NCERT Solutions for Class 11 Maths Chapter 2 Exercise 2.3 - Relations and Functions

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,
As 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, 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, 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 domain. For Example: f(x) = x + 1, now if we put x = 0, f(x) = 1. This 1 is a value of Range that we obtained.

Since f(x) is defined for $xϵR$, 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 - x² ≥ 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,
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,
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,
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 $t(C)=\frac{9C}{5}+32$ t (0)

Answer:

Given function is
$t(C)=\frac{9C}{5}+32$
Now,
$t(0)=\frac{9(0)}{5}+32=0+32=32$
Therefore,
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 $t(C)=\frac{9C}{5}+32$ 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,
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 $t(C)=\frac{9C}{5}+32$ 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,
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,
$212=\frac{9(C)}{5}+32$
$212\times 5=9(C)+160$
$9(C)=1060-160$
$C=\frac{900}{9}=100$
Therefore,
When t(C) = 212 , value of C is 100

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

$f(x)=2-3x,xϵR,x>0.$

Answer:

Given function is

$f(x)=2-3x,xϵR,x>0.$
It is given that $x>0$
Now,
$\Rightarrow 3x>0$
$\Rightarrow -3x<0$
Add 2 on both the 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 the 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