Let P-x-x- and 7x-2 1 and Q-x- x- and -x-1- 2- Then nP - Q is

7x+2 1 ⇒7x+2 1 gt;0 ⇒5 xx+2 0 ⇒x 5x+2 0 ⇒ x∈ ( 2,5) ∴ P={1,2,3,4} |x 1| 2 ⇒ 2 x 1 2 ⇒ 1 x 3 ∴ Q={1,2} n(P Δ Q)=n(P)+n(Q) 2n(P∩ Q) =4+2 2×2 =2 ...

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Standard XIII

Mathematics

Properties of Symmetric Difference of Sets

Let P=x:x∈ℕ a...

Question

Let $P = {x : x \in N and \frac{7}{x + 2} > 1}$ and $Q = {x : x \in N and | x - 1 | < 2} .$ Then $n (P Δ Q)$ is

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Solution

The correct option is C $2$
$\frac{7}{x + 2} > 1$
$\Rightarrow \frac{7}{x + 2} - 1 > 0$
$\Rightarrow \frac{5 - x}{x + 2} > 0$
$\Rightarrow \frac{x - 5}{x + 2} < 0$
$\Rightarrow x \in (- 2, 5)$
$∴ P = {1, 2, 3, 4}$

$| x - 1 | < 2$
$\Rightarrow - 2 < x - 1 < 2$
$\Rightarrow - 1 < x < 3$
$∴ Q = {1, 2}$

$n (P Δ Q) = n (P) + n (Q) - 2 n (P \cap Q)$
$= 4 + 2 - 2 \times 2$
$= 2$

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Let P={x:x∈N and 7x+2>1} and Q={x:x∈N and |x−1|<2}. Then n(PΔQ) is

The correct option is C 27x+2>1 ⇒7x+2−1>0 ⇒5−xx+2>0 ⇒x−5x+2<0 ⇒x∈(−2,5) ∴P={1,2,3,4} |x−1|<2 ⇒−2<x−1<2 ⇒−1<x<3 ∴Q={1,2} n(PΔQ)=n(P)+n(Q)−2n(P∩Q) =4+2−2×2 =2

Let $P = {x : x \in N and \frac{7}{x + 2} > 1}$ and $Q = {x : x \in N and | x - 1 | < 2} .$ Then $n (P Δ Q)$ is