Title SOLVING PROBLEMS ON FINDING domain of functions.pdf ✅

What you already know • Definition of a function • Domain and range of a function • Types of functions What you will learn • Finding the domain of various functions Domain To find domain of basic types of functions: ( ) ( ) ( ) ( ) ( ) ( ) If then If then If then 1 y= f x 0 f x y = f x f x 0 1 y = f x 0 f x ≠ ≥ > Method to find the domain if addition or subtraction of two or more functions are given, ( ) ( ) ( ) where and are domains o f and ( ) ( ) ( ) 1 2 1 2 h x = f x ± g x D = D D D, D D h x , f x g x . ↓ ↓↓  Find the domain of the following functions: Solution 1. ( ) 2 2 1 f x = x - x - 20 + x - 5x -14 ( )( ) ( ] [ ) ( )( ) ( ) ( ) 2 2 x - x - 20 0 x - 5 x + 4 0 x -4 x 5 x - , -4 5, x - 5x - 14 > 0 x - 7 x + 2 > 0 x < -2 x > 7 x - , -2 7, ≥ ⇒ ≥ ⇒ ≤ ≥ ⇒∈∞ ∞ ⇒ ⇒ ⇒∈ ∞ ∞     Case 1 : Case 2 : By taking intersection of Case 1 and Case 2, we get domain as follows: x - , - 4 7, ∈∞ ∞ ( ] ( ) NOTES RELATIONS AND FUNCTIONS MATHEMATICS SOLVING PROBLEMS ON FINDING DOMAIN OF FUNCTIONS © 2021, BYJU'S. All rights reserved
2. ( ) 1 fx = x - x 3. ( ) 1 fx = x- x 4. ( ) 2 f x = sin x + 16 - x Analytical method |x| - x > 0 When x ≥ 0, When x < 0, ⇒ x > x ⇒ - x > x Not possible ⇒ 2x < 0 ⇒ x < 0 x - |x| > 0 ⇒ |x| < x From the graph, we can see that the green line does not exceed the yellow line at any point. x = Φ sin x ≥ 0 is possible if y = sinx, is above the X-axis. 16 - x2 ≥ 0 ⇒ x2 - 16 ≤ 0 ⇒ (x - 4)(x + 4) ≤ 0 ⇒ -4 ≤ x ≤ 4 By taking intersection of above two sets of x we get domain as x ∈ [-4, -π] U [0, π] |x| > x |x| > x Let x = 3, 3 > 3 Not possible Let x = -3, 3 > -3 True Let x = 3.1, 3.1 > 3.1 Not possible Let x = -3.1, 3.1 > -3.1 We can see that the inequality holds for negative values of x ⇒ x ∈ (-∞, 0) ⇒ x ∈ (-∞, 0) ⇒ x ∈ (-∞, 0) Graphical method Hit and trial Note For the composite function f(g(x)) we have the rule, Domain of f(g(x)) = Domain of g(x), if Domain of function f is R. Y y = -x y = x y = x X Y y = |x| y = x X O -1 1 y = sin x -2π -4 -π π 4 2π Y X © 2021, BYJU'S. All rights reserved 02
Find the domain of the following functions: Solution ( ) ( ) ( ) ( ( )) ( ) ( ) ( ( )) ( ) f x = sin x - 1 + 6 - x f x = sin g x ; g x = x - 1 + 6 - x f x = f g x = g x x - 1 + 6 - x , x-1 0 x 1 6-x 0 x 6 x ⇒ ≥⇒≥ ≥⇒≤ 1. Domain of We know, Domain of Domain of Domain of By taking intersection of above two sets of we get  domain as x 1, 6 ∈[ ] ( ) ( ) ( ) ( ) [ ] [ ] f x = cos sinx cos sinx 0 f x = cos ; = sinx cos 0 cos X . = sin x -1, 1 ; -1, 1 . ≥ θ θ ⇒ θ≥ θ θ ∈ θ For function to exist, is that part of which is above - axis Also, i.e, is bounded