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 $\frac{f(1.1)-f(1)}{(1.1-1)}$