Class 12 Relations And Functions · Very Long Answer5 Markstypes_of_functions

Class 12 Maths · Relations And Functions · Very Long Answer #195

Prove f:[0,∞)→[−5,∞), f(x)=4x²+4x−5, is bijective and obtain f⁻¹.

✅ Step-by-Step Solution

Final Answer: f⁻¹(y)=(−1+√(y+6))/2, y≥−5.

For x≥0, f(x)+5=4x(x+1)≥0, so f maps into [−5,∞).
If f(u)=f(v), then 4(u−v)(u+v+1)=0; the second factor is positive, so u=v and f is injective.
Write y=(2x+1)²−6. For y≥−5, the nonnegative-domain solution is x=(−1+√(y+6))/2≥0.
Substitution gives f(x)=(√(y+6))²−6=y, proving that every codomain value has a preimage.
Thus f is bijective and f⁻¹(y)=(−1+√(y+6))/2 on [−5,∞).

[1 mark] For x≥0, f(x)+5=4x(x+1)≥0, so f maps into [−5,∞).
[1 mark] If f(u)=f(v), then 4(u−v)(u+v+1)=0; the second factor is positive, so u=v and f is injective.
[1 mark] Write y=(2x+1)²−6. For y≥−5, the nonnegative-domain solution is x=(−1+√(y+6))/2≥0.
[1 mark] Substitution gives f(x)=(√(y+6))²−6=y, proving that every codomain value has a preimage.
[1 mark] Thus f is bijective and f⁻¹(y)=(−1+√(y+6))/2 on [−5,∞).

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