11th NCERT Relatins and functions.
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## Exercise 2.3 Questions 5

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)}
The relation is many to one, so it is a function.
Domain = { 2, 5, 8, 11, 14, 17}
Range = {1}
(ii) (2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
As the relation is one to one, it is a function.
Domain = { 2, 4, 6, 8, 10, 12, 14}
Range = { 1, 2, 3, 4, 5, 6, 7}
(iii) {(1, 3), (1, 5), (2, 5)}
The relation is many to many. it is not the function.

Question (2)

Find the domain and range of the following real function:
(i) f(x) = –|x| (ii)$f(x) = \sqrt {9 - {x^2}}$

Solution

(i) f(x) = –|x|
x ∈ R, so domain of f(x) is R.
Since x ∈ R, then |x| ≥ 0
∴ - |x| ≤ 0
So range is (- ∞ , 0]
(ii) $f(x) = \sqrt {9 - {x^2}}$ As we can not find square root of negative number , $9 - {x^2} \ge 0$ ${x^2} \le 9$ $\Rightarrow x \le 3\,or\;x \ge - 3$ $\Rightarrow - 3 \le x \le 3$ So domain of f(x) = { x : -3≤ x ≤ 3}
$- 3 \le x \le 3$ $\Rightarrow 0 \le {x^2} \le 9$ $\Rightarrow 0 \ge - {x^2} \ge - 9$ $\Rightarrow 9 - 0 \ge 9 - {x^2} \ge 9 - 9$ $\Rightarrow 9 \ge 9 - {x^2} \ge 0$ $\Rightarrow \sqrt 9 \ge \sqrt {9 - {x^2}} \ge \sqrt 0$ $0 \le f(x) \le 3$ So the range of function f = { x : 0 ≤ x ≤3 }

Question (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

(i) f(0)
f(x) = 2x – 5.
f(0) = 2(0) - 5
= 0 - 5 = - 5
(ii) f(7)
f(x) = 2x – 5.
f(7) = 2 (7) - 5
= 14 - 5 = 9
(iii) f(-3)
f(x) = 2x – 5.
f(-3) = 2 ( - 3) - 5
= - 6 - 5 = -11

Question (4)

The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by. $t\left( C \right) = \frac{9}{5}C + 32$ Find (i) t (0) (ii) t (28) (iii) t (–10) (iv) The value of C, when t(C) = 212

Solution

(i) t(0)
$t\left( C \right) = \frac{9}{5}C + 32$ $t\left( 0 \right) = \frac{9}{5}\left( 0 \right) + 32$ $= 0 + 32 = 32$ (ii)t(28) $t\left( C \right) = \frac{9}{5}C + 32$ $t\left( {28} \right) = \frac{9}{5}\left( {28} \right) + 32$ $= \frac{{252}}{5} + 32$ $= \frac{{252 + 160}}{5}$ $= \frac{{412}}{5}$ (iii) t(-10) $t\left( C \right) = \frac{9}{5}C + 32$ $t\left( { - 10} \right) = \frac{9}{5}\left( { - 10} \right) + 32$ $= - 18 + 32$ $= 14$ (iv) The value of C, when t(C) = 212
$t\left( C \right) = \frac{9}{5}C + 32$ $212 = \frac{9}{5}\left( C \right) + 32$ $212 - 32 = \frac{9}{5}\left( C \right)$ $180 = \frac{9}{5}\left( C \right)$ $C = \frac{{180 \times 5}}{9} = 100$

Question (5)

Find the range of each of the following functions.
(i) f(x) = 2 – 3x, x ∈ R, x > 0.
(ii) f(x) = x2 + 2, x, is a real number.
(iii) f(x) = x, x is a real number

Solution

(i) f(x) = 2 – 3x, x ∈ R, x > 0.
x > 0
∴ -3x < 0 [ multiply by -3]
∴ 2 - 3x < 2 + 0 [ add 2 on both sides ]
∴ f(x) < 2
So range of f(x) is ( - ∞, 2)
(ii) f(x) = x2 + 2, x, is a real number.
x ∈ R
∴ x2 ≥ 0
∴ x2 + 2 ≥ 0 + 2 [ adding 2]
∴ f(x) ≥ 2
So range of f(x) is [ 2, ∞)
(iii) f(x) = x, x is a real number
x ∈ R, f(x) = x ∈ R.
So range of f(x) is R.