11th NCERT Relatins and functions.

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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) = x

(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) = x

x ∈ R

∴ x

∴ x

∴ 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.