NCERT Solutions for Class 11-commerce Maths CBSE Chapter 2: Relations and Functions

Relations and Functions Exercise Ex. 2.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Relations and Functions Exercise Ex. 2.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Relations and Functions Exercise Ex. 2.3

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Relations and Functions Exercise Misc. Ex.

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

$Given colon space straight f open parentheses straight x close parentheses equals fraction numerator straight x squared over denominator 1 plus straight x squared end fraction Let space straight y equals fraction numerator straight x squared over denominator 1 plus straight x squared end fraction rightwards double arrow straight y plus yx squared equals straight x squared rightwards double arrow straight y equals straight x squared minus yx squared rightwards double arrow straight y equals straight x squared open parentheses 1 minus straight y close parentheses rightwards double arrow fraction numerator straight y over denominator 1 minus straight y end fraction equals straight x squared rightwards double arrow straight x equals square root of fraction numerator straight y over denominator 1 minus straight y end fraction end root straight x space is space defined space only space if space fraction numerator straight y over denominator 1 minus straight y end fraction greater or equal than 0 rightwards double arrow straight y greater or equal than 0 space and space 1 minus straight y greater or equal than 0 rightwards double arrow straight y greater or equal than 0 space and space 1 greater or equal than straight y straight y not equal to 1 Therefore comma space straight y greater or equal than 0 space and space 1 greater than straight y rightwards double arrow 0 less or equal than straight y less than 1 rightwards double arrow straight y element of left square bracket 0 comma space 1 right parenthesis rightwards double arrow fraction numerator straight x squared over denominator 1 plus straight x squared end fraction element of left square bracket 0 comma space 1 right parenthesis Hence comma space the space range space of space straight f space is space left square bracket 0 comma space 1 right parenthesis.$

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Class 11-commerce NCERT Solutions Maths Chapter 2: Relations and Functions

Relations and Functions Exercise Ex. 2.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Relations and Functions Exercise Ex. 2.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Relations and Functions Exercise Ex. 2.3

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Relations and Functions Exercise Misc. Ex.

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

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