NCERT Solutions for Class 11 Maths Chapter 2 Relations And Functions

Question:1 If $(\frac{x}{3}+1,y-\frac{2}{3})=(\frac{5}{3},\frac{1}{3})$ , find the values of x and y.

Answer:
It is given that
$(\frac{x}{3}+1,y-\frac{2}{3})=(\frac{5}{3},\frac{1}{3})$
Since the ordered pairs are equal, the corresponding elements will also be equal
Therefore,
$\frac{x}{3}+1=\frac{5}{3}\text{and}y-\frac{2}{3}=\frac{1}{3}$
$\Rightarrow \frac{x}{3}=\frac{5}{3}-1\text{and}y=\frac{1}{3}+\frac{2}{3}$
$\Rightarrow \frac{x}{3}=\frac{5-3}{3}\text{and}y=\frac{1+2}{3}$
$\Rightarrow \frac{x}{3}=\frac{2}{3}\text{and}y=\frac{3}{3}$
$\Rightarrow x=2\text{and}y=1$

Therefore, the values of x and y are 2 and 1, respectively.

Question:2: If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in $(A\times B)$.

Answer:
It is given that set A has 3 elements and the elements in set B are 3, 4, and 5
Therefore, the number of elements in set B is 3
Now,
Number of elements in $(A\times B)$
= ( Number of elements in set A ) $\times$ ( Number of elements in set B)
= 3 $\times$ 3
= 9
Therefore, the number of elements in $(A\times B)$ is 9.

Question:3 If G = {7, 8} and H = {5, 4, 2}, find $G\times HandH\times G$

Answer:

It is given that
G = {7, 8} and H = {5, 4, 2}
We know that the cartesian product of two non-empty sets P and Q is defined as
P $\times$ Q = {(p,q) , where p $\in$ P , q $\in$ Q }
Therefore,
G $\times$ H = {(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)}
And
H $\times$ G = {(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)}

Question:4 (i) State whether each of the following statements is true or false.
If the statement is false, rewrite the given statement correctly.
If P = {m, n} and Q = { n, m}, then $P\times Q$ = {(m, n),(n, m)}.

Answer:

This statement is FALSE.
If P = {m, n} and Q = {n, m}
Then,
$P\times Q=\{(m,m),(m,n),(n,m),(n,n)\}$

Question:4(ii): State whether each of the following statements is true or false. If the statement is false, rewrite the given statement correctly.
If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that $x\in A$ and $y\in B$

Answer:

It is a TRUE statement.

$\because$ If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that $x\in A$ and $y\in B$

Question:4 (iii) State whether each of the following statements is true or false. If the statement is false, rewrite the given statement correctly.
If A = {1, 2}, B = {3, 4}, then $A\times (B\cap \varphi )=\varphi$

Answer:

This statement is TRUE.

$\because$ If A = {1, 2}, B = {3, 4}, then

$B\cap \varphi =\varphi$

There for

$A\times (B\cap \varphi )=\varphi$

Question:5 If A = {–1, 1}, find $A\times A\times A$

Answer:

It is given that
A = {–1, 1}
A is a non-empty set.
Therefore,
First, find $A\times A$
$A\times A=\{-1,1\}\times \{-1,1\}=\{(-1,-1),(-1,1),(1,-1),(1,1)\}$
Now,
$A\times A\times A=A\times (A\times A)=\{-1,1\}\times \{(-1,-1),(-1,1),(1,-1),(1,1)\}$ $=\{(-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1),(1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1)\}$

Question:6 If $A\times B$ = {(a, x),(a , y), (b, x), (b, y)}. Find A and B.

Answer:

It is given that
$A\times B$ = {(a, x),(a , y), (b, x), (b, y)}
We know that the cartesian product of two non-empty sets P and Q is defined as:

$P\times Q=\{(p,q):p\in P,q\in Q\}$
Now, we know that A is the set of all first elements and B is the set of all second elements
Therefore,
$A=\{a,b\}andB=\{x,y\}$

Question:7 (i) Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that $A\times (B\cap C)=(A\times B)\cap (A\times C)$.

