# Ex 2.3 sets ncert maths solution class 11

# Â EXERCISE 2.3 (Class 11)

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

In this relation no two ordered pair have same first element. Hence this relation is function

Here, domain = {2, 5, 8, 11, 14, 17} and range = {1}

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

In this relation no two ordered pair have same first element. Hence this relation is function

Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}

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

In this relation (1, 3), (1, 5) have first element same, Therefore this relation is not a function.

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

**(i)Â f(x) = â€“|x| (ii) f(x) = âˆš(9 â€“ x^{2})Â **

**Solution:** **(i)**Â Given,

*f*(*x*) = â€“|*x*|,Â *x*Â âˆˆÂ R

AsÂ *f*(*x*) is defined forÂ *x*Â âˆˆÂ R, the domain ofÂ *f*Â =Â R(Real number).

We know that

|x| âˆˆ [0, âˆž) â‡’ 0 â‰¤ |x |< âˆž

â‡’ â€“âˆž < -|x| â‰¤ 0

â‡’ -|x| âˆˆ (-âˆž, 0]

â‡’ f(x) âˆˆ (-âˆž, 0]

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

**(ii)Â **f(x) = âˆš(9 â€“ x^{2})

We know that

9 â€“ xÂ² â‰¥ 0 â‡’ (3 – x)(3 + x) â‰¥ 0

â‡’ 3 â€“ x â‰¥ 0 and 3 + x â‰¥ 0

â‡’ x â‰¤ 3 andÂ x â‰¤ -3

â‡’ x âˆˆ [-3, 3]

So, the domain ofÂ *f*(*x*) is {*x*: â€“3 â‰¤Â *x*Â â‰¤ 3} or [â€“3, 3].

Now,

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

Therefore, the range ofÂ *f*(*x*) is {*x*: 0 â‰¤Â *x*Â â‰¤ 3} or [0, 3].

**3. A functionÂ fÂ is defined byÂ f(x) = 2xÂ â€“ 5. Write down the values of**

**(i)Â f(0), (ii)Â f(7), (iii)Â f(â€“3)**

**Solution: **Given,

Function,Â *f*(*x*) = 2*x*Â â€“ 5

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

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

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

**4. The function â€˜ tâ€™, which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by **.

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

**Solution: Given function**

**(i)**

**(ii)**

**(iii)**

**(iv)**Â t(c) = 212

Therefore

Therfore the value of t(c) when t(c) = 212 is 100

**5. Find the range of each of the following functions:**

**(i)Â f(x) = 2 â€“ 3x,Â xÂ âˆˆÂ R,Â xÂ > 0**

**(ii)Â f(x) =Â x^{2}Â + 2,Â xÂ is a real number**

**(iii)Â f(x) =Â x,Â xÂ is a real number**

**Solution: (i)** Given,

f(x) = 2 â€“ 3*x*,Â *x*Â âˆˆÂ R,Â *x*Â > 0

Since, Â x > 0

â‡’ -3x < 0

â‡’Â 2 – 3x < 2

â‡’ f(x) < 2

Range of f = (â€“âˆž, 2)

**(ii)** Given,

*f*(*x*) =Â *x*^{2}Â + 2,Â *x*Â is a real number

We know that,

*x*^{2}Â â‰¥ 0

*â‡’ x*^{2}Â + 2 â‰¥ 2 [Adding 2 on both sides]

â‡’ f(x) â‰¥ 2

Hence, Range = [2, âˆž)

**(iii)** Given,

*f*(*x*) =Â *x, x*Â is a real number

Clearly, theÂ range ofÂ *f*Â is the set of all real numbers.

Thus,

Range ofÂ *f*Â =Â R