Differentiation (Part-38) | S N De

 

In the previous chapter,  we faced few questions of Relation and Mapping of S.N.Dey Maths and solved them. You can check them out in part-1 and part-2 of those solutions. Now , in this article we'll solve few more problems related to Relation and Mapping . So, without further ado, let's get started.

Very Short Answer Type Questions: 

3. Let, the function $\,f: \mathbb R \to \mathbb R\,\,$ be defined by :  

$f(x)= 1 , \text{for}\,\,x\in \mathbb Q \wedge -1 , \text{for}\,\,x\not\in \mathbb Q $

Sol. From the given definition of $\,f\,$ we have,   

$f(x)= 1 , \text{for}\,\,x\in \mathbb Q \wedge -1 , \text{for}\,\,x\not\in \mathbb Q $

Now, we know that $\,\,2,2.\overline{23} \in \mathbb Q, $ and           

$ \sqrt{2},\pi,e \not \in \mathbb Q$ and  hence $f(2)=f(2.\overline{23})=1,$ and  $f(\sqrt{2})=f(\pi)=f(e)=-1$ 

4. Find the domain for which the functions $\,\,f(x)=3x^2-2x,\,\,g(x)=9x-6\,\,$ are equal.

Sol. By the condition, $\,f(x)=g(x) \\ \Rightarrow 3x^2-2x=9x-6 \\ \Rightarrow 3x^2-9x-2x+6=0 \\ \Rightarrow 3x(x-3)-2(x-3)=0 \\ \Rightarrow (x-3)(3x-2)=0 \\ \Rightarrow x=3, \frac23$

Hence,the domain for which the functions $\,\,f(x)=3x^2-2x,\,\,g(x)=9x-6\,\,$ are equal is

$\,\, \{\frac23,3\}$

5. Let $\,\,\mathbb C\,$ be the set of complex numbers and the functions $\,\,f: \mathbb R \to \mathbb R,\,\, g: \mathbb C \to \mathbb C\,\, $ be defined by $\,\,f(x)=x^2,\,g(x)=x^2.$  

State the reason whether $\,\,f=g\,$ or not.

Sol. Since $\,x \in \mathbb R, f(x)=x^2 \geq 0,$ always.  

But  for $\,g: \mathbb C \to \mathbb C,$ let us take $\,x=i \in \mathbb C\,$ so that $g(x)=g(i)=i^2=-1.$ 

So, $g(x)\,$ is not always positive for $x \in \mathbb C.$  

Hence, $\, f \neq g $ 

6. Functions $\,\,f ,g\,$ are defined as follows : 

$\, f: \mathbb R-\{2\} \to \mathbb R,\,\,$ where $\,f(x)=\frac{x^2-4}{x-2}\,\,$ and $\,g : \mathbb R \to \mathbb R,$ where $\,g(x)=x+2.$

State with reasons whether $\,f=g\,$ or not. 

Sol. We know  $\,f=g\,$ only when Domain of $\,f\,=$ Domain of $\,g$   

and Range of $\,f=$Range of $\,g.$

Here, we see $\,2 \not \in \text{Domain of} \,f$ whereas $\,\,2 \in \text{Domain of}\,\, g$ .

Hence ,Domain of $\,f\, \neq$ Domain of $\,g$  

Similarly, $\,f(x)=\frac{(x+2)(x-2)}{(x-2)}=x+2$ but since $2 \not \in $ Domain of $\,f$, $4 \not \in $ Range of $f$ but $\,\,4 \in \text{Range of}\,\, g$   

as $g(2)=2+2=4$

Hence , Range of $\,f\, \neq$ Range of $\,g$  

So, we can conclude $\,f \neq g\,$

7. Let $\,f\,$ be a subset of $\,\,\mathbb Z \times \mathbb Z$ such that $\,\,f=\{(xy,x-y): x,y \in \mathbb Z\}.$  Is $\,f\,$ a mapping from $\,\,\mathbb Z\,\,$ into $\,\,\mathbb Z\,$ ? Give reasons for your answer.

Sol. Since $\,\, 0 \in \mathbb Z, \,\, (0 \times 0, 0-0)=(0,0) \in f. $  

Again, since $\,\,1,0 \in \mathbb Z,\,\,(1 \times 0, 1-0)=(0,1) \in f.$  

Thus we see that the ordered pairs $\,\,(0,0), (0,1)$ have the same first entry and they belong to $\,\,f.$ 

Therefore, $\,\,f\,\,$ does not define a function from $\,\,\mathbb Z \,\,$ into $\,\,\mathbb Z \,\,$.

