Relations and Functions Test - 67

We have, \(f(x)=\frac{x^2-4}{x-2}, x \neq 2\).

\(\Rightarrow  f(x)=\frac{(x-2)(x+2)}{x-2}=x+2\) for all \(x \neq 2\).

\(\Rightarrow  f(x)=g(x)\) for all \(x \neq 2\).

Thus, \(f(x)=g(x)\) for all \(x \in R-\{2\}\). 

But, \(f(x)\) and \(g(x)\) have different domains.

Infact, domain of \(f=R-\{2\}\) and domain of \(g=R\). 

Therefore, \(f \neq g\).

\(f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}\)

Domain of function \(f(x)=\frac{(x+1)^2}{x^2-6 x-2 x+12}\)

\(\Rightarrow f(x)=\frac{(x+1)^2}{x(x-6)-2(x-6)}\)

\(\Rightarrow f(x)=\frac{(x+1)^2}{(x-2)(x-6)}\)

It can be seen that the function \(f\) is defined for all real numbers except at \(\mathrm{x}=6\) and \(\mathrm{x}=2\).

Hence, the domain of the function is \(\mathrm{R}-\{2,6\}\).

We have,

\(f(x)=\sqrt{4-x^2}\)

Clearly, \(f(x)\) assumes real values for all \(x\) satisfying

\(4-x^2 \geq 0\) 

\(\Rightarrow-\left(x^2-4\right) \geq 0\) 

\(\Rightarrow x^2-4 \leq 0\) 

\(\Rightarrow(x-2)(x+2) \leq 0\) 

\(\Rightarrow x \in[-2,2]\)

Hence, Domain \((f)=\{-2,2\}\)

We have, \(A=\{a, b\}\) and \(B=\{1,2,3\}\)

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

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

Clearly, \((A \times B) \cap(B \times A)=\phi\).

Since no two ordered pairs in \(g\) have the same first component. 

So, \(g\) is a function such that \(g(1)=1, g(2)=3, g(3)=5\) and \(g(4)=7\).

It is given that \(g(x)=\alpha x+\beta\).

\(\therefore g(1)=1\) and \(g(2)=3\) 

In case of \(g(1)= \alpha \times 1 + \beta \)

\(\Rightarrow \alpha+\beta=1\cdots(i)\)

and in case of \(g(2)=\alpha \times 2 + \beta\)

\(\Rightarrow 2 \alpha+\beta=3 \cdots(ii)\)

On solving eq.(i) and (ii)

\( \alpha=2, \beta=-1\)

We have, \(f(x)=\frac{3}{2-x^2}\)

For \(f(x)\) to be real, we must have

\(2-x^2 \neq 0 \Rightarrow x \neq \pm \sqrt{2} \)

\(\therefore \text { Domain }(f)=R-\{-\sqrt{2}, \sqrt{2}\}\)

Let \(f(x)=y\). 

Then,

\(y=f(x) \Rightarrow y=\frac{3}{2-x^2}\) 

\(\Rightarrow 2 y-x^2 y=3\) 

\(\Rightarrow x^2 y=2 y-3\) 

\(\Rightarrow x= \pm \sqrt{\frac{2 y-3}{y}}\)

We observe that \(x\) will take real values other than \(-\sqrt{2}\) and \(\sqrt{2}\), if

\(\frac{2 y-3}{y}>0\) 

\(\Rightarrow y \in(-\infty, 0) \cup[\frac{3}{2}, \infty)\)

So, range \(f=(-\infty, 0) \cup[\frac{3}{2}, \infty)\)

Given:

\(A=\{1,2,3, \ldots, 14\}\) and \(R\) is a relation defined on \(A\) such that:

\(R=\{(x, y): 3 x-y=0\) where \(x, y \in A\}\)

When \(x=1 \in\), then:

\(y=3 x\)

\(=3 \times 1\)

\(=3 \in A\)

\(\Rightarrow(1,3) \in R\)

When \(x=2 \in\) A, then:

\(y=3 x\)

\(=3 \times 2\)

\(=6 \in A\)

\(\Rightarrow(2,6) \in R\)

When \(x=3 \in\) A, then:

\(y=3 x\)

\(=3 \times 3\)

\(=9 \in A\)

\(\Rightarrow(3,9) \in R\)

When \(x=4 \in\) A, then:

\(y=3 x\)

\(=3 \times 4\)

\(=12 \in A\)

\(\Rightarrow(4,12) \in R\)

So,

\(R=\{(1,3),(2,6),(3,9),(4,12)\}\)

As, we know that,

Range \((R)=\{b:(a, b) \in R\}\)

\(\Rightarrow\) Range \((R)=\{3,6,9,12\}\)

We have,

\(f(x)=x^2+1\).

\(\therefore  f(-1)=(-1)^2+1=2\), 

\(f(0)=0^2+1=1\), 

\(f(2)=2^2+1=5\) and \(f(4)=4^2+1=17\)

So, \(f=\{(x, f(x)): x \in A\}=\{(-1,2),(0,1),(2,5),(4,17)\}\).

Hence, Range of \((f)=\{2,1,5,17\}\)

We have, 

\(A=\{1,2,3\}\) 

And, \(B=\{x: x \in N, x\) is prime less than 5\(\}=\{2,3\}\)

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

\(=\{(1,2),(1,3),(2,2),(2,3),(3,2),(3,3)\} \)

Given:

\(A\) and \(B\) are the domain and range respectively for the relation \(R\) such that

\(R=\{(x, x+5): x \in\{0,1,2,3,4,5\}\}\)

So, the relation R can be re-written as:

\(R=\{(0,5),(1,6),(2,7),(3,8),(4,9),(5,10)\}\)

As we know that,

Domain of \((R)=\{a:(a, b) \in R\}\)

\(\Rightarrow {A}=\{0,1,2,3,4,5\}\)

We also know that,

Range of \((R)=\{b:(a, b) \in R\}\)

\(\Rightarrow B=\{5,6,7,8,9,10\}\)