Last updated at May 29, 2018 by Teachoo

Transcript

Ex 2.3, 4 The function t which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by . t(C) = 9/5 C+ 32 . Find (i) t (0) (ii) t (28) (iii) t (-10) Given t(C) = 9/5 C + 32 Putting C = 0, t(0) = (9 )/5 (0) + 32 = 0 + 32 = 32 We know that t(C) = 9/5 C + 32 Putting C = 28, t(28) = (9 )/5 (28) + 32 = (252 )/5 + 32 = (252 + 32 5)/5 = (252 + 160)/5 = 412/5 Given t(C) = 9/5 C + 32 Putting C = -10, t(0) = (9 )/5 ( 10) + 32 = 9 ( 2) + 32 = 18 + 32 = 14 Ex 2.3, 4 The function t which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by . t(C) = 9/5 C+ 32 . Find (iv) The value of C, when t(C) = 212 Given t(C) = 9/5 C+ 32 Putting t(C) = 212 212 = 9/5 C+ 32 212 32 = 9/5 C 9/5 C = 212 32 9/5 C= 180 9C = 180 5 C = (180 5)/9 C = 100

Finding values at certain points

Chapter 2 Class 11 Relations and Functions

Concept wise

- Finding Cartesian Product
- Equality of 2 ordered pairs
- Number of elements
- Operations on sets+ cartesian product
- Relations - Definition
- Finding Relation - Set-builder form given
- Finding Relation - Arrow Depiction given
- Number of Relations
- Functions - Definition
- Finding values at certain points
- Different Functions and their graphs
- Finding Domain and Range - By drawing graphs
- Finding Domain and Range - General Method
- Algebra of real functions

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.