Let X-x x-x- n 3 -2n-1-n- R and Y- x-x-3 n 2 -7-n- R then X- Y is a subset of

The correct option is C { x:x=7n 1,n∈ R} For x∈ (X∩ Y) ⇒ n ^ 3 +2n+1= 3n ^ 2 +7 ⇒ n ^ 3 3n ^ 2 +2n 6=0 ⇒ (n 3)( n ^ 2 +2)=0 as n ...

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Byju's Answer

Standard VI

Mathematics

Equal Sets

Let X=x x:x...

Question

Let $X = x {x : x = n^{3} + 2 n + 1, n \in R}$ and $Y = {x : x = 3 n^{2} + 7, n \in R}$ then $X \cap Y$ is a subset of

{x : x = 3 n + 5, n \in R}

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{x : x = n^{2} + n + 1, n \in R}

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{x : x = 7 n - 1, n \in R}

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None of the above.

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Solution

The correct option is C ${x : x = 7 n - 1, n \in R}$
For $x \in (X \cap Y)$
$\Rightarrow n^{3} + 2 n + 1 = {3 n}^{2} + 7 \Rightarrow n^{3} - {3 n}^{2} + 2 n - 6 = 0 \Rightarrow (n - 3) (n^{2} + 2) = 0$
as $n$ belongs to $R$
$\Rightarrow n = 3$
$x = n^{3} + 2 n + 1 = {3 n}^{2} + 7 \Rightarrow x = 3^{3} + 2 (3) + 1 = {3 (3)}^{2} + 7 \Rightarrow x = 34$
In option $C$ for $n = 5$
$x = 7 (5) - 1 = 34$
In $(a)$ and $(b)$ $x \neq 34$ for any value of $n$
So option $C$ is correct.

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Let X=x{x:x=n3+2n+1,n∈R} and Y={x:x=3n2+7,n∈R} then X∩Y is a subset of

The correct option is C {x:x=7n−1,n∈R}For x∈(X∩Y)⇒n3+2n+1=3n2+7⇒n3−3n2+2n−6=0⇒(n−3)(n2+2)=0as n belongs to R⇒n=3x=n3+2n+1=3n2+7⇒x=33+2(3)+1=3(3)2+7⇒x=34In option C for n=5x=7(5)−1=34In (a) and (b) x≠34 for any value of nSo option C is correct.

Let $X = x {x : x = n^{3} + 2 n + 1, n \in R}$ and $Y = {x : x = 3 n^{2} + 7, n \in R}$ then $X \cap Y$ is a subset of