Let [.] denote greatest integer function. If f(x)=[x] and g(x)=3[(x/3)], then the set of all real x such that f(x)=g(x) is

Question

Let [.] denote greatest integer function. If f(x)=[x] and g(x)=3[(x/3)], then the set of all real x such that f(x)=g(x) is (A) R (B)  x ∈ R / x=3 k, k ∈ Z  (C)  x ∈ R / 3 k-1<x ≤ 3 k, k ∈ Z  (D)  x ∈ R / 3 k ≤ x<3 k+1, k ∈ Z

Tardigrade Admin · Accepted Answer

We have, f(x)=[x] and g(x)=3[(x/3)] Given, f(x)=g(x) ∴ [x]=3[(x/3)] Here, [x] and [(x/3)] is an integers Let [(x/3)]=k ∵[x]=3 k ∵ x ∈[3 k, 3 k+1) x ∈ R / 3 k ≤ x<3 k+1, k ∈ Z

Q. Let $[.]$ denote greatest integer function. If $f (x) = [x]$ and $g (x) = 3 [\frac{x}{3}]$ , then the set of all real $x$ such that $f (x) = g (x)$ is

$R$

${x \in R / x = 3 k, k \in Z}$

${x \in R /3 k - 1 < x \leq 3 k, k \in Z}$

${x \in R /3 k \leq x < 3 k + 1, k \in Z}$

We have, $& f (x) = [x]$ and $g (x) = 3 [\frac{x}{3}]$
Given, $f (x) = g (x)$
$∴ [x] = 3 [\frac{x}{3}]$
Here, $[x]$ and $[\frac{x}{3}]$ is an integers
Let $[\frac{x}{3}] = k$
$∵ [x] = 3 k$
$∵ x \in [3 k, 3 k + 1)$
${x \in R /3 k \leq x < 3 k + 1, k \in Z}$

Q. Let [.] denote greatest integer function. If f(x)=[x] and g(x)=3[3x​], then the set of all real x such that f(x)=g(x) is

R

{x∈R/x=3k,k∈Z}

{x∈R/3k−1<x≤3k,k∈Z}

{x∈R/3k≤x<3k+1,k∈Z}

We have, &f(x)=[x] and g(x)=3[3x​] Given, f(x)=g(x) ∴[x]=3[3x​] Here, [x] and [3x​] is an integers Let [3x​]=k ∵[x]=3k ∵x∈[3k,3k+1) {x∈R/3k≤x<3k+1,k∈Z}

Q. Let $[.]$ denote greatest integer function. If $f (x) = [x]$ and $g (x) = 3 [\frac{x}{3}]$ , then the set of all real $x$ such that $f (x) = g (x)$ is

$R$

${x \in R / x = 3 k, k \in Z}$

${x \in R /3 k - 1 < x \leq 3 k, k \in Z}$

${x \in R /3 k \leq x < 3 k + 1, k \in Z}$

We have, $& f (x) = [x]$ and $g (x) = 3 [\frac{x}{3}]$
Given, $f (x) = g (x)$
$∴ [x] = 3 [\frac{x}{3}]$
Here, $[x]$ and $[\frac{x}{3}]$ is an integers
Let $[\frac{x}{3}] = k$
$∵ [x] = 3 k$
$∵ x \in [3 k, 3 k + 1)$
${x \in R /3 k \leq x < 3 k + 1, k \in Z}$