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}

$|x|\left\{\begin{matrix} x &if \ x> 0 \\ -x& if \ x<0 \end{matrix}\right.$
$\Rightarrow f(x)=-|x|\left\{\begin{matrix} -x &if \ x> 0 \\ x& if \ x<0 \end{matrix}\right.$

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 $x \ \epsilon \ R$, 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 $(-\infty , 0]$

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]
$\Rightarrow 3^2-x^2= (3-x)(3+x)\geq 0$
$\Rightarrow -3\leq x\leq 3$
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 $t ( C ) = \frac{9 C }{5} + 32$ t (0)

Answer:

Given function is
$t ( C ) = \frac{9 C }{5} + 32$
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 $t ( C ) = \frac{9 C }{5} + 32$ t (28)

Answer:

Given function is
$t ( C ) = \frac{9 C }{5} + 32$
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 $t ( C ) = \frac{9 C }{5} + 32$ t (-10)

Answer:

Given function is
$t ( C ) = \frac{9 C }{5} + 32$
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 $t ( C ) = \frac{9 C }{5} + 32$The value of C, when t(C) = 212.

Answer:

Given function is
$t ( C ) = \frac{9 C }{5} + 32$
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.

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

Answer:

Given function is

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

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,
$\Rightarrow x^2 \geq 0$
Add 2 on both the sides
$\Rightarrow x^2+2 \geq 0+2$
$\Rightarrow f(x) \geq 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because f(x) = x^2+2)$
Therefore,
Range of function $f ( x ) = x ^2 +2$ is $[2,\infty)$

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