Class 11 Maths Ncert Solutions Chapter 2 Ex 2.4 Relations & Functions PDF

# Class 11 Maths Ncert Solutions Ex 2.4

## Class 11 Maths Ncert Solutions Chapter 2 Ex 2.4

Q-1:  The relation ‘m’ is defined by:

m (y) = y2,  0y5$0\leq y\leq 5$

= 5y,  5y30$5\leq y\leq 30$

The relation ‘n’ is defined by

n (y) = y2,  0y4$0\leq y\leq 4$

= 5y,  4y30$4\leq y\leq 30$

Now, prove that ‘m’ is a function and ‘n’ is not a function.

Sol:

Here,

m (y) = y2,  0y5$0\leq y\leq 5$

= 5y, 5y30$5\leq y\leq 30$

Now, 0y5$0\leq y\leq 5$,   m(y) = y2

And, 5y30$5\leq y\leq 30$,   m(y) = 5y

Now, at y = 5, m(y) = 52 = 25 or m(y) = 5 × 5 = 25

i.e., at y = 5, m(y) = 25

Therefore, for 0y30$0\leq y\leq 30$ , the images of m(y) are unique. Thus, the given relation is a function.

Now, n(y) = y2,   0y4$0\leq y\leq 4$

= 5y,   4y30$4\leq y\leq 30$

Now, at y = 4, m (y) = 42 = 16 or m(y) = 5 × 4 = 20

Thus, element 4 of the domain 0y30$0\leq y\leq 30$ of relation ‘n’ has 2 different images i.e., 16 and 20

Therefore, this relation is not a function.

Q-2: If g(y) = y2 then, Find g(1.2)g(1)(1.21)$\frac{g(1.2) – g(1)}{(1.2 – 1)}$

Sol:

Here, g(y) = y2

Therefore, g(1.2)g(1)(1.21)$\frac{g(1.2) – g(1)}{(1.2 – 1)}\\$

=(1.2)2(1)2(1.21)=1.4410.2$= \frac{(1.2)^{2} – (1)^{2}}{(1.2 – 1)}= \frac{1.44 – 1}{0.2}$ =0.440.2=2.2$= \frac{0.44}{0.2} = 2.2$

Q-3: Find the domain for the function given below:

g(y)=y22y+1y29y+20$g(y) = \frac{y^{2} – 2y + 1}{y^{2} – 9y + 20}$

Sol:

Here, g(y)=y22y+1y29y+20$g(y) = \frac{y^{2} – 2y + 1}{y^{2} – 9y + 20}\\$

=y22y+1(y5)(y4)$= \frac{y^{2} – 2y + 1}{(y – 5)(y – 4)}\\$

Now, it clear from above equation that the function ‘g’ is defined for all real numbers except ‘y = 4’ and ‘y = 5’.

Therefore, the required domain is R: {4, 5}

Q-4: Find the range and domain of the function given below:

g(y)=(y5)$g(y) = \sqrt{(y – 5)}$

Sol:

Here, g(y)=(y5)$g(y) = \sqrt{(y – 5)}$ is the given function.

So, it is clear that the function is defined for y5$y\geq 5$.

So, the domain will be the set of all real numbers greater than or equal to 5. i.e., The domain for g(y) is [5,)$[5,\infty )$

Now, for range of the given function, we have:

y5$y\geq 5$ (y5)0$\Rightarrow (y – 5)\geq 0$ (y5)0$\Rightarrow \sqrt{(y – 5)} \geq 0\\$

Therefore, the range of g(y) is the set of all real numbers greater than or equal to 0.

i.e, the range of g(y) is [0,)$[0,\infty )$.

Q-5: Find the range and domain of the function:   g(y) = |y – 4|

Sol:

Here |y – 4| is the given function.

So, it is clear that the function is defined for all the real numbers.

The domain for g(y) is R

Now, for range of the given function, we have:

yϵR,$y\epsilon R ,$, |y – 4| assumes for all real numbers.

Therefore, the range of g(y) is the set of all non- negative real numbers.

Q-6: A function from ‘R into R’ is given below. Find the range of ‘g’.

g=[(y,y21+y2);yϵR]$g = \left [ \left ( y, \frac{y^{2}}{1 + y^{2}} \right ); y\epsilon R \right ]$

Sol:

g=[(y,y21+y2);yϵR]$g = \left [ \left ( y, \frac{y^{2}}{1 + y^{2}} \right ); y\epsilon R \right ]\\$

=[(0,0),(±0.5,15),(±1,12),(±1.5,913),(±2,45),(3,910),.]$[(0,0), (\pm 0.5, \frac{1}{5}), (\pm 1, \frac{1}{2}),(\pm 1.5, \frac{9}{13}),(\pm 2, \frac{4}{5}), (3, \frac{9}{10}),…….]\\$

Thus, the range of ‘g’ is the set of all 2nd elements. It can be seen that all these elements are:    0but<1$\geq 0\;but\;< 1$.

Therefore, the range of ‘g’ is = [0,1)

Q-7: Assume that function ‘m’ and ‘n’ is defined from: R$\rightarrow$R.

m (y) = y + 2, n(y) = 3y – 2

Find m + n, m – n and mn$\frac{m}{n}$

Sol:

Here, m(y) = y + 2 and n(y) = 3y – 2 are defined from R$\rightarrow$R.

