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

          = 5y,  5y30

The relation ‘n’ is defined by

n (y) = y2,  0y4

         = 5y,  4y30

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

 

Sol:

Here,

m (y) = y2,  0y5

= 5y, 5y30

Now, 0y5,   m(y) = y2

And, 5y30,   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 , the images of m(y) are unique. Thus, the given relation is a function.

Now, n(y) = y2,   0y4

= 5y,   4y30

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

Thus, element 4 of the domain 0y30 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)

 

Sol:

Here, g(y) = y2

Therefore, g(1.2)g(1)(1.21)

=(1.2)2(1)2(1.21)=1.4410.2 =0.440.2=2.2

 

 

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

g(y)=y22y+1y29y+20

Sol:

Here, g(y)=y22y+1y29y+20

=y22y+1(y5)(y4)

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)

Sol:

Here, g(y)=(y5) is the given function.

So, it is clear that the function is defined for y5.

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,)

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

y5 (y5)0 (y5)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,).

 

 

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 – 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]

 

Sol:

g=[(y,y21+y2);yϵR]

=[(0,0),(±0.5,15),(±1,12),(±1.5,913),(±2,45),(3,910),.]

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

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

 

 

Q-7: Assume that function ‘m’ and ‘n’ is defined from: RR.

m (y) = y + 2, n(y) = 3y – 2

Find m + n, m – n and mn

 

Sol:

Here, m(y) = y + 2 and n(y) = 3y – 2 are defined from RR.

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

Therefore, (mn)(y)=y+23y2,3y20,3y2

(mn)(y)=y+23y2,y23

 

 

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 u+v=1 (0,1)ϵgg(0)=1u×0+v=1 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 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.

2.(x,x)ϵf,forallxϵN

3.(x,y)ϵf(y,x)ϵf

Also justify your answer.

 

Sol:

f = {(x,y): x,yϵN and x = y2}

(1). Now, take (4,2)ϵf,(25,5)ϵfbecause4,2,25,5ϵN and 4 = 22 and 16 = 42.

Therefore, the given statement is true.

(2). Now, let 3ϵNbut332=9

Therefore, the statement is false.

(3). (16,4)ϵNbecause16,4ϵN and 16 = 42.

Now, 4162=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.

Justify your answer.

 

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}. Is ‘g’ a function from ‘Z to Z’, also justify your answer.

 

Sol:

Here,

f = {(xy, x + y): x,yϵZ}

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],[(4)×(12),4+(12)]ϵg

i.e.  [ (48, 16), (48, -16) ] ϵg

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’: XN 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: XN 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.

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