Answer:

It is given that
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
Now,
$B\cap C=\{1,2,3,4\}\cap \{5,6\}=\mathrm{\Phi }$
Now,
$A\times (B\cap C)=A\times \varphi =\varphi -(i)$

$A\times B=\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)\}$
And
$A\times C=\{(1,5),(1,6),(2,5),(2,6)\}$
Now,
$(A\times B)\cap (A\times C)=\varphi -(ii)$
From equations (i) and (ii), it is clear that,
$L.H.S.=R.H.S.$
Hence,
$A\times (B\cap C)=(A\times B)\cap (A\times C)$

Question:7 (ii) Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that $A\times C$ is a subset of $B\times D$

Answer:

It is given that
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
Now,
$A\times C=\{(1,5),(1,6),(2,5),(2,6)\}$
And
$B\times D=\{(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7),(3,8),(4,5),(4,6),(4,7),(4,8)\}$
We can observe that all the elements of the set $A\times C$ are the elements of the set $B\times D$.
Therefore, $A\times C$ is a subset of $B\times D$.

Question:8 Let A = {1, 2} and B = {3, 4}. Write $A\times B$. How many subsets will $A\times B$ have? List them.

Answer:

It is given that
A = {1, 2} and B = {3, 4}
Then,
$A\times B=\{(1,3),(1,4),(2,3),(2,4)\}$
$\Rightarrow n(A\times B)=4$
Now, we know that if C is a set with $n(C)=m$
Then,
$n[P(C)]=2^m$
Therefore,
The set $A\times B$ has $2^4=16$ subsets.

Question:9 Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in $A\times B$, find A and B, where x, y and z are distinct elements.
Answer:
It is given that
n(A) = 3 and n(B) = 2 and If (x, 1), (y, 2), (z, 1) are in A × B.

By definition of the Cartesian product of two non-empty sets P and Q, we get:
$P\times Q=\{(p,q):p\in P,q\in Q\}$
Now, we can see that
P = set of all first elements
And
Q = set of all second elements
Now,
$\Rightarrow$ (x, y, z) are elements of A and (1,2) are elements of B
As n(A) = 3 and n(B) = 2
Therefore,
A = {x, y, z} and B = {1, 2}

Question:10 The Cartesian product $A\times A$ has 9 elements among which are found (–1, 0) and (0,1). Find the set A and the remaining elements of $A\times A$

Answer:

It is given that the Cartesian product (A × A) has 9 elements, among which are found (–1, 0) and (0,1).
Now,
Number of elements in (A× B) = (Number of elements in set A) × (Number of elements in B)
$n(A\times A)=n(A)\times n(A)$
It is given that $n(A\times A)=9$
Therefore,
$n(A)\times n(A)=9$
$\Rightarrow n(A)=3$
Now,
By definition A × A = {(a, a): a ? A}
Therefore,
-1, 0 and 1 are the elements of set A
Now, because, n(A) = 3 therefore, A = {-1, 0, 1}
Therefore, the remaining elements of the set (A × A) are
(-1,-1), (-1,1), (0,0), (0, -1), (1,1), (1, -1) and (1, 0).

Question:1 Let A = {1, 2, 3,...,14}. Define a relation R from A to A by $R=\{(x,y):3x-y=0,wherex,y\in A\}$ .Write down its domain, co-domain and range.

Answer:
It is given that
$A=\{1,2,3,...,14\}andR=\{(x,y):3x-y=0,wherex,y\in A\}$
Now, the relation R from A to A is given as
$R=\{(x,y):3x-y=0,wherex,y\in A\}$
Therefore,
the relation in roaster form is, $R=\{(1,3),(2,6),(3,9),(4,12)\}$
Now,
We know that Domain of R = set of all first elements of the order pairs in the relation
Therefore,
Domain of $R=\{1,2,3,4\}$
And
Co-domain of R = the whole set A
i.e., Co-domain of $R=\{1,2,3,...,14\}$
Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
range of $R=\{3,6,9,12\}$

Question:2 Define a relation R on the set N of natural numbers by $R=(x,y):y=x+5,x$ is a natural number less than $4;x,y\in N\}$. Depict this relationship using roster form. Write down the domain and the range.

