Sets Test - 47

Given:

\(A=\{1,3,5\}, B=\{2,4,6\}\) and \(C=\{0,2,4,6,8,10\}\)

As we know that, any set \(U\) is said to be a universal set which contains all the elements such that all other sets are its subset.

From the given options, we can see that, the set \(\{0,1,2,3,4,5,6,8,10\}\) contains all the elements of the set \(A, B\) and \(C\) such that \(A, B\) and \(C \subseteq\{0,1,2,3,4,5,6,8,10\}\)

Thereforee, for the given set \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\), set \(\{0,1,2,3,4,5,6,8\) \(10\}\) is the universal set.

This problem can be solved using the Venn diagram as:

So, \((A-B) \cup(B-A) \cup(A \cap B)=A \cup B\).

Thus, the answer is \(A \cup B\).

Given:

\(B=\{2,4,8,16,32,64,128,256,512,1024,2048\}\)

As we can see that, all elements are mutilplied by 2 with previous elements.

So, the required set-builder form of set \(B\) will be:

\(\left\{x: x=2^{n}, n \in N\right.\) and \(1 \leq \mathrm{n} \leq 11\}\)

Let \(A\) and \(B\) be the sets of students who have taken Mathematics and Computer Science respectively.

Given:

\(n(A \cup B)=25, n(A)=12\) and \(n(A \cap B)=8\)

We know that:

\(n(A \cup B)=n(A)+n(B)-n(A \cap B)\)

\(\Rightarrow 25=12+n(B)-8\)

\(\Rightarrow 25=4+n(B)\)

\(\Rightarrow n(B)=25-4=21\)

And, the number of students who have taken computer science but NOT mathematics is:

\(n(B-A)=n(B)-n(A \cap B)\)

\(=21-8\)

\(=13\)

Given:

\(A=\left\{x: x^{2}-2 x-3=0\right\}\)

So,

\(x^{2}-2 x-3=0\)

\(\Rightarrow x^{2}-3 x+x-3=0\)

\(\Rightarrow x(x-3)+1(x-3)=0\)

\(\Rightarrow(x+1)(x-3)=0\)

\(\Rightarrow x=-1 \text { or } 3\)

As we know that, roaster form of a set is described by listing all the elements, separated by commas, within the braces \(\{\}\).

Therefore, the required roaster form of set \(A=\{-1,3\}\).

We know that:

A set which has no elements is called an empty set.

Then:

(C) \(\{x: x \in N, x<5\) and \(x>8\}\)

As we know that, there is no natural number \(x\) such that \(x<\) 5 and \(x>8\)

Therefore, \(\{x: x \in N, x<5\) and \(x>8\}\) is an empty set.

Given:

\(A=\) The set of lines which are parallel to the \(x\)-axis

and, \(B=\) The set of numbers which are multiples of 5

As we know that, we can draw infinite number of lines parallel to the \(x\)-axis.

So, \(A\) is an infinite set.

Similarly, there are \(n\) numbers that can be a multiple of 5 i.e.,

\(B=\{5 n: n \in Z\}\)

We can see that, \(B\) is also an infinite set.

Therefore,both \(A\) and \(B\) are infinite sets.

Given that:

\(\tan \theta=1\)

\(\Rightarrow \tan \theta=1=\tan \frac{\pi}{4}\)

We know that:

If \(\tan\theta=\tan\alpha\), then:

\(\theta=n\pi+\alpha\), where \(n=1, 2, \ldots\)

\(\Rightarrow \theta=n \pi+\frac{\pi}{4}\), where \(n \in Z\)

\(\Rightarrow \theta=\left\{x: x \in\left(n \pi+\frac{\pi}{4}\right), n \in Z\right\}\)

Therefore, the solution set of the given equation is an infinite set.

Given: 

\(A=\left\{x \in R: x^{2}-10 x+9=0\right\}, B=\left\{y \in R: y^{2}-3 y+2=0\right\}\) and \(U=\{z \in N: 1 \leq z \leq 10\}\) 

\(\Rightarrow x^{2}-10 x+9=0\) (Given)

\(\Rightarrow x^{2}-x-9 x+9=0\) 

\(\Rightarrow(x-1)(x-9)=0\) 

\(\Rightarrow x=1 \text { or } 9\) 

\(\Rightarrow A=\{1,9\}\)

Similarly, 

\(y^{2}-3 y+2=0\) 

\(\Rightarrow y^{2}-y-2 y+2=0\) 

\(\Rightarrow(y-1)(y-2)=0\) 

\(\Rightarrow y=1 \text { or } 2\) 

\(\Rightarrow B=\{1,2\}\)

The shaded region represents the set, \(A-B=\{x: x \in A\) and \(x \notin B\}=\{9\}\)

The possible Venn diagram will be:

Given:

\(n(X \cup Y)=70, n(X)=38, n(Y)=42\)

By using the formula,

\(n(X \cup Y)=n(X)+n(Y)-n(X \cap Y) \)

\(\Rightarrow n(X \cap Y)=n(X)+n(Y)-n(X \cup Y)\)

\(\Rightarrow n(X \cup Y)=38+42-70\)

\(\therefore n(X \cup Y)=10\)