C and D are two sets such that C-p-p- R- 1- p -7 and set D-q- q- R -2-q- 4- then find C- D-

Given: Set C={p:p∈ R, 1≤ p lt; 7}and set D={q:q∈ R, 2 lt;q≤ 4} In Interval form it can be written as: C = [1,7); D = ( 2,4] Now, C∩ D will be the elemen ...

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Standard XII

Mathematics

Intersection of Sets

C and D are t...

Question

$C$ and $D$ are two sets such that $C = {p : p \in R, 1 \leq p < 7}$ and set $D = {q : q \in R - 2 < q \leq 4},$ then find $C \cap D .$

(

0, 4)

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(1, 7)

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[1, 4]

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(- 2, 4]

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Solution

The correct option is C $[1, 4]$
Given: Set $C = {p : p \in R, 1 \leq p < 7}$ and set $D = {q : q \in R, - 2 < q \leq 4}$

In Interval form it can be written as:
$C = [1, 7); D = (- 2, 4]$

Now, $C \cap D$ will be the elements common to both the sets.
$\Rightarrow C \cap D = [1, 4]$

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C and D are two sets such that C={p:p∈R,1≤p<7} and set D={q:q∈R−2<q≤4}, then find C∩D.

The correct option is C [1,4]Given: Set C={p:p∈R,1≤p<7}and set D={q:q∈R,−2<q≤4} In Interval form it can be written as: C=[1,7);D=(−2,4] Now, C∩D will be the elements common to both the sets. ⇒C∩D=[1,4]

$C$ and $D$ are two sets such that $C = {p : p \in R, 1 \leq p < 7}$ and set $D = {q : q \in R - 2 < q \leq 4},$ then find $C \cap D .$

The correct option is C $[1, 4]$
Given: Set $C = {p : p \in R, 1 \leq p < 7}$ and set $D = {q : q \in R, - 2 < q \leq 4}$

In Interval form it can be written as:
$C = [1, 7); D = (- 2, 4]$

Now, $C \cap D$ will be the elements common to both the sets.
$\Rightarrow C \cap D = [1, 4]$