Answer:

It is given that
$R=(x,y):y=x+5,x$ is a natural number less than $4;x,y\in N\}$
As x is a natural number which is less than 4.
Therefore,
the relation in roaster form is, $R=\{(1,6),(2,7),(3,8)\}$
Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R=\{1,2,3\}$
Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
the range of $R=\{6,7,8\}$

Therefore, domain and the range are $\{1,2,3\}and\{6,7,8\}$ respectively

Question:3: A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by $R=\{(x,y):\text{the difference between x and y is odd};x\in A,y\in B\}$. Write R in roster form.

Answer:

It is given that
A = {1, 2, 3, 5} and B = {4, 6, 9}
And
$R=\{(x,y):$ the difference between $x$ and $y$ is odd $;x\in A,y\in B\}$
Now, it is given that the difference should be odd. Let us take all possible differences.
(1 - 4) = - 3, (1 - 6) = - 5, (1 - 9) = - 8(2 - 4) = - 2, (2 - 6) = - 4, (2 - 9) = - 7(3 - 4) = - 1, (3 - 6) = - 3, (3 - 9) = - 6(5 - 4) = 1, (5 - 6) = - 1, (5 - 9) = - 4
Taking the odd differences, we get,

Therefore,
the relation in roaster form, $R=\{(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)\}$

Question:4 (i) The Fig2.7 shows a relationship between the sets P and Q.
Write this relation in set-builder form.

Answer:
It is given in the figure that
P = {5,6,7}, Q = {3,4,5}
Therefore,
the relation in set builder form is,
$R=\left\{(x,y):y=x-2;x\in P\right\}$
Or,
$R=\left\{(x,y):y=x-2;forx=5,6,7\right\}$

Question:4 (ii) The Fig2.7 shows a relationship between the sets P and Q.
Write this relation roster form. What is its domain and range?

Answer:
From the given figure. we observe that

P = {5,6,7}, Q = {3,4,5}
And the relation in roaster form is , $R=\left\{(5,3),(6,4),(7,5)\right\}$

Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R=\left\{5,6,7\right\}$
Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
the range of $R=\left\{3,4,5\right\}$

Question:5 (i) Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by $\{(a,b):a,b\in A,\text{b is exactly divisible by a}\}$. Write R in roster form.

Answer:

It is given that
A = {1, 2, 3, 4, 6}
And$R=\{(a,b):a,b\in A,b$ is exactly divisible by $a\}$

Therefore,
the relation in roaster form is, $R=\left\{(1,1),(1,2),(1,3),(1,4),(1,6),(2,2),(2,4),(2,6),(3,3),(3,6),(4,4),(6,6)\right\}$

Question:5 (ii): Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by $\{(a,b):a,b\in A,\text{b is exactly divisible by a}\}$. Find the domain of R.

Answer:

It is given that
A = {1, 2, 3, 4, 6}
And
$R=\{(a,b):a,b\in A,bisexactlydivisiblebya\}$
Now,
Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R=\left\{1,2,3,4,6\right\}$

Question:5 (iii) Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by $\{(a,b):a,b\in A,b$ is exactly divisible by $a\}$. Find the range of R.

