Class 11 Maths Ncert Solutions Ex 2.3

Class 11 Maths Ncert Solutions Chapter 2 Ex 2.3

Q.1: Which of the given relations are functions? Answer with reason. If it is a function, find its range and domain.

(i) {(12, 11), (15, 11), (18, 11), (1, 11), (4, 11), (7, 11)}

(ii) {(12, 11), (14, 12), (16, 13), (18, 14), (0, 15), (2, 16), (4, 17)}

(iii) {(11, 13), (11, 15), (12, 15)}

 

Sol:

(i).   {(12, 11), (15, 11), (18, 11), (1, 11), (4, 11), (7, 11)}

Since, 12, 15, 18, 1, 4 and 7 are the elements of the domain of the given relation which have unique images.

Therefore, the given relation is a function.

Domain = {12, 15, 18, 1, 4, 7}

Range = {11}

 

(ii).   {(12, 11), (14, 12), (16, 13), (18, 14), (0, 15), (2, 16), (4, 17)}

Since, 12, 14, 16, 18, 0, 2 and 4 are the elements of the domain of the given relation which have unique images.

Therefore, the given relation is a function.

Domain = {12, 14, 16, 18, 0, 2, 4}

Range = {11, 12, 13, 14, 15, 16, 17}

(iii).  {(11, 13), (11, 15), (12, 15)}

Since, the same first element that is 11 corresponds to two images that is 13 and 15.

Therefore, the given relation is not a function.

 

 

Q.2: Find the range and domain of the given real function:

(i) f(y) = -|y|

(ii) f(y) = 9y2

 

Sol:

(i) f(y) = -|y|, y R

As we know:

|y| = {y,ify0y,ify<0

Therefore,

f(y) = -|y| = {y,ify0y,ify<0

Since, f(y) is defined for y R, the domain of ‘f’ is R.

The range of f(y) = -|y|is all the real number except positive real number.

Therefore, range of ‘f’ is (,0].

 

(ii) f(y) = 9y2

9y2 is defined for the real number which are greater than or equal to -3 and less than or equal to 3.

Therefore, Domain of f(y) = {y: -3 y 3} or [-3, 3].

For any value of ‘y’ such that 3y3, the value of f(y) will lie between 0 and 3.

Therefore, the Range of f(y) = {y: 0 y 3} or [0, 3]

 

 

Q.3: A function f is f(y) = 3y – 6. Find the values of the following:

(i) f(1)

(ii) f(8)

(iii) f(-2)

Sol:

The function of ‘f’ is:

f(y) = 3y – 6

(i) f(1) = (3 × 1) – 6

= 3 – 6 = -3

 

(ii) f(8) = (3 × 8) – 6

= 24 – 6= 18

 

(iii) f(-2)  = (3 × -2) – 6

= -6 – 6= -12

 

 

Q.4: The function ‘f’ which shows temperature in degree Celsius into temperature in degree Fahrenheit is expressed as: 9C5+32.

Find for the following values:

(i) f(0)

(ii) f(28)

(iii) f(-10)

(iv) The value of C, when f(C) = 212

 

Sol:

F = 9C5+32

(i) f(0):

= 9×05+32

= 0 + 32= 32

 

(ii) f(28):

= 9×285+32

= 252+1605

= 4125

 

(iii) f(-10):

= 9×105+32

= 9×(2)+32

= -18 + 32= 14

 

(iv) The value of C, when f(C) = 212:

212=9×C5+32

9×C5=21232

9×C5=180

9×C=180×5

C=180×59

i.e.  C=100

Therefore, The value of f, when f(C) = 212 is 100.

 

 

Q.5: Calculate range of the given functions:

(i) f(y) = 2 – 3y, y R, y > 0.

(ii) f(y) = y2+2, is a real no.

(iii) f(y)  = y, y is a real no.

Sol:

(i)  f(y)=23y,yR,y>0

We can write the value of f(y) for different real numbers x > 0 in tabular form as:

y 0.01 0.1 0.9 1 2 2.5 4 5
f(y) 1.97 1.7

0.7

1

4

5.5

10

13

Thus, we can clearly observe that the range of ‘f’ forms the set for all real numbers which are less than 2. 

i.e. Range of f = (,2)

Alternative:

Let, y > 0

3y > 0

2 – 3y < 2

i.e f(y) < 2

Therefore, Range of f = (,2)

 

(ii) f(y)=y2+2, y, is a real number.

We can write the value of f(y) for different real numbers x, in tabular form as:

y 0 ±0.3 ±0.8 ±1 ±2 ±3 . . .
f(y) 2 2.09 2.64 3 6 11 . . .

Thus, we can clearly observe that the range of f forms the set for all real numbers which are less than 2. 

i.e. Range of f = (,2)

Alternative:

Let ‘y’ be any real number. Then,

y20=>y2+20+2=>y2+22f(y)2

Therefore, Range of f = [2,)

 

(iii) f(y) = y, where y is a real number

Here, we can see that the range of f is the set of all the real numbers.

Therefore, Range of f = R