# Ex 1.3 , 8

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex1.3 , 8 (Method 1) Consider f: R+ [4, given by f(x) = x2 + 4. Show that f is invertible with the inverse f 1 of given f by f-1 (y) = y 4 , where R+ is the set of all non-negative real numbers. f(x) = x2 + 4 Step 1 Put f(x) = y y = x2 + 4 y 4 = x2 x2 = y 4 x = 4 Since f: R+ [4, ) x R+ , so x is positive Hence we cannot take x = 4 x = 4 Let g(y) = 4 where g: [4, R+ Step 2: gof = g(f(x)) = g(x2 + 4) = x2 + 4 4 = 2 = 1 2 2 = 1 = x Hence, gof(x) = IX = x Step 3: fog = f(g(y)) = f( 4 ) = ( 4 )2 + 4 = ( 4) 1 2 2 + 4 = ( 4) 1 + 4 = y 4 + 4 = y Hence, fog(y) = IY = y Since gof = IX and fog = IY, f is invertible & Inverse of f = g(y) = 4 So, f-1 = Ex1.3 , 8 (Method 2) Consider f: R+ [4, given by f(x) = x2 + 4. Show that f is invertible with the inverse f 1 of given f by f-1 (y) = y 4 , where R+ is the set of all non-negative real numbers. f(x) = x2 + 4 f is invertible if f is one-one and onto Checking one-one f (x1) = (x1)2 + 4 f (x2) = (x2)2 + 4 Putting f (x1) = f (x2) (x1)2 + 4 = (x2)2 + 4 (x1)2 = (x2)2 x1 = x2 or x1 = x2 Since f: R+ [4, So x R+ i.e. x is always positive, Hence x1 = x2 is not true So, x1 = x2 f is one-one Check onto f(x) = x2 + 4 Let f(x) = y such that y [4, y = x2 + 4 y 4 = x2 x2 = y 4 x = 4 Since f: R+ [4, x R+ , so x is positive Hence we cannot take x = 4 x = 4 Since, y [4, i.e. y is greater than or equal to 4 i.e. y 4 y 4 0 Hence the value inside root is positive Hence 4 0 x 0 Hence x is a real number which is greater than or equal to 0. x R+ Hence, the function f is onto Since the function is one-one and onto It is invertible Calculating inverse For finding inverse, we put f(x) = y and find x in terms of y We have done that while proving onto x = 4 Let g(y) = 4 where g: [4, R+ Inverse of f = g(y) = 4 So, f-1 =

Ex 1.2, 5
Important

Ex 1.2 , 10 Important

Example 23 Important

Example 25 Important

Ex 1.3, 3 Important

Ex 1.3 , 6 Important

Ex 1.3 , 8 Important You are here

Ex 1.3 , 9 Important

Ex 1.3 , 13 Important

Ex 1.3 , 14 Important

Ex 1.4, 11 Important

Misc 3 Important

Misc. 4 Important

Misc 14 Important

Misc 18 Important

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.