Answer:

It is given that
A = {1, 2, 3, 4, 6}
And
$R=\{(a,b):a,b\in A,b$ is exactly divisible by $a\}$
Now,
As the range of R = set of all second elements of the order pairs in the relation.
Therefore,
Range of $R=\left\{1,2,3,4,6\right\}$

Question:6 Determine the domain and range of the relation R defined by $R=\{(x,x+5):x\in \{0,1,2,3,,4,5\}\}$

Answer:
It is given that
$R=\{(x,x+5):x\in \{0,1,2,3,,4,5\}\}$

Therefore,
the relation in roaster form is, $R=\left\{(0,5),(1,6),(2,7),(3,8),(4,9),(5,10)\right\}$

Now,
Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R=\left\{0,1,2,3,4,5\right\}$

Now,
As Range of R = set of all second elements of the order pairs in the relation.
Range of $R=\left\{5,6,7,8,9,10\right\}$
Therefore, the domain and range of the relation R is $\{0,1,2,3,4,5\}and\left\{5,6,7,8,9,10\right\}$ respectively

Question: 7: Write the relation $R=\{(x,x^3):x$ is a prime number less than 10$\}$ in roster form.

Answer:
It is given that
$R=\{(x,x^3):x$ is a prime number less than 10$\}$
Now,
As we know, the prime numbers less than 10 are 2, 3, 5 and 7.
Therefore,
the relation in roaster form is, $R=\left\{(2,8),(3,27),(5,125),(7,343)\right\}$

Question:8 Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.

Answer:

It is given that
A = {x, y, z} and B = {1, 2}
Now,
$A\times B=\left\{(x,1),(x,2),(y,1),(y,2),(z,1),(z,2)\right\}$
Therefore,
$n(A\times B)=6$
Then, the number of subsets of the set $(A\times B)=2^n=2^6$

Therefore, the number of relations from A to B is $2^6$.

Question:9 Let R be the relation on Z defined by $R=\{(a,b):a,b\in Z,a-bisaninteger\}$
Find the domain and range of R.

Answer:

It is given that
$R=\{(a,b):a,b\in Z,a-bisaninteger\}$
Now, as we know, the difference between any two integers is always an integer.
And
Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
The domain of R = Z
Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore, the range of R = Z
Therefore, the domain and range of R is Z and Z respectively

Question:1 (i) Which of the following relations are functions? Give reasons.
If it is a function, determine its domain and range.
{(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}

Answer:
Since 2, 5, 8, 11, 14 and 17 are the elements of domain R having their unique images. Hence, this relation R is a function.
Now,
Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R=\left\{2,5,8,11,14,17\right\}$
Now,
As Range of R = set of all second elements of the order pairs in the relation.
Therefore,
Range of $R=\left\{1\right\}$
Therefore, the domain and range of R are $\left\{2,5,8,11,14,17\right\}and\left\{1\right\}$, respectively.

Question:1 (ii) Which of the following relations are functions? Give reasons.
If it is a function, determine its domain and range.
{(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}

Answer:
Since 2, 4, 6, 8, 10,12 and 14 are the elements of domain R having their unique images. Hence, this relation R is a function.
Now,
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R=\left\{2,4,6,8,10,12,14\right\}$
Now,
As Range of R = set of all second elements of the order pairs in the relation.
Therefore,
Range of $R=\left\{1,2,3,4,5,6,7\right\}$

Therefore, the domain and range of R are $\left\{2,4,6,8,10,12,14\right\}and\left\{1,2,3,4,5,6,7\right\},$ respectively

Question:1 (iii) Which of the following relations are functions? Give reasons.
If it is a function, determine its domain and range.
{(1,3), (1,5), (2,5)}.

Answer:
Since the same first element 1 corresponds to two different images 3 and 5.
Hence, this relation is not a function.

Question:2(i) Find the domain and range of the following real functions:
$f(x)=-|x|$

Answer:
Given function is
$f(x)=-|x|$
Now, we know that,
$|x|\{\begin{array}{cc}x& ifx>0\\ -x& ifx<0\end{array}$
$\Rightarrow f(x)=-|x|\{\begin{array}{cc}-x& ifx>0\\ x& ifx<0\end{array}$

Now, for a function f(x),
Domain: The values that can be put in the function to obtain real value. For example f(x) = x, now we can put any value in place of x and we will get a real value.
Hence, the domain of this function will be Real Numbers.
Range: The values that we obtain of the function after putting the value from the domain. For Example: f(x) = x + 1, now if we put x = 0, f(x) = 1.
This 1 is a value of the Range that we obtained.

