Class 12 Relations And Functions · Very Long Answer5 Markstypes_of_functions

Class 12 Maths · Relations And Functions · Very Long Answer #194

For f:[−4,4]→[0,4], f(x)=√(16−x²), prove it is onto but not one-one. Find all a for which f(a)=√7.

✅ Step-by-Step Solution

Final Answer: Onto, not one-one; a=−3 or 3.

For −4≤x≤4, 0≤16−x²≤16, so the square root lies in [0,4].
For any y∈[0,4], choose x=√(16−y²)∈[0,4]⊆[−4,4]; then f(x)=y, proving onto.
The distinct inputs 1 and −1 have equal image √15, disproving injectivity.
The equation √(16−a²)=√7 implies 16−a²=7, hence a²=9; both a=−3 and 3 lie in the domain.
Therefore f is onto but not one-one, and the requested inputs are a=±3.

[1 mark] For −4≤x≤4, 0≤16−x²≤16, so the square root lies in [0,4].
[1 mark] For any y∈[0,4], choose x=√(16−y²)∈[0,4]⊆[−4,4]; then f(x)=y, proving onto.
[1 mark] The distinct inputs 1 and −1 have equal image √15, disproving injectivity.
[1 mark] The equation √(16−a²)=√7 implies 16−a²=7, hence a²=9; both a=−3 and 3 lie in the domain.
[1 mark] Therefore f is onto but not one-one, and the requested inputs are a=±3.

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