हिंदी

A function f : [– 4, 4] → [0, 4] is given by f(x) = 16-x2. Show that f is an onto function but not a one-one function. Further, find all possible values of 'a' for which f(a) = 7.

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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)`.

योग
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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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