Since f(x) is defined for $x\in R$, the domain of f is R.

It can be observed that the range of f(x) = -|x| is all real numbers except positive real numbers. Because will always get a negative number when we put a value from the domain.
Therefore, the range of f is $(-\mathrm{\infty },0]$

Question:2 (ii) Find the domain and range of the following real functions:
$f(x)=\sqrt{9-x^2}$

Answer:

Given function is
$f(x)=\sqrt{9-x^2}$
Now,
Domain: These are the values of x for which f(x) is defined.
For the given f(x), we can say that f(x) should be real and for that,9 - x 2 ≥ 0 [Since a value less than 0 will give an imaginary value]
$\Rightarrow 3^2-x^2=(3-x)(3+x)\ge 0$
$\Rightarrow -3\le x\le 3$
Therefore,
The domain of f(x) is $[-3,3]$
Now,
If we put the value of x from $[-3,3]$ we will observe that the value of function $f(x)=\sqrt{9-x^2}$ varies from 0 to 3
Therefore,
Range of f(x) is $[0,3]$

Question:3(i) A function f is defined by f(x) = 2x –5. Write down the values of f (0),

Answer:
Given function is
$f(x)=2x-5$
Now,
$f(0)=2(0)-5=0-5=-5$
Therefore, the value of f(0) is -5.

Question:3 (ii) A function f is defined by f(x) = 2x –5. Write down the values of f (7)

Answer:
Given function is
$f(x)=2x-5$
Now,
$f(7)=2(7)-5=14-5=9$
Therefore, the value of f(7) is 9.

Question:3(iii) A function f is defined by f(x) = 2x –5. Write down the values of f (-3)

Answer:

Given function is
$f(x)=2x-5$
Now,
$f(-3)=2(-3)-5=-6-5=-11$
Therefore, the value of f(-3) is -11.

Question:4(i) The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t(C)=\frac{9C}{5}+32$
Find t(0).

Answer:
Given function is
$t(C)=\frac{9C}{5}+32$
Now,
$t(0)=\frac{9(0)}{5}+32=0+32=32$
Therefore, the Value of t(0) is 32.

Question:4(ii) The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t(C)=\frac{9C}{5}+32$
Find t (28).

Answer:
Given function is
$t(C)=\frac{9C}{5}+32$
Now,
$t(28)=\frac{9(28)}{5}+32=\frac{252}{5}+32=\frac{252+160}{5}=\frac{412}{5}$
Therefore, the value of t(28) is $\frac{412}{5}$.

Question:4(iii) The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t(C)=\frac{9C}{5}+32$
Find t (-10).

Answer:
Given function is
$t(C)=\frac{9C}{5}+32$
Now,
$t(-10)=\frac{9(-10)}{5}+32=\frac{-90}{5}+32=-18+32=14$
Therefore, the value of t(-10) is 14.

Question:4(iv) The function ‘t’ which maps temperature in degree Celsius into temperature in
degree Fahrenheit is defined by $t(C)=\frac{9C}{5}+32$
The value of C, when t(C) = 212.

Answer:
Given function is
$t(C)=\frac{9C}{5}+32$
Now,
$212=\frac{9(C)}{5}+32$
$\Rightarrow 212\times 5=9(C)+160$
$\Rightarrow 9(C)=1060-160$
$\Rightarrow C=\frac{900}{9}=100$
Therefore,
When t(C) = 212, the value of C is 100.

Question:5 (i) Find the range of each of the following functions.

$f(x)=2-3x,x\in R,x>0.$

Answer:
Given function is:
$f(x)=2-3x,x\in R,x>0.$
It is given that $x>0$
Now,
$\Rightarrow 3x>0$
$\Rightarrow -3x<0$
Add 2 on both sides,
$\Rightarrow -3x+2<0+2$
$\Rightarrow 2-3x<2$
$\Rightarrow f(x)<2(\because f(x)=2-3x)$
Therefore,
Range of function $f(x)=2-3x$ is $(-\mathrm{\infty },2)$

Question:5 (ii) Find the range of each of the following functions:
$f(x)=x^2+2$, x is a real number.

