Class 12 Relations And Functions · Very Long Answer5 Markstypes_of_functions

Class 12 Maths · Relations And Functions · Very Long Answer #189

Prove f:(−∞,0)→(−1,0), f(x)=x/(1+|x|), is bijective and obtain its inverse.

✅ Step-by-Step Solution

Final Answer: f⁻¹(y)=y/(1+y), −1<y<0.

For x<0, |x|=−x and f(x)=x/(1−x)=−1+1/(1−x). Since 1−x>1, −1<f(x)<0.
If u/(1−u)=v/(1−v), cross multiplication gives u−uv=v−uv, hence u=v: f is injective.
For any −1<y<0 solve y=x/(1−x) to get x=y/(1+y). The denominator is positive and numerator negative, so x<0.
For this input 1−x=1/(1+y), hence f(x)=y; f is surjective onto (−1,0).
Thus f is bijective, with inverse f⁻¹(y)=y/(1+y) on (−1,0).

[1 mark] For x<0, |x|=−x and f(x)=x/(1−x)=−1+1/(1−x). Since 1−x>1, −1<f(x)<0.
[1 mark] If u/(1−u)=v/(1−v), cross multiplication gives u−uv=v−uv, hence u=v: f is injective.
[1 mark] For any −1<y<0 solve y=x/(1−x) to get x=y/(1+y). The denominator is positive and numerator negative, so x<0.
[1 mark] For this input 1−x=1/(1+y), hence f(x)=y; f is surjective onto (−1,0).
[1 mark] Thus f is bijective, with inverse f⁻¹(y)=y/(1+y) on (−1,0).

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