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

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

Answer:

It is given that
$f (x) = \left\{\begin{matrix} x^2 & 0 \leq x\leq 3 \\ 3x &3 \leq x \leq 10 \end{matrix}\right.$
Now,
$f(x) = x^2 \ for \ 0\leq x\leq 3$
And
$f(x) = 3x \ for \ 3\leq x\leq 10$

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 $0\leq x\leq 10$, f(x) has unique images.

Therefore, by definition of a function, the given relation is a function.

Now,
It is given that
$g(x) = \left\{\begin{matrix} x^2 & 0 \leq x\leq 2 \\ 3x &2 \leq x \leq 10 \end{matrix}\right.$
Now,
$g(x) = x^2 \ for \ 0\leq x\leq 2$
And
$g(x) = 3x \ for \ 2\leq x\leq 10$

At x = 2, $g(x) = x^2 = 2^2 = 4$
Also, at x = 2, $g(x) = 3x = 3\times2 = 6$
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 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:

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

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

Answer:

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

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

Answer:

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

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 for 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. $[0,\infty)$

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

Answer:

The given function is
$f = \left \{ \left ( x , \frac{x^2}{1+ x^2} \right ) : x \epsilon R \right \}$
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 Range of the function.
Let's take
$y = \frac{x^2}{1+x^2}$
$\Rightarrow y(1+x^2)=x^2$
$\Rightarrow y+yx^2=x^2$
$\Rightarrow y=x^2(1-y)$
$\Rightarrow x^2= \frac{y}{1-y}$
$\Rightarrow x= \pm \sqrt{\frac{y}{1-y}}$
Now, 1-y should be greater than zero and y should be greater than or equal to zero for x to exist, because other than those values the x will be imaginary
Thus, $1 - y > 0 , y < 1 \ and \ y \geq 0$
Therefore,
Range of given function is $[0,1)$

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

Answer:

It is given that
$f,g : R \rightarrow R$
$f(x)=x+1 \ \ and \ \ g(x) = 2x - 3$
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,
$\left ( \frac{f}{g} \right )x = \frac{f(x)}{g(x)} , g(x)\neq 0$
$=\frac{x+1}{2x-3} \ , x \neq \frac{3}{2}$
Therefore, values of $(f+g)x,(f-g)x \ and \ \left ( \frac{f}{g} \right )x$ 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, 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 will 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 $R = \left \{ ( a,b): a,b \epsilon N \: \:and \: \: a = b ^ 2 \right \}$ . Are the following true?

$( a,a ) \epsilon R ,$ for all $a \epsilon N$

Answer:

It is given that
$R = \left \{ ( a,b): a,b \epsilon N \: \:and \: \: a = b ^ 2 \right \}$
And
$( a,a ) \epsilon R ,$ for all $a \epsilon N$
Now, it can be seen that $2 \ \epsilon \ N$ But, $2 \neq 2^ 2 = 4$
Therefore, this statement is FALSE

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

$( a,a ) \epsilon R ,$ implies (b,a) $\epsilon$ R

Answer:

It is given that
$R = \left \{ ( a,b): a,b \epsilon N \: \:and \: \: a = b ^ 2 \right \}$
And
$( a,b ) \ \epsilon \ R ,$ implies $(b,a) \ \epsilon \ R$
Now , it can be seen that $( 2,4 ) \ \epsilon \ R ,$ and $4 = 2^2 = 4$ , But $2 \neq 4^2 =16$
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 $R = \left \{ ( a,b): a,b \epsilon N \: \:and \: \: a = b ^ 2 \right \}$ . Are the following true?

(a,b) $\epsilon$ R, (b,c) $\epsilon$ R implies (a,c) $\epsilon$ R.

Answer:

It is given that
$R = \left \{ ( a,b): a,b \epsilon N \: \:and \: \: a = b ^ 2 \right \}$
And
$( a,b ) \ \epsilon \ R , (b,c) \ \epsilon \ R$ implies $(a,c) \ \epsilon \ R$
Now, it can be seen that $(16,4) \ \epsilon \ R , ( 4,2 ) \ \epsilon \ R$ because $16 = 4^2 = 16$ and $4 = 2^2 = 4$ , But $16 \neq 2^2 =4$
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 $f = \{ (ab, a + b) : a, b \epsilon Z \}$ . Is f a function from Z to Z? Justify your answer.

Answer:

It is given that
$f = \{ (ab, a + b) : a, b \epsilon Z \}$
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 $\epsilon \ Z$
$f = \left \{ (2 \times 6, 2 + 6), (-2 \times -6, -2 - 6), (2 \times -6, 2 - 6), (-2 \times 6, -2 + 6) \right \}$
$\Rightarrow f = \left \{ (12, 8), (12, -8), (-12, -4), (-12, 4) \right \}$
Now, we can observe that 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$\rightarrow$ 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

Prime factor of 11 = 11

Prime factor of 12 = 2,3

Prime factor of 13 = 13

f(n) = the highest prime factor of n.

Hence,

f(9) = the highest prime factor of 9 = 3

f(10) = the highest prime factor of 10 = 5

f(11) = the highest prime factor of 11 = 11

f(12) = the highest prime factor of 12 = 3

f(13) = the highest prime factor of 13 = 13

As the range of f is the set of all f(n), where $n \ \epsilon \ A$

Therefore, the range of f is: {3, 5, 11, 13}.