Answer:
Given function is
$f(x)=x^2+2$
It is given that x is a real number
Now,
$\Rightarrow x^2\ge 0$
Add 2 on both sides
$\Rightarrow x^2+2\ge 0+2$
$\Rightarrow f(x)\ge 2(\because f(x)=x^2+2)$
Therefore,
Range of function $f(x)=x^2+2$ is $[2,\mathrm{\infty })$

Question:5 (iii) Find the range of each of the following functions.
f (x) = x, x is a real number

Answer:

Given function is
$f(x)=x$
It is given that x is a real number
Therefore, Range of function $f(x)=x$ is R.

Question:1 The relation f is defined by $f(x)=\{\begin{array}{cc}{x}^2& 0\le x\le 3\\ 3x& 3\le x\le 10\end{array}$ The relation g is defined by $g(x)=\{\begin{array}{cc}{x}^2& 0\le x\le 2\\ 3x& 2\le x\le 10\end{array}$ Show that f is a function and g is not a function.

Answer:
It is given that
$f(x)=\{\begin{array}{cc}{x}^2& 0\le x\le 3\\ 3x& 3\le x\le 10\end{array}$
Now,
$f(x)=x^{2}for0\le x\le 3$
And
$f(x)=3xfor3\le x\le 10$

At x = 3, $f(x)=x^2=3^2=9$
Also, at x = 3, $f(x)=3x=3\times 3=9$
We can see that for $0\le x\le 10$, f(x) has unique images.
Therefore, by definition of a function, the given relation is a function.
Now,
It is given that
$g(x)=\{\begin{array}{cc}{x}^2& 0\le x\le 2\\ 3x& 2\le x\le 10\end{array}$
Now,
$g(x)=x^{2}for0\le x\le 2$
And
$g(x)=3xfor2\le x\le 10$

At x = 2, $g(x)=x^2=2^2=4$
Also, at x = 2, $g(x)=3x=3\times 2=6$
We can clearly see that element 2 of the domain of relation g(x) corresponds to two different images, i.e. 4 and 6. Thus, f(x) does not have unique images
Therefore, by the definition of a function, the given relation is not a function.

Hence proved.

Question:2 If $f(x)=x^2$ find $\frac{f(1.1)-f(1)}{(1.1-1)}$

Answer:
Given function is:
$f(x)=x^2$
Now,
$\frac{f(1.1)-f(1)}{(1.1-1)}=\frac{(1.1{)}^2-1^2}{(1.1-1)}=\frac{1.21-1}{0.1}=\frac{0.21}{0.1}=2.1$

Therefore, the value of $\frac{f(1.1)-f(1)}{(1.1-1)}$ is 2.1.

Question:3 Find the domain of the function $f(x)=\frac{x^2+2x+1}{x^2-8x+12}$

Answer:

Given function is
$f(x)=\frac{x^2+2x+1}{x^2-8x+12}$
Now, we will simplify it into
$f(x)=\frac{x^2+2x+1}{x^2-8x+12}$
$=\frac{x^2+2x+1}{x^2-6x-2x+12}$
$=\frac{x^2+2x+1}{x(x-6)-2(x-6)}$
$=\frac{x^2+2x+1}{(x-2)(x-6)}$
Now, we can clearly see that $x\ne 2,6$.
Therefore, the Domain of f(x) is $(R-\{2,6\})$.

