The roster form of the set x - x -2 r -3- where r - I and -2- r - 1A- 0-1-3-5-7-9B- 1-3-5-7-9- 0-1-3-5-7D- 1-3-5

The correct option is D {1,3,5} Given 2 lt;r ≤ 1 and r ∈ I So, r = 1, 0, 1. Also x=2r+3 Substituting r = 1, 0, 1 we get, x=2( 1)+3=1 x=2(0)+3=3 x=2(1)+3=5 ...

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Standard VII

Mathematics

Roster Method

The roster fo...

Question

The roster form of the set ${x : x = 2 r + 3, where r \in I and - 2 < r \leq 1}$ .

${1, 3, 5, 7, 9}$

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${0, 1, 3, 5, 7, 9}$

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${1, 3, 5}$

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${0, 1, 3, 5, 7}$

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Solution

The correct option is C
${1, 3, 5}$

Given $- 2 < r \leq 1$ and $r \in I$

So, r = -1, 0, 1. Also $x = 2 r + 3$

Substituting r = -1, 0, 1 we get,

$x = 2 (- 1) + 3 = 1$

$x = 2 (0) + 3 = 3$

$x = 2 (1) + 3 = 5$

$\Rightarrow$ The set is {1, 3, 5,}

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Similar questions

The roster form of the set ${x : x = 2 r + 3, where r \in I and - 2 < r \leq 3}$ .

The roster form of the set ${x : x = 2 r + 3, where r ϵ I and - 2 < r \leq 3}$ .

Q. Write the following set in roster form.
A =

{x : x = 2 r + 3, where r ϵ I and - 2 < r \leq 3}

A = {x : x \in R, x^{2} = 16 and 2 x = 6}

. Represent the set in roster form.

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The roster form of the set {x:x=2r+3, where r∈I and −2<r≤1}.

The correct option is C {1,3,5} Given −2<r≤1 and r∈I So, r = -1, 0, 1. Also x=2r+3 Substituting r = -1, 0, 1 we get, x=2(−1)+3=1 x=2(0)+3=3 x=2(1)+3=5 ⇒ The set is {1, 3, 5,}

The roster form of the set ${x : x = 2 r + 3, where r \in I and - 2 < r \leq 1}$ .

The correct option is C
${1, 3, 5}$

Given $- 2 < r \leq 1$ and $r \in I$

So, r = -1, 0, 1. Also $x = 2 r + 3$

Substituting r = -1, 0, 1 we get,

$x = 2 (- 1) + 3 = 1$

$x = 2 (0) + 3 = 3$

$x = 2 (1) + 3 = 5$

$\Rightarrow$ The set is {1, 3, 5,}