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प्रश्न
A function f : [– 4, 4] `rightarrow` [0, 4] is given by f(x) = `sqrt(16 - x^2)`. Show that f is an onto function but not a one-one function. Further, find all possible values of 'a' for which f(a) = `sqrt(7)`.
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उत्तर
We have, f : [– 4, 4] `rightarrow` [0, 4] defined as f(x) = `sqrt(16 - x^2)`
(i) One-One
f(x1) = f(x2)
`\implies sqrt(16 - x_1^2) = sqrt(16 - x_2^2)`
`\implies 16 - x_1^2 = 16 - x_2^2`
`\implies x_1^2 = x_2^2`
`\implies x_1^2 - x_2^2` = 0
`\implies` (x1 + x2) (x1 – x2) = 0
Here, x1 + x2 = 0 is also possible.
As if x1 = 4 and x2 = – 4.
Then, x1 + x2 = 0 is also possible.
∴ x1 = – x2
But for one-one,
x1 = – x2
so, f(x) is not one-one.
(ii) Onto
Let, y = `sqrt(16 - x^2)`
`\implies` y2 = 16 – x2
`\implies` x2 = 16 – y2
`\implies` x = `sqrt(16 - y^2) ∈ [0, 4]`
So, f(x) is onto.
For f(a) = `sqrt(7)`, we have
f(a) = `sqrt(16 - a^2)`
`\implies sqrt(7) = sqrt(16 - a^2)`
On equating both sides
`\implies` 7 = 16 – a2
`\implies` a2 =16 – 7
`\implies` a2 = 9
`\implies` a = ± 3
Hence, possible values of a are 3 and – 3.
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