## 8. Let $\,\,A=\{-2,-1,1,2\},\,\,B=\{3,4,5,6,7\}$ and $\,\,f,g,h\,\,$ be the rules associating elements of $\,\,A\,\,$ tp elements of $\,\,B\,\,$ as follows : $f(-1)=4, f(-2)=7, f(2)=5; \\ g(-1)=6, g(-2) = 7, g(1) = 3, g(2)=5, \\ g(1) = 4 \,\,\text{and}\,\, h(-1)=3, h(-2)=6, h(1) = 3, \\h(2)=4 .$ Prove that, $\,f\,$ and $\,g\,$ do not represent mappings but $\,h\,$ represents a mapping. Sol. From the given problem , we notice that $f(-1)=4, f(-2)=7, f(2)=5; \,\,$ which means $\,\,f(1)\,$ does not exist. Hence $\,f: A \to B\,$ does not represent mapping. Also, $\,g(-1)=6, g(-2) = 7, g(1) = 3,\\ g(2)=5, g(1) = 4 \, \\ \Rightarrow g(1)=3 \neq 4=g(1).$ So, $\,1\,$ has two different image under $\,\,g$ which is not possible by the definition of mapping. Hence, $\,g:A \to B\,$ does not represent mapping. But $h(-1)=3, h(-2)=6, h(1) = 3, h(2)=4 $ shows that each different element of $\,A\,$ has an image which are elements of the set $\,B\,$ under $\,h.\,$ So, $\,h:A \to B\,$ represents mapping.

9. Let $\,\,A=\{-2,-1,0,1,2\}$ and $\,\,f:A \to Z$ be defined by $\,\,f(x)=2x+1\, \forall x \in A.$ Find $\,\,f\,\,$ as a set of ordered pairs and find its range.

Sol. Since, $\,\,f(x)=2x+1\, \forall x \in A,$ we have , $\,f(-2)=2(-2)+1=-3 ,\, f(-1)=2(-1)+1=-1 ,\, f(0)=2(0)+1=1,\\f(1)=2 \times 1+1=3, f(2)= 2\times 2 +1=5 .$

So, $\,f\,$ as a set of ordered pairs is : $\,\,\{ (-2,-3),(-1,-1),(0,1),(1,3),(2,5)\}.$  

Range of $f$ is: $\{-3,-1,1,3,5\}$ 

10. Let $\,A=\{0,1,2,3,4\}$ and $\,\,f : A \to Z$ be defined by $\,\,f(x)=x^2-5x+2;$  

Find (i) the range of $\,f;$ (ii) pre-images of $\,\,(-2),1,2.$

Sol.  Since $\,\,f(x)=x^2-5x+2;\\ \Rightarrow f(0)=0^2-5(0)+2=2,\\ f(1)=1^2-5(1)+2=-2,\\f(2)=2^2-5(2)+2=-4,\\ f(3)=3^2-5(3)+2=-4,\\ f(4)=4^2-5(4)+2=-2.$  

Hence , the range $\,\,f=\{2,-2,-4\}. $   

To find the pre-image of $\, (-2),\,$  

We see $\,\,-2=x^2-5x+2 \\ \Rightarrow x^2-5x+4=0 \\ \Rightarrow (x-1)(x-4)=0 \\ \Rightarrow x=1,4$  

Hence , $\,\,1,4\, $  are the pre-images of $\,(-2).$

To find the pre-image of $\, 1,\,$  

We see $\,\,1=x^2-5x+2 \\ \Rightarrow x^2-5x+1=0  \cdots (1)$  

Hence , from (1), we can say  there is no  pre-images of $\,1.$

To find the pre-image of $\, 2,\,$  

We see $\,\,2=x^2-5x+2 \\ \Rightarrow x^2-5x=0 \\ \Rightarrow  x(x-5)=0 \\ \Rightarrow  x=0, [\text{Since,}\,\, x \neq 5 \not \in A]. \cdots (2)$  

Hence ,  from (2), we can say  $\,\,0\, $  is the pre-image of $\,2.$

11. Let $\,\,A=\{-2,-1,0,1,\frac32,2\},\,\,B=\{-6,-5,-3,0,3,4\}\,\,$ and $\,\,f : A \to B\,\,$ be defined by $\,\,f(x)=2x^2+x-6.\,$ Find $\,\,f(A).$ Is $\,\,f(x)=B\,\,?$ 

 Sol. #

$x (x \in A)$	$f(x)=2x^2+x-6$
$-2$	$0$
$-1$	$-5$
$0$	$-6$
$\frac{3}{2}$	$0$
$2$	$4$