Question:4 Find the domain and the range of the real function f defined by $f(x)=\sqrt{(x-1)}$

Answer:

Given function is
$f(x)=\sqrt{(x-1)}$
We can clearly see that f(x) is only defined for the values of x, $x\ge 1$
Therefore,
The domain of the function $f(x)=\sqrt{(x-1)}$ is $[1,\mathrm{\infty })$
Now, as
$\Rightarrow x\ge 1$
$\Rightarrow x-1\ge 1-1$
$\Rightarrow x-1\ge 0$
Taking the square root on both sides, we get,
$\Rightarrow \sqrt{x-1}\ge 0$
$\Rightarrow f(x)\ge 0(\because f(x)=\sqrt{x-1})$
Therefore,
Range of function $f(x)=\sqrt{(x-1)}$ is $[0,\mathrm{\infty })$

Question:5 Find the domain and the range of the real function f defined by $f(x)=|x-1|$

Answer:
Given function is
$f(x)=|x-1|$
As the given function is defined of all real numbers.
The domain of the function $f(x)=|x-1|$ is R
Now, as we know that the mod function always gives only positive values.
Therefore,
Range of function $f(x)=|x-1|$ is all non-negative real numbers i.e. $[0,\mathrm{\infty })$

Question:6 Let $f=\{(x,\frac{x^2}{1+x^2}):x\in R\}$ R be a function from R into R. Determine the range of f.

Answer:
Given function is
$f=\{(x,\frac{x^2}{1+x^2}):x\in R\}$
Range of any function is the set of values obtained after the mapping is done in the domain of the function. So every value of the codomain that is being mapped is the Range of the function.
Let's take
$y=\frac{x^2}{1+x^2}$
$\Rightarrow y(1+x^2)=x^2$
$\Rightarrow y+y{x}^2=x^2$
$\Rightarrow y=x^2(1-y)$
$\Rightarrow x^2=\frac{y}{1-y}$
$\Rightarrow x=\pm \sqrt{\frac{y}{1-y}}$
Now, 1 - y should be greater than zero, and y should be greater than and equal to zero for x to exist because other than those values, the x will be imaginary
Thus, $1-y>0,y<1andy\ge 0$
Therefore, Range of the given function is $[0,1)$.

Question:7 Let f, g : R $\to$ R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find (f + g), (f – g) and $\frac{f}{g}$

Answer:
It is given that
$f,g:R\to R$
$f(x)=x+1andg(x)=2x-3$
Now,
$(f+g)x=f(x)+g(x)$
$=(x+1)+(2x-3)$
$=3x-2$
Therefore,
$(f+g)x=3x-2$
Now,
$(f-g)x=f(x)-g(x)$
$=(x+1)-(2x-3)$
$=x+1-2x+3$
$=-x+4$
Therefore,
$(f-g)x=-x+4$
Now,
$\left(\frac{f}{g}\right)x=\frac{f(x)}{g(x)},g(x)\ne 0$
$=\frac{x+1}{2x-3},x\ne \frac{3}{2}$
Therefore, values of $(f+g)x,(f-g)xand\left(\frac{f}{g}\right)x$ are $(3x-2),(-x+4)and\frac{x+1}{2x-3}$ respectively.

Question:8 Let f = {(1,1), (2,3), (0,–1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a and b.

Answer:
It is given that
$f=\left\{(1,1),(2,3),(0,-1),(-1,-3)\right\}$
And
$f(x)=ax+b$
Now,
At x = 1 , $f(x)=1$
$\Rightarrow f(1)=a(1)+b$
$\Rightarrow a+b=1-(i)$
Similarly,
At $x=0$ , $f(x)=-1$
$\Rightarrow f(0)=a(0)+b$
$\Rightarrow b=-1$
Now, put this value of b in equation (i), we get,
$a=2$
Therefore, the values of a and b are 2 and -1, respectively.

Question:9 (i) Let R be a relation from N to N defined by $R=\{(a,b):a,b\in Nanda=b^2\}$. Are the following true? $(a,a)\in R,$ for all $a\in N$

Answer:
It is given that
$R=\{(a,b):a,b\in Nanda=b^2\}$
And
$(a,a)\in R,$ for all $a\in N$
Now, it can be seen that $2\in N$ But, $2\ne 2^2=4$
Therefore, this statement is FALSE.

