Ex 2.3 sets ncert maths solution class 11

  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:  (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) f(x) = –|x| (ii) f(x) = √(9 – x2

Solution: (i) Given,

f(x) = –|x|, x ∈ R

As f(x) is defined for x ∈ R, the domain of f  =  R(Real number).

We know that

|x| ∈ [0, ∞) ⇒ 0 ≤ |x |< ∞

⇒ –∞ < -|x| ≤ 0

⇒ -|x| ∈ (-∞, 0]

⇒ f(x) ∈ (-∞, 0]

Therefore, the range of f is given by (–∞, 0].

(ii) f(x) = √(9 – x2)

We know that

9 – x² ≥ 0 ⇒ (3 – x)(3 + x) ≥ 0

⇒ 3 – x ≥ 0 and 3 + x ≥ 0

⇒ x ≤ 3 and  x ≤ -3

⇒ x ∈ [-3, 3]

So, the domain of f(x) is {x: –3 ≤ x ≤ 3} or [–3, 3].

Now,

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

Therefore, the range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3].

3. A function f is defined by f(x) = 2x – 5. Write down the values of

(i) f(0), (ii) f(7), (iii) f(–3)

Solution: Given,

Function, f(x) = 2x – 5

(i) f(0) = 2 × 0 – 5 = 0 – 5 = –5

(ii) f(7) = 2 × 7 – 5 = 14 – 5 = 9

(iii) f(–3) = 2 × (–3) – 5 = – 6 – 5 = –11

4. The function ‘t’, which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t(C) = 9\frac{9c}{5} + 32.

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

Solution: Given function

t(C) = 9\frac{9c}{5} + 32

(i) t(0) = \frac{9\times 0}{5}+32

= 0 + 32 = 32

(ii) t(28) = \frac{9\times 28}{5}+32

= \frac{252+160}{5} = \frac{412}{5}

(iii) t(-10) = \frac{9\times (-10)}{5}+32

= 9\times (-2) + 32

= -18 + 32 = 14

(iv)  t(c) = 212

Therefore

212 = \frac{9 c}{5}+32

\Rightarrow \frac{9c}{5}= 212-32

Rightarrow \frac{9c}{5} =  180

\Rightarrow  9c = 180\times 5

\Rightarrow c = \frac{180\times 5}{9}

= 100

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

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

(i) f(x) = 2 – 3xx ∈ R, x > 0

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

(iii) f(x) = xx is a real number

Solution: (i) Given,

f(x) = 2 – 3xx ∈ R, x > 0

Since,  x > 0

⇒ -3x < 0

⇒  2 – 3x < 2

⇒ f(x) < 2

Range of f = (–∞, 2)

 

(ii) Given,

f(x) = x2 + 2, x is a real number

 

We know that,

x2 ≥ 0

⇒ x2 + 2 ≥ 2 [Adding 2 on both sides]

⇒ f(x) ≥ 2

Hence, Range = [2, ∞)

(iii) Given,

f(x) = x, x is a real number

Clearly, the range of f is the set of all real numbers.

Thus,

Range of f = R


Ex 2.2 sets ncert maths solution class 11

 

Class 12 relation and functions multiple choice

Team Gmath

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