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a question solve fastt.
Question---:-find the range --- attachment file
--class 11th
chapter---relation n functions
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Answers
Answered by
65
[1]
=> f(x) = 2-3x
let's
=> f(x) = y
=> 2 - 3x = y
=> 2 - y = 3x
=> X = (2 - y)/3
Given tha "x>0"
so
=> (2-y)/3 > 0
=> 2 - y > 0
=> 2 > y
=> y < 2
HENCE
Range of f(x) is = (-infinity, 2)
______________________[ANSWER]
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[2]
=> f(x) = x² + 2
X is a real no..
Given that X is a real no. so square of a real no. always be positive or 0 .
So we can say that :-
=> x² ≥ 0
Adding 2 both Side
=> x² + 2 ≥ 0 + 2
=> x² + 2 ≥ 2
=> f(x) ≥ 2
Hence
The range of f(x) is = [2 , infinity ]
______________________[ANSWER]
=================================
DEVIL_KING ▄︻̷̿┻̿═━一
=> f(x) = 2-3x
let's
=> f(x) = y
=> 2 - 3x = y
=> 2 - y = 3x
=> X = (2 - y)/3
Given tha "x>0"
so
=> (2-y)/3 > 0
=> 2 - y > 0
=> 2 > y
=> y < 2
HENCE
Range of f(x) is = (-infinity, 2)
______________________[ANSWER]
=====================================
[2]
=> f(x) = x² + 2
X is a real no..
Given that X is a real no. so square of a real no. always be positive or 0 .
So we can say that :-
=> x² ≥ 0
Adding 2 both Side
=> x² + 2 ≥ 0 + 2
=> x² + 2 ≥ 2
=> f(x) ≥ 2
Hence
The range of f(x) is = [2 , infinity ]
______________________[ANSWER]
=================================
DEVIL_KING ▄︻̷̿┻̿═━一
Deepsbhargav:
theku Dimpy.. xD
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