Now, $\,\,f(A)=\{f(x): x \in A\} \\ \Rightarrow f(A)=\{0,-5,-6,4\}$  

 Clearly, $\,\,f(A) \neq B$

12. Let the function $\,\,f: \mathbb R \to \mathbb R\,\,$ be defined by, $\,\,f(x)=a^x \,( a >0, a \neq 1).$     

Find (i) range of $\,f\,$ (ii) $\,\,\{x: f(x)=1\}$ 

Also show that , $\,\,f(x+y)=f(x)f(y), \quad \forall x,y \in \mathbb R$

Sol. (i) Clearly , $\,\,a^x >0, \quad \forall x \in \mathbb R \,\,[ a >0] \\ \Rightarrow f(x) >0 \\ $

Hence, the range of $\,f\,=\{f(x) : x \in \mathbb R \}\\=\{y: y >0\}\\=(0,\infty)\\=\text{the set of all positive real numbers}$  

(ii) $\,\,f(x)=1 \\ \Rightarrow a^x=1 \\ \Rightarrow x=0.$         

Therefore, $\,\,\{x: f(x)=1\}=\{0\} \\ f(x+y)=a^{x+y}=a^x.a^y=f(x).f(y)$

13. Let $\,\,A=\{0,1\},B=\{2,6\}\,$ and $\,\,f: A \to B\,\,$ be given by ,$\,\,f(x)=6-4x\,\,\text{and}\,\, g: A \to B\,\,$ be given by, $\,\,g(x)=x^2-5x+6.\,$ State whether $\,\,f=g\,$ or not.

Sol. We see that $\,\,f(0)=6-4(0)=6,\,\,f(1)=6-4(1)=6-4=2 \\ g(0)=0^2-5(0)+6=6,\,\,g(1)=1^2-5(1)+6=2.$  

Hence, $\,\,f(0)=g(0)=6,\,\,f(1)=g(1)=2$ and so $\,\,f=g$

14. Find the image set of the domain of each of the following functions:  

(i) $\,f: \mathbb R \to \mathbb R\,\,$ given by,$\,\,f(x)=\cos{x}\,\,\forall x \in \mathbb R $

Sol. Here $\,f(x)=\cos x$  

Since $\,\,-1 \leq \cos x \leq 1, \text{so,}$  

 the range of $f(x) =[-1,1] .$  

Alternatively, we can write , the range of $\,\,f(x)=\{f(x)\in \mathbb R: |f(x)|\leq 1\}$ 

(ii) $\,f: \mathbb R \to \mathbb R\,\,$ given by,$\,\,f(x)=\csc{x}\,\,\forall x \in \mathbb R \\ (x \neq n\pi, n \in \mathbb Z)$

Sol. Here by $\,\,\csc x\,\,$ we mean cosec x.  

Now, $\,f(x)=\csc x =\frac{1}{\sin x}.$  

Since the range of $\,\,\sin x\,$ is $\,[-1,1],\,$  

therefore the range of $f(x) \\=(-\infty,-1) \cup (1, +\infty) \\ \implies \text{range of}\,\,\, f(x): \{f(x) \in \mathbb R: |f(x)|\geq 1\}$ 

(iii) $\,f: \mathbb R \to \mathbb R\,\,$ given by,$\,\,f(x)=\tan{x}\,\,\forall x \in \mathbb R $

Sol. Here $\,f(x)=\tan x$  

Since $\,\,\tan x \in (-\infty,+\infty) \,\,\,\,\text{so,}$  

 the range of $f(x) =(-\infty,+\infty).$  

Alternatively, we can write , the range of $\,\,f(x)=\mathbb R,\,\,\text{the set of all real numbers.}$ 

(iv) $\,g: \mathbb R \to \mathbb R\,\,$ given by,$\,\,g(x)=x^2+3\,\,\forall x \in \mathbb R $

Sol. Here $\,g(x)=x^2+3$  

Since $\,\,x^2+3 \geq 3 \,\,\,\,\text{as,}\, x^2\geq 0\,\,\forall x \in \mathbb R$  

 the range of $g(x) =[3,\infty)$  

Alternatively, we can write , the range of $\,\,g(x)=\{g(x) \in \mathbb R: g(x) \geq 3\}$

(v) $\,h: \mathbb R^{+} \to \mathbb R\,\,$ given by,$\,\,h(x)=\log x\,\,\forall x \in \mathbb R $

Sol. Since $\forall x\in \mathbb R^{+},\, \log x \in (-\infty,+\infty),$ 

the range of $\,\,\,\,h(x)=\mathbb R.$

15. Show that, $\,\,f= \{(1, 1), (-1, −5), (2, 4), (3, 7)\}\,\,\,$ defines a mapping. If this mapping is described by the rule $\,\,f(x) = px +q,$ then what values should be assigned to $\,p\,$ and $\,q\,$?

