Domain of f x -cos sin x -log x x - where - denotes the fractional part is

The correct option is D ( 0,1 ) For #160; sin x ϵ[ 1,1] cos x≥0 since x 1 as base of log _ x { x} #160; will not be defined and ...

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Standard XII

Mathematics

Definition of Functions

Domain of f...

Question

Domain of $f (x) = \sqrt{cos (sin x)} + \sqrt{{log}_{x} {x}}$ ; where {.} denotes the fractional part is

[1, π)

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(0, 2 π) \sim [1, π)

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(0, \frac{π}{2}) \cup {1}

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(0, 1)

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Solution

The correct option is D $(0, 1)$
For $sin x ϵ [- 1, 1]$
$cos x \geq 0$
since $x \neq 1$ as base of ${log}_{x} {x}$ will not be defined and
${x} ϵ (0, 1)$
so $x ϵ (0, 1)$
to make ${log}_{x} {x}$ as positive.
So, domain of $f (x)$ is $(0, 1)$

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Domain of f(x)=√cos(sinx)+√logx{x} ; where {.} denotes the fractional part is

The correct option is D (0,1)For sinx ϵ[−1,1]cosx≥0since x≠1 as base of logx{x} will not be defined and {x} ϵ(0,1)so x ϵ(0,1)to make logx{x} as positive.So, domain of f(x) is (0,1)

Domain of $f (x) = \sqrt{cos (sin x)} + \sqrt{{log}_{x} {x}}$ ; where {.} denotes the fractional part is

The correct option is D $(0, 1)$
For $sin x ϵ [- 1, 1]$
$cos x \geq 0$
since $x \neq 1$ as base of ${log}_{x} {x}$ will not be defined and
${x} ϵ (0, 1)$
so $x ϵ (0, 1)$
to make ${log}_{x} {x}$ as positive.
So, domain of $f (x)$ is $(0, 1)$