Question

find range of f(x)= x^2/1+x^2
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Answer 1

Answer :

Range (f) = [0,1)

Note :

Real valued function : A function whose domain and co-domain (and so as range) are subsets of R is called real valued function .

Solution :

• Given : f(x) = x²/(1 + x²)
• To find : Range (f) = ?

Let y = f(x) .

Thus ,

=> y = x²/(1 + x²)

=> y(1 + x²) = x²

=> y + x²y = x²

=> y = x² - x²y

=> y = x²(1 - y)

=> x² = y/(1 - y)

=> x = √[y/(1 - y)]

For x to be real , the following conditions must be satisfied simultaneously :

1) y/(1 - y) ≥ 0

and

2) 1 - y ≠ 0

Condition 1 :

y/(1 - y) ≥ 0

Here ,

Two cases arises :

1a) y ≥ 0 and 1 - y ≥ 0

OR

1b) y ≤ 0 and 1 - y ≤ 0

Case 1a :

=> y ≥ 0 and 1 - y ≥ 0

=> y ≥ 0 and 1 ≥ y

=> y ≥ 0 and y ≤ 1

=> y € [0,1]

OR

Case 1b :

=> y ≤ 0 and 1 - y ≤ 0

=> y ≤ 0 and 1 ≤ y

=> y ≤ 0 and y ≥ 1

=> y € ∅ { °•° There exist no real number which is less than 0 and greater than 1 respectively }

Thus ,

Condition1 will be satisfied if ;

=> y € [0,1] or y € ∅

=> y € [0,1] U ∅ { Union of solutions found in case1 and case2 }

=> y € [0,1]

Condition 2 :

=> 1 - y ≠ 0

=> 1 ≠ y

=> y ≠ 1

Thus ,

For x to be real ,

=> y/(1 - y) ≥ 0 and 1 - y ≠ 0

=> y € [0,1] and y ≠ 1

=> y € [0,1] - {1}

=> y € [0,1)

=> Range = [0,1)

Hence , Range (f) = [0,1)