to Part 3. Step 1 : Step 2 : Step 3 : cos -1, 1 [ ] X x . θ ∈ of in is already above - axis. Hence,  ( ) ( ) ( ) ( ) ( ) [ ] - x - 1 + 6 - x -g x g x = x - 1 + 6 - x x-1 0 x 1 6-x 0 x 6 x x 1, 6 f x = 4 f x = 4 ; x - 1 + 6 - x , ≥⇒≥ ≥⇒≤ ∈ 2. Domain of By taking intersection of above two sets of we get domain as Note ( ( )) ( ( )) ( ( )) ( ) ( ( )) ( ( )) ( ( )) ( ) 11 1 4 6 Even 11 1 3 5 Odd f x , f x , ... , f x = f x f x , f x , ... , f x = f x 1. Domain of Domain of 2. Domain of Domain of - -1 0 1 Y X π 2 π 2 © 2021, BYJU'S. All rights reserved 03
Find the domain of the following function: ( )       1 1 - x 12 f x = 5 + x Solution ( ) ( ( )) ( ) ( ] 1 12 1 Even 1 - x f x = 5 + x f x = f x 1 - x 1 - x 0 5 + x 5 + x x -5, 1       ⇒ ⇒ ≥ ⇒ ∈ Domain of Domain of Domain of Find the domain of the following functions: ( ) ( )             1. 2. 2 10 3 1 100x 3 2log x + 1 log log x + 10x + 5 f x = log -x Solution ( ) 2 3 1 3 log log x + 10x + 5       1. Step 1: ( ( )) ( ) 3 2 1 3 2 2 2 2 log g x , = 3 > 0, 3 1 : log x + 10x + 5 > 0 x + 10x + 5 < 1 x + 10x + 4 < 0 x + 10x + 4 = 0, -10 ± 100 - 16 x = 2 -10 ± 2 21 x = x = -5 ± 21 2 x + 10x + 4 < ≠ ⇒ ⇒ ⇒ ⇒ ⇒ For Base Also, Now, For For Step 2 ( ) 0, x -5 - 21, -5 + 21 ∈ Step 3: 1 ( ( )) 3 log h x 1 1 = > 0, , 1 3 3 ≠ For Base Also Step 4: ( ) 2 2 x + 10x + 5 = 0, -10 ± 100 - 20 x = 2 -10 ± 2 20 x = 2 x = -5 ± 20 x + 10x + 5 > 0 x - -5 - 20, -5 + 20 ⇒ ⇒ ⇒ ∈ Now, for So, for  Step 5: x -5 - 21, -5 - 20 -5 + 20, -5 + 21 ∈ ∪ ( ) ( ) -5 1 Logarithmic Functions © 2021, BYJU'S. All rights reserved 04
2. ( ) For function to exist, a. Base Also b. 10 100x 10 10 10 10 10 2log x + 1 f x = log -x 1 = 100x > 0; x > 0, 100x 1; x 100 2log x + 1 2log x + 1 > 0 < 0 -x x 2log x + 1 < 0 2log x < -1 log x <       ≠ ≠ ⇒ ⇒ ⇒ ⇒ ( ) ( ) ( ) c. In Considering a , b and c , 1 - 2 10 1 1 - x < 10 x < 2 10 2log x + 1, x > 0 1 1 1 x 0, , 100 100 10 ⇒ ⇒     ∈          Find the domain of the following functions: ( ) ( )     1. 2. 3. 3 2 10 10 10 10 log 1 + x f x = log 4x - 1 log 1 - log x - 5x + 16 Solution { } 3 10 3 3 log 1 + x = 10 0, 10 > 1 1 + x 0 x -1 x -1 x - -1 ≠ ≠ ⇒ ≠ ⇒≠ ⇒ ∈ 1. For function to exist, a. Base also b.  ( ) 10 f x = log 4x - 1 = 10 0, 10 > 1 1 4x - 1 0 x 4 1 x - 4 ≠ ≠ ⇒≠   ⇒ ∈     For function to 2 exist, a. Base also b. .  ( ) ( ) ( ) ( )( ) x 3 For funct . ion to e ist, a.Base also, b. 2 10 10 2 10 2 2 10 2 log 1 - log x - 5x + 16 = 10 0, 10 > 1 1 - log x - 5x + 16 > 0 log x - 5x + 16 < 1 x - 5x + 16 < 10 x - 5x + 6 < 0 x - 2 x - 3 < 0     ≠ ⇒ ⇒ ⇒⇒ ⇒ ( ) c. 2 2 < x < 3 x - 5x + 16 > 0 D = 25 - 4 × 16 = -ve x x 2, 3 ⇒ ⇒ ∈ ⇒ ∈  © 2021, BYJU'S. All rights reserved 05