Sol. By the problem, $\,\,f(1)=1 \Rightarrow p+q=1 \cdots(1) \\ f(-1)=-5 \Rightarrow -p+q=-5\cdots(2).$

Solving (1) and (2), we get $\,\,p=3,\,\,q=-2$

16. If the mapping $\,\,\{(1, 2), (-1, 6), (2, 3), (3, 6)\}\,\,\,$ is described by the rule $\,\,f(x) = ax² + bx+c,\,\,$ then find the assigned values of $\,\,a,b,c.$

Sol.  From the given problem , $f(1)=2 \Rightarrow a+b+c=2 \cdots (1)\\ f(-1)=6 \Rightarrow a-b+c=6\cdots (2) \\f(2)=3 \Rightarrow 4a+2b+c=3 \cdots (3)\\f(3)=6 \Rightarrow 9a+3b+c=6 \cdots (4).$

Hence, solving (1),(2),(3),(4), we get $\,\,a=1,\,b=-2,\,c=3.$

17. Let $\,\,f\,\,$ be the subset of $\,\mathbb Z \times \mathbb Z\,\,$ such that $\,\,f=\{(xy,x-y): x ,y \in \mathbb Z\}.\,\,$ Is $\,f\,$ a mapping from $\,\mathbb Z\,$ into $\,\mathbb Z.\,$  

Give reasons for your answer. 

Sol. We see $\,\,f(3,4)=(4 \times 3,4-3)=(12,1) \\ f(6,2)=(6 \times 2,6-2)=(12,4).$ 

So, we see that the ordered pairs $\,\,\,(12,1),\,\,(12,4)\,\,$ have the same first entry and they belong to $\,\,f.$  

Thus, $\,f\,$ does not define a function from $\,\,\mathbb Z$ into $\,\,\mathbb Z$.

18.If $\,\,A=\{1,2,3,4,5\},\,\,$ show that the relation $\,f=\{(x,y):x+y=6\}, \forall x,y \in A\,$ , defines a mapping from $\,A\,$ to itself , but the relation $\,\,g=\{(x,y): y<x\}\,\,$ does not define a mapping in set $\,\,A$. 

Sol. Since $\,f=\{(x,y):x+y=6\}, \forall x,y \in A\, \\ f(1)=5,f(2)=4,f(3)=3,f(4)=2,f(5)=1.\cdots (1)$   

But  $\,\,g(1,2)\, $  does not exist as $\,\,2 \not<1 \cdots (2)$   

So, from (1) and (2), we can conclude that  the relation $\,f=\{(x,y):x+y=6\}, \forall x,y \in A\,$ , defines a mapping from $\,A\,$ to itself , but the relation $\,\,g=\{(x,y): y<x\}\,\,$ does not define a mapping in set $\,\,A$. 

19. Let $\,\,A = \{1, 2, 3, 4\}\,\,$ and $\,\,B = \{7, 8, 9\};\,\,$ then which of following relations is not a mapping from set $\,A\,$ to set $\,B\,$ 

(i)$\, R_1 = \{(1, 7), (2, 8), (1, 8), (4, 9)\} $

(ii)$\, R_2 = \{(1, 7), (2, 7), (3, 7), (4, 7)\}$

(iii)$\, R_3 = \{(1, 8), (2, 9), (3, 8), (4,8)\}$

(iv)$\,R_4=\{(1,7), (2, 8), (3, 7), (4, 8)\}$

Sol. For a relation to be  a mapping from set $\,A\,$ to set $\,B,\,$ each element of $\,A\,$ should be associated to some element of $\,B,\,$ .

(i) Dom. of $\,\,(R_1)=\{x:(x,y) \in R_1\}=\{1,2,4\}.$  

Clearly, $\,\,3 \in A$ does not have any image in $\,\,B\,$. So, $\,R_1\,$ is not mapping. 

(ii) Dom. of $\,\,(R_2)=\{x:(x,y) \in R_2\}=\{1,2,3,4\}=A.$  

Clearly, each element $ \in A$ does  have an image in $\,\,B\,$. So, $\,R_2\,$ is  mapping. 

(iii) Dom. of $\,\,(R_3)=\{x:(x,y) \in R_3\}=\{1,2,3,4\}=A.$  

Clearly, each element $ \in A$ does  have an image in $\,\,B\,$. So, $\,R_3\,$ is  mapping. 

(iv) Dom. of $\,\,(R_4)=\{x:(x,y) \in R_4\}=\{1,2,3,4\}=A.$  

Clearly, each element $ \in A$ does  have an image in $\,\,B\,$. So, $\,R_4\,$ is  mapping. 

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