Question:9 (ii) Let R be a relation from N to N defined by $R=\{(a,b):a,b\in Nanda=b^2\}$. Are the following true? $(a,a)\in R,$ implies (b,a) $\in$ R

Answer:
It is given that
$R=\{(a,b):a,b\in Nanda=b^2\}$
And
$(a,b)\in R,$ implies $(b,a)\in R$
Now , it can be seen that $(2,4)\in R,$ and $4=2^2=4$ , But $2\ne 4^2=16$
Therefore, $(2,4)\notin N$
Therefore, the given statement is FALSE.

Question:9 (iii) Let R be a relation from N to N defined by $R=\{(a,b):a,b\in Nanda=b^2\}$. Are the following true? (a,b) $\in$ R, (b,c) $\in$ R implies (a,c) $\in$ R.

Answer:
It is given that
$R=\{(a,b):a,b\in Nanda=b^2\}$
And
$(a,b)\in R,(b,c)\in R$ implies $(a,c)\in R$
Now, it can be seen that $(16,4)\in R,(4,2)\in R$ because $16=4^2=16$ and $4=2^2=4$ , But $16\ne 2^2=4$
Therefore, $(16,2)\notin N$
Therefore, the given statement is FALSE.

Question:10(i) Let A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)}. Are the following true?
f is a relation from A to B. Justify your answer

Answer:
It is given that
$A=\left\{1,2,3,4\right\}$
$B=\left\{1,5,9,11,15,16\right\}$
and $f=\left\{(1,5),(2,9),(3,1),(4,5),(2,11)\right\}$
Now,
$A\times B=\{(1,1),(1,5),(1,9),(1,11),(1,15),(1,16),(2,1),(2,5),(2,9),(2,11),(2,15),(2,16),(3,1),(3,5),(3,9),(3,11),(3,15),(3,16),(4,1),(4,5),(4,9),(4,11),(4,15),(4,16)\}$
Now, a relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B
And we can see that f is a subset of $A\times B$
Hence, f is a relation from A to B.
Therefore, the given statement is TRUE.

Question:10 (ii) Let A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11).} Are the following true?
f is a function from A to B. Justify your answer

Answer:
It is given that
$A=\left\{1,2,3,4\right\}$
$B=\left\{1,5,9,11,15,16\right\}$
and $f=\left\{(1,5),(2,9),(3,1),(4,5),(2,11)\right\}$
Now,
$A\times B=\{(1,1),(1,5),(1,9),(1,11),(1,15),(1,16),(2,1),(2,5),(2,9),(2,11),(2,15),(2,16),(3,1),(3,5),(3,9),(3,11),(3,15),(3,16),(4,1),(4,5),(4,9),(4,11),(4,15),(4,16)\}$
As we can observe, the same first element, i.e., 2, corresponds to two different images, that is, 9 and 11.
Hence, f is not a function from A to B.
Therefore, the given statement is FALSE.

Question:11: Let f be the subset of $Z\times Z$ defined by $f=(ab,a+b):a,b\in Z$.
Is f a function from Z to Z? Justify your answer.

Answer:
It is given that
Now, we know that relation f from a set A to a set B is said to be a function only if every element of set A has a unique image in set B
Now, for value 2, 6, -2, -6 $\in Z$$f=(2\times 6,2+6),(-2\times -6,-2-6),(2\times -6,2-6),(-2\times 6,-2+6)$$\Rightarrow f=\{(12,8),(12,-8),(-12,-4),(-12,4)\}$
Now, we can observe that the same first element, i.e. 12, corresponds to two different images that are 8 and -8.
Thus, f is not a function.

Question:12 Let A = {9,10,11,12,13} and let f: A $\to$ N be defined by f (n) = the highest prime factor of n. Find the range of f.

Answer:
It is given that
A = {9,10,11,12,13}
And
f: A → N be defined by f(n) = the highest prime factor of n.
Now,
Prime factor of 9 = 3

Prime factor of 10 = 2,5