Class 11 NCERT Solutions- Chapter 2 Relation And Functions – Exercise 2.3

Last Updated : 13 Jan, 2021

Question 1. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.

(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

(iii) {(1, 3), (1, 5), (2, 5)}

Solution:

(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

Here, each element in domain is having unique/distinct image. So, the given relation is a function.

Domain = {2, 5, 8, 11, 14, 17}

Range of the function = {1}

(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

Here, each element in domain is having unique/distinct image. So, the given relation is a function.

Domain = {2, 4, 6, 8, 10, 12, 14}

Range of function = {1, 2, 3, 4, 5, 6, 7}

(iii) {(1, 3), (1, 5), (2, 5)}

This relation is not a function since an element 1 corresponds to two elements/images i.e, 3 and 5.

Hence, this relation is not a function.

Question 2.Find the domain and range of the following real function:

(i) f(x) = â€“|x|

(ii) f(x) = âˆš(9 â€“ x2

Solution:

(i) Given,

f(x) = â€“|x|, x âˆˆ R

We know that,  |x| =

Here f(x) = -x =

As f(x) is defined for x âˆˆ R, the domain of f is R.

It is also seen that the range of f(x) = â€“|x| is all real numbers except positive real numbers.

Therefore, the range of f is given by (â€“âˆž, 0].

(ii) f(x) = âˆš(9 â€“ x2)

As âˆš(9 â€“ x2) is defined for all real numbers that are greater than or equal to â€“3 and less than or equal to 3, for 9 â€“ x2 â‰¥ 0.

|x| <=3

So, the domain of f(x) is {x: â€“3 â‰¤ x â‰¤ 3} or Domain of f = [â€“3, 3].

For any value of x in the range [â€“3, 3], the value of f(x) will lie between 0 and 3.

Therefore, the range of f(x) is {x: 0 â‰¤ x â‰¤ 3} or we can say Range of f = [0, 3].

(i) f(0), (ii) f(7), (iii) f(â€“3)

Solution:

Given, function, f(x) = 2x â€“ 5.

(i) f(0) = 2 Ã— 0 â€“ 5 = 0 â€“ 5 = â€“5

(ii) f(7) = 2 Ã— 7 â€“ 5 = 14 â€“ 5 = 9

(iii) f(â€“3) = 2 Ã— (â€“3) â€“ 5 = â€“ 6 â€“ 5 = â€“11

Find (i) t (0)      (ii) t (28)     (iii) t (â€“10)     (iv) The value of C, when t(C) = 212

Solution :

Here in ques , it is given that :

t(C) = 9C / 5 +32

So, (i) t(0) = 9(0) / 5 + 32

= 0 + 32

= 32

(ii) t(28) = 9(28) / 5 + 32

Taking LCM and solving ,

= ( 252 +160 ) / 5

= 412 / 5

(iii) t(-10) = 9(-10) / 5 + 32

= -18 + 32

= 14

(iv) Here , in this ques we have to find the value of C.

Given that , t(C) = 212,

9C / 5 + 32 = 212

9C / 5 = 180

9C = 180 X 5

C = 100

The value of C is 100.

Question 5. Find the range of each of the following functions.

(i) f(x) = 2 â€“ 3x, x âˆˆ R, x > 0.

(ii) f(x) = x2 + 2, x is a real number.

(iii) f(x) = x, x is a real number.

Solution:

(i) Given f (x) = 2 â€“ 3x, x âˆˆ R, x > 0

âˆµ x > 0 â‡’ -3x < 0 (Multiplying both sides by -3)

â‡’ 2 â€“ 3x < 2 + 0 â‡’ f (x) < 2

âˆ´ Hence, The range of f (x) is (-âˆž , 2).

(ii) Given f (x) = x2+ 2, x is a real number

We know x2â‰¥ 0 â‡’ x2+ 2 â‰¥ 0 + 2

â‡’ x2 + 2 > 2 âˆ´ f (x) â‰¥ 2

âˆ´ Hence, The range of f (x) is [2, âˆž).

(iii) Given f (x) = x, x is a real number.

Let y = f (x) = x â‡’ y = x

âˆ´ Range of f (x) = Domain of f (x)

âˆ´ Hence, Range of f (x) is R. (f(x) takes all real values)

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