Relations and Functions Test - 69

We have, \(f(x)=x^2=x-2\).

\(\therefore  f(-1)=(-1)^2-(-1)-2=0\), 

\(f(0)=0^2-0-2=-2\), 

\(f(2)=2^2-2-2=0\)

\(f(5)=5^2-5-2=18\), 

\(f(6)=6^2-6-2=28\) 

and \(f(11)=11^2-11-2=108\)

Hence, \(f(A)=\{f(x): x \in A\}=\{f(-1), f(0), f(2), f(5), f(6), f(11)\}\)

\(f(A)=\{0,-2,18,28,108\}\)

Given:

\(f(x)=\sqrt{20-x^{2}}\)

It is defined only when \(20-x^{2} \geq 0\)

So, \(y \geq 0\)\(\quad\).........(i)

Let, \(y=f(x)\)

\(\Rightarrow y=\sqrt{20-x^{2}}\)

Squaring both sides, we get,

\(y^{2}=20-x^{2}\)

\(\Rightarrow x^{2}=20-y^{2}\)

\(\Rightarrow x=\sqrt{20-y^{2}}\)

It is defined only when \(20-y^{2} \geq 0\)

\(\Rightarrow y^{2} \leq 20\)

\(\Rightarrow y^{2}-20 \leq 0\)

\(\Rightarrow(y-2 \sqrt{5})(y+2 \sqrt{5}) \leq 0\)

\(\Rightarrow y \in[-2 \sqrt{5}, 2 \sqrt{5}]\)\(\quad\)........(ii)

From equations (i) and (ii), we get:

\(y \in[0,2 \sqrt{5}]\)

Since \((-1,0) \in A \times A\) and \((0,1) \in A \times A\). Therefore,

\(\therefore (-1,0) \in A \times A \Rightarrow-1,0 \in A\) 

and \((0,1) \in A \times A \Rightarrow 0,1 \in A \)

\(-1,0,1 \in A\)

It is given that \(A \times A\) has 9 elements. Therefore, \(A\) has exactly three elements.

Hence, \(A=\{-1,0,1\}\).

Let \(y=\frac{x^2}{1+x^2}\)

\(\Rightarrow y+x^2 y=x^2\)

\(\Rightarrow y=x^2(1-y) \)

\(\Rightarrow x^2=\frac{y}{1-y} \)

\(\Rightarrow x=\sqrt{\frac{y}{1-y}}\)

Since, \({x}\) is real

\(\Rightarrow \frac{y}{1-y} \geq 0\)

\(\Rightarrow \frac{y(1-y)}{(1-y)^2} \geq 0\)

\(\Rightarrow y(1-y) \geq 0 \text { and }(1-y)^2>0\)

\(\Rightarrow 0 \leq y \leq 1 \text { and }-y>-1\)

\(\Rightarrow 0 \leq y \leq 1 \text { and } y<1\)

Hence, \(0 \leq \mathrm{y}<1\)

So, the range of \(f\) is \(\{0,1\}\).

Given:

\(A=\{2,3,4\} ; B=\{5,6\}\)

We know that:

For any two non-empty sets \({A}\) and \({B}\), we have:

I. \(A \times B=\{(a, b) \mid a \in A\) and \(b \in B\}\)

II. \(B \times A=\{(b, a) \mid a \in A\) and \(b \in B\}\)

III. Any two ordered pairs \((a, b)=(c, d)\) if and only if \(a=c\) and \(b=d\).

Then,

\({A} \times {B}=\{(2,5),(2,6),(3,5),(3,6),(4,5),(4,6)\}\)

The number of elements in \(A \times B\) i.e.,

\(n(A \times B)=6\)

Therefore, the number of subsets of\(A \times B=2^{n}\)

\(=2^{6}\)

\(=64\)

Therefore, there are 64 subsets for \(A \times B\).

Given:

\((2,4),(4,6)\) and \((5,6)\) are three elements in \(A \times B\).

Also, \(A \times B\) has 6 elements.

As we know that,

\(n(A \times B)=n(A) \times n(B)\)

\(\Rightarrow {n}({A} \times {B})=6\)

\(=3 \times 2\)

By the definition of Cartesian Product,

\(A \times B=\{(a, b): a \in A\) and \(b \in B)\}\)

\(\Rightarrow {n}({A})=3\) and \({n}({B})=2\)

Thus, the elements of set \({A}\) is \(\{2,4,5\}\) and set \({B}\) is \(\{4,6\}\).

Now,

\(B \times A=\{4,6\} \times\{2,4,5\}\)

\(=\{(4,2),(4,4),(4,5),(6,2),(6,4),(6,5)\}\)

Therefore,

\({B} \times {A}=\{(4,2),(4,4),(4,5),(6,2),(6,4),(6,5)\}\)

\(\mathrm{f}=\{(1,1),(2,3),(0,-1),(-1,-3)\} \)

\(\mathrm{f}(\mathrm{x})=\mathrm{ax}+\mathrm{b}\)

\((1,1) \in \mathrm{f}\)

\(\Rightarrow \mathrm{f}(1)=1\)

\(\Rightarrow \mathrm{a} \times 1+\mathrm{b}=1\)

\( \Rightarrow \mathrm{a}+\mathrm{b}=1 \ldots . .(1)\)

\( (0,-1) \in \mathrm{f}\)

\(\Rightarrow \mathrm{f}(0)=-1\)

\(\Rightarrow \mathrm{a} \times 0+\mathrm{b}=-1\)

\(\Rightarrow \mathrm{b}=-1\)

On substituting \(b=-1\) in eqn (1), we get

\( a+(-1)=1 \)

\(\Rightarrow a=1+1=2\)

Thus the respective values of \(\mathrm a\) and \(\mathrm b\) are \(2\) and \(-1\).

We know that:

Let \(A\) and \(B\) be any two sets.

\(A \times B=\{(x, y)\) where \(x\) in \(A\) and \(y\) in \(B\}\)

Let \({A}\) and \({B}\) are two non-empty sets.

Consider, \(A=\{1,2,3\}\) and \(B=\{2,3\}\)

Here, in \({A}\) and \({B}\), there are 2 common elements.

Now,

\({A} \times {B}=\{(1,2),(1,3),(2,2),(2,3),(3,2),(3,3)\}\)

And, \({B} \times {A}=\{(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\}\)

In \(A \times B\) and \(B \times A\), there are \(4(=2^{2})\) i.e., common elements \(=n^{2}\)

Therefore, if \({A}\) and \({B}\) are two non-empty sets having \({n}\) elements in common, then \({n}^{2}\) common elements in the sets \(A \times B\) and \(B \times A\).

It is given that \((x, y) \in R \Leftrightarrow x

For the elements of the given sets \(A\) and \(B\), we find that

\(1<4,1<5,2<4,2<5,3<4,3<5\) and \(4<5 \)

\(\therefore  (1,4) \in R,(1,5) \in R,(2,4) \in R,(2,5) \in R,(3,4) \in R,(3,5) \in R \text { and }(4,5) \in R \)

Thus, \( R=\{(1,4),(1,5),(2,4),(2,5),(3,4),(3,5),(4,5)\}\)

\(f(x)\) is defined for all \(x\) satisfying

\(4-x \geq 0\) and \(x^2-1>0 \)

\(\Rightarrow  x-4 \leq 0 \text { and }(x-1)(x+1)>0 \)

\(\Rightarrow  x \leq 4 \text { and }(x<-1 \text { or } x>1)\)

\(\Rightarrow x \in(-\infty,-1) \cup(1,4]\)

Hence, Domain \((f)=(-\infty,-1) \cup(1,4]\).