Which of the following function(s) have no domain?

Question

Which of the following function(s) have no domain? (A) f(x)= log x-1(2-[x]-[x]2) where [x] denotes the greatest integer function. (B) g(x)= cos -1(2- x ) where x denotes the fractional part function. (C) h ( x )= ln ln ( cos x ) (D) f(x)=(1/ sec -1( operatornamesgn(e-x)))

Tardigrade Admin · Accepted Answer

(A) obviously x>1 for 1< x <2,[ x ]=1 ⇒ N =0 for x ≥ 2, N <0 ⇒ no domain (B) 0 ≤ x <1 hence -1<- x ≤ 0 hence 1<2- x ≤ 2 ⇒ cos -1(2- x ) does not exist. ]

Q. Which of the following function(s) have no domain?

$f (x) = lo g_{x - 1} (2 - [x] - [x]^{2})$ where $[x]$ denotes the greatest integer function.

$g (x) = cos^{- 1} (2 - {x})$ where ${x}$ denotes the fractional part function.

$h (x) = ln ln (cos x)$

$f (x) = \frac{1}{s e c ^{- 1} ( sgn ( e ^{- x} ) )}$

(A) obviously $x > 1$
for $1 < x < 2, [x] = 1 \Rightarrow N = 0$
for $x \geq 2, N < 0 \Rightarrow$ no domain
(B) $0 \leq {x} < 1$ hence $- 1 < - {x} \leq 0$
hence $1 < 2 - {x} \leq 2$
$\Rightarrow cos^{- 1} (2 - {x})$ does not exist. $]$

Q. Which of the following function(s) have no domain?

f(x)=logx−1​(2−[x]−[x]2) where [x] denotes the greatest integer function.

g(x)=cos−1(2−{x}) where {x} denotes the fractional part function.

h(x)=lnln(cosx)

f(x)=sec−1(sgn(e−x))1​

(A) obviously x>1 for 1<x<2,[x]=1⇒N=0 for x≥2,N<0⇒ no domain (B) 0≤{x}<1 hence −1<−{x}≤0 hence 1<2−{x}≤2 ⇒cos−1(2−{x}) does not exist. ]

$f (x) = lo g_{x - 1} (2 - [x] - [x]^{2})$ where $[x]$ denotes the greatest integer function.

$g (x) = cos^{- 1} (2 - {x})$ where ${x}$ denotes the fractional part function.

$h (x) = ln ln (cos x)$

$f (x) = \frac{1}{s e c ^{- 1} ( sgn ( e ^{- x} ) )}$

(A) obviously $x > 1$
for $1 < x < 2, [x] = 1 \Rightarrow N = 0$
for $x \geq 2, N < 0 \Rightarrow$ no domain
(B) $0 \leq {x} < 1$ hence $- 1 < - {x} \leq 0$
hence $1 < 2 - {x} \leq 2$
$\Rightarrow cos^{- 1} (2 - {x})$ does not exist. $]$