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**Hint:**In this problem, we have to find the element of Set A and Set B. After that, we have to analyze that both Set A and Set B are subsets of Set U. The complement of set refers to elements, not in that set.

**Complete step-by-step solution:**We have been given that;

$Set U=1,2,3,4,5,6,7,8,9$

$A=\{x:x\in N,30< {{x}^{2}}< 70\}$

Here x is a natural number.

$\Rightarrow 30< {{x}^{2}}< 70$

We can split it into two cases:

Case 1: when $30< {{x}^{2}}$

$\begin{align}

&\Rightarrow {{x}^{2}}> 30 \\

& \Rightarrow {{x}^{2}}-30 >0 \\

& \Rightarrow (x+\sqrt{30})(x-\sqrt{30})> 0 \\

\end{align}$

Now, we will use a wavy curve method to get the values of x. According to the wavy curve method for quadratic polynomial, if a quadratic polynomial is given as $\left( x-\alpha \right)\left( x-\beta \right)$ , where $\alpha $ and $\beta $ are the zeroes, then the value of the polynomial $\forall x\in \left( \alpha ,\beta \right)$ will be less than zero and the value of the polynomial $\forall x\in \left( -\infty ,\alpha \right)\cup \left( \beta ,\infty \right)$ is greater than zero.

Now by using wavy curve method;

$(x+\sqrt{30})(x-\sqrt{30}) >0$

So, we have to find such value of x for which $(x+\sqrt{30})(x-\sqrt{30}) >0$, so using wavy curve, if we take any number from interval $(-\sqrt{30},\sqrt{30})$, we have $(x+\sqrt{30})(x-\sqrt{30})< 0$and if we take any value from interval $(-\infty ,-\sqrt{30})$ and $(\sqrt{30},\infty )$, we have $(x+\sqrt{30})(x-\sqrt{30}) >0$

So, we can say that for $(x+\sqrt{30})(x-\sqrt{30}) > 0$, we have

$\Rightarrow x>\sqrt{30}$ and $x< -\sqrt{30}............(1)$

Case 2: when ${{x}^{2}}< 70$

$\begin{align}

&\Rightarrow {{x}^{2}} <70 \\

& \Rightarrow {{x}^{2}}-70 <0 \\

& \Rightarrow (x+\sqrt{70})(x-\sqrt{70}) <0 \\

\end{align}$

By using wavy curve method;

So, we have to find such value of x for which $(x+\sqrt{70})(x-\sqrt{70}) <0$, so using wavy curve, if we take any number from interval $(-\infty ,-\sqrt{70})$ and $(\sqrt{70},\infty )$, we have $(x+\sqrt{70})(x-\sqrt{70})> 0$ and if we take any value from interval $(-\sqrt{70},\sqrt{70})$, we have $(x+\sqrt{70})(x-\sqrt{70}) <0$

So, we can say that for, we have

$-\sqrt{70}< x <\sqrt{70}........(2)$

Now we have two considering both case 1 and case 2. By wavy curve method, we have to take the intersection of equation (1) and (2) to get the value of x.

$\Rightarrow -\sqrt{70}< x <-\sqrt{30}$ and $\sqrt{30}< x < \sqrt{70}$

Here we have given that x is a natural number. So, we have to neglect all negative values of x.

$\therefore \sqrt{30}< x < \sqrt{70}$

Here given that x is a natural number. So, the possible value of x is 6, 7 and 8.

$ x\in [6,7,8]$

$\Rightarrow A=\{6,7,8\}$

$B=\{x:x\text{ is a prime number 10}\}$

The prime numbers less than 10 are 2, 3, 5 and 7.

$\begin{align}

& x\in [2,3,5,7] \\

& \Rightarrow B=\{2,3,5,7\} \\

\end{align}$

Here we can see that all elements of Set A and Set B belong to Set U. so we can say that Set A and Set B are subsets of Set U.

Now, we have to find A’ which means that elements belong to Set U but not Set A.

$\Rightarrow U=\{1,2,3,4,5,6,7,8,9\}$ and $A=\{6,7,8\}$

$\Rightarrow A'=U-A=\{1,2,3,4,5,9\}$

Now, we have to find B’ which means that elements belong to Set U but not Set B.

$\Rightarrow U=\{1,2,3,4,5,6,7,8,9\}$ and $B=\{2,3,5,7\}$

$\Rightarrow B'=U-B=\{1,4,6,8,9\}$

Now, we have to find $(A\cup B)$ which means elements either belonging to Set A or Set B.

$\Rightarrow A=\{6,7,8\}$ And $B=\{2,3,5,7\}$

$\Rightarrow A\cup B=\{2,3,5,6,7,8\}$

Now, we have to find $(A'\cap B')$ which means element common to both Set A’ and Set B’.

$\Rightarrow A'=\{1,2,3,4,5,9\}$ and $B'=\{1,4,6,8,9\}$

$\Rightarrow (A'\cap B')=\{1,4,9\}$

Now, we have to find$(A-B)'$. This can also be written as; $U-(A-B).........(3)$

Now, we have to find $(A-B)$ which means element present in Set A but not in Set B.

$\Rightarrow A=\{6,7,8\}$ and $B=\{2,3,5,7\}$

$\Rightarrow (A-B)=\{6,8\}$

We are putting the value of \[(A-B)\] in equation (3).

$(A-B)'=U-(A-B)$

We get;

$\begin{align}

& (A-B)'=U-\{6,8\} \\

& \Rightarrow \mathsf{U}=\{1,2,3,4,5,6,7,8,9\} \\

&\Rightarrow (A-B)'=\{1,2,3,4,5,6,7,8,9\}-\{6,8\} \\

& \Rightarrow (A-B)'=\{1,2,3,4,5,7,9\} \\

\end{align}$

Hence,

$\begin{align}

& A'=\{1,2,3,4,5,9\} \\

&\Rightarrow B'=\{1,4,6,8,9\} \\

&\Rightarrow (A\cup B)=\{2,3,5,6,7,8\} \\

&\Rightarrow (A'\cap B')=\{1,4,9\} \\

&\Rightarrow (A-B)'=\{1,2,3,4,5,7,9\} \\

\end{align}$

**Note:**In this problem, the biggest mistake is when we don’t analyze that Set A and Set B is a subset of Set U. We can directly find the value of Set A by putting some natural numbers i.e. 4, 5, 6, 7, 8, 9 and so on. So, we can see that only 6, 7, and 8 holds for Set A.

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