡x)>0\ln \left( {\cos x} \right) > 0
Assuming ln to be ln⁡e{\ln _e}
That is, ln⁡e(cos⁡x)>0{\ln _e}\left( {\cos x} \right) > 0
= cos⁡x<e0\cos x < {e^0}
So, cos⁡x=1\cos x = 1 (As e0=1{e^0} = 1) .
But, cos⁡x\cos xis always less than or equal to 1.
That means, h(x)=ln⁡ln⁡(cos⁡x)h(x) = \ln \ln \left( {\cos x} \right) has no solution.
(D) f(x)=1sec⁡−1(sgn(e−x))f\left( x \right) = \dfrac{1}{{{{\sec }^{ - 1}}({\mathop{\rm sgn}} \left( {{e^{ - x}}} \right))}}
As, we all know the definition of sgn function that is sign function which means an odd mathematical function that extracts a real number.
By definition, we can say that sgn(x)=−1{\mathop{\rm sgn}} \left( x \right) = - 1 for x<0x < 0
=0= 0 for x=0x = 0
=1= 1 for x>0x > 0
Therefore, we can calculate for sgn(e−x)=1{\mathop{\rm sgn}} \left( {{e^{ - x}}} \right) = 1 for all x as e−x>0{e^{ - x}} > 0 for all values of x.
So, f=1sec⁡−1(1)=10f = \dfrac{1}{{{{\sec }^{ - 1}}(\left. 1 \right)}} = \dfrac{1}{0} (As we know, sec⁡−1(−1)=0{\sec ^{ - 1}}\left( { - 1} \right) = 0)
Therefore, f doesn’t exist for any values of x.
So, it has no domain.
f(x)=1sec⁡−1(sgn(e−x))f\left( x \right) = \dfrac{1}{{{{\sec }^{ - 1}}({\mathop{\rm sgn}} \left( {{e^{ - x}}} \right))}} is correct.

Note: In this question, we should remember the meaning of functions [x]\left[ x \right] denotes the greatest integer function & \left\\{ x \right\\} denotes the fractional part function. As, well as the definition of signum or sign function.