Now, (m + n) (y) = m(y) + n(y) = (y + 2) + (3y – 2) = 4y

Therefore, (m + n) (y) = 4y

(m – n) (y) = m(y) – n(y) = (y + 2) – (3y – 2) = -2y + 4

Therefore, (m – n) (y) = -2y + 4

(mn)(y)=m(y)n(y),n(y)0,yϵR$(\frac{m}{n})(y) = \frac{m(y)}{n(y)}, n(y)\neq 0, y\epsilon R\\$

Therefore, (mn)(y)=y+23y2,3y20,3y2$(\frac{m}{n})(y) = \frac{y + 2}{3y – 2}, 3y – 2 \neq 0, 3y\neq 2\\$

(mn)(y)=y+23y2,y23$(\frac{m}{n})(y) = \frac{y + 2}{3y – 2}, y\neq \frac{2}{3}$

Q-8: Let g = {(1, 1), (2, 3), (0, -1), (-1, -3)} be a function from ‘Z to Z’ defined by g(y) = uy + v, for some integers u,v. Find u,v.

Sol:

g = {(1, 1), (2, 3), (0, -1), (-1, -3)} and g(y) = uy + v

(1,1)ϵgg(1)=1u×1+v=1$(1, 1) \;\epsilon g \Rightarrow g(1) = 1 \Rightarrow u\times 1 + v = 1$ u+v=1$\Rightarrow u + v = 1$ (0,1)ϵgg(0)=1u×0+v=1$(0,-1)\;\epsilon g \Rightarrow g(0) = -1 \Rightarrow u\times 0 + v = -1$ v=1$\Rightarrow v = -1$

By putting v = -1 in u + v = 1, we get

u = 2.

Therefore, u = 2 and v = -1.

Q-9: Let ‘f’ be a relation from ‘N to N’ defined by f = {(x,y): x,yϵN$x,y \epsilon N$ and x = y2}. Find out which of the following is true and which one is false.

1.(x,y)ϵf,(y,z)ϵf(x,z)ϵf$(x,y) \epsilon f, (y,z)\epsilon f \Rightarrow (x,z)\epsilon f$.

2.(x,x)ϵf,forallxϵN$(x,x) \epsilon f, \;for\;all\;x\epsilon N$

3.(x,y)ϵf(y,x)ϵf$(x,y) \epsilon f \Rightarrow (y,x)\epsilon f$

Sol:

f = {(x,y): x,yϵN$x,y \epsilon N$ and x = y2}

(1). Now, take (4,2)ϵf,(25,5)ϵfbecause4,2,25,5ϵN$(4, 2)\epsilon f, (25, 5)\epsilon f\;because\; 4,2,25,5\epsilon N$ and 4 = 22 and 16 = 42.

Therefore, the given statement is true.

(2). Now, let 3ϵNbut332=9$3\epsilon N\;but\;3\neq 3^{2} = 9$

Therefore, the statement is false.

(3). (16,4)ϵNbecause16,4ϵN$(16, 4)\epsilon N\;because\;16,4\epsilon N$ and 16 = 42.

Now, 4162=256;$4\neq 16^{2} = 256;$ therefore (4, 16) does not belongs to N.

Therefore, the statement is false.

Q-10: Assume U = {1, 2, 3, 4}, V = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2,9), (3,1), (4, 5), (2, 11)}. Find out which of the following is true and which one is false.

(1). ‘f’ is a function from U to V.

(2). ‘f’ is a relation from U to V.

Sol:

Here, U = {1, 2, 3, 4} and V = {1, 5, 9, 11, 15, 16}

Therefore, U x V = {(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)}

f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

(1). As, the 1st member i.e. ‘2’ corresponds to 2 different images i.e. ‘9’ and ‘11’, Therefore, ‘f’ is not a function.

(2). A relation of non-empty set ‘U’ to a non-empty set ‘V’ is a subset of the Cartesian product U x V.

Also it can be seen that ‘f’ is a subset of U x V.

Therefore, ‘f’ is a relation from U to V.

Q-11: Assume ‘g’ be the subset of ‘Z to Z’ defined by f = {(xy, x + y): x,yϵZ$x, y \epsilon Z$}. Is ‘g’ a function from ‘Z to Z’, also justify your answer.

Sol:

Here,

f = {(xy, x + y): x,yϵZ$x, y \epsilon Z$}

As we know that a relation g from set X to a set Y is said to be a function only if every element of set X is having a unique images in set Y.

Since, 4,12,4,12ϵZ$4, 12, -4, -12 \;\epsilon \;Z$

[4×12,4+12],[(4)×(12),4+(12)]ϵg$[4\times 12, 4 + 12], [(-4)\times (-12),-4 + (-12)]\;\epsilon \;g$

i.e.  [ (48, 16), (48, -16) ] ϵg$\epsilon\; g$

Here, the same 1st member ‘48’ is having 2 images ‘16’ and ‘-16’.

Therefore, relation ‘g’ is not a function from ‘Z to Z’.

Q-12: Assume ‘X’ = {5,7,9, 10, 11, 12, 13} and let ‘g’: X$\rightarrow$N be defined by g(n) = The highest prime factor of ‘n’. Find the range of ‘g’.

Sol:

X = {5, 7, 9, 10, 11, 12, 13, 14, 15};

g: X$\rightarrow$N be defined by g(n) = The highest prime factor of ‘n’

Prime factor of 5 = 5

Prime factor of 7 = 7

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

Prime factor of 14 = 2, 7

Prime factor of 15 = 3, 5

Therefore,

g(5) = The highest prime factor of 5 = 5

g(7) = The highest prime factor of 7 = 7

g(9) = The highest prime factor of 9 = 3

g(10) = The highest prime factor of 10 = 5

g(11) = The highest prime factor of 11 = 11

g(12) = The highest prime factor of 12 = 3

g(13) = The highest prime factor of 13 = 13

g(14) = The highest prime factor of 14 = 7

g(15) = The highest prime factor of 15 = 5

Thus, the range of ‘g’ is the set of all ‘g(n)’, where nϵX$n\epsilon X$.

Therefore, Range of g = {3, 5, 7, 11, 13}