A function f:[-4,4] → [0,4] is given by f(x) = √16 - x²

A function f:[-4,4] → [0,4] is given by f(x) = √16 – x²

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Question:

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.

A function f:[-4,4] → [0,4] is given by f(x) = √16 - x²

Solution:- A function f:[-4, 4] → [0, 4] is given that f(x) = \sqrt{16-x^2}

One-one :- Let x_1 = -2 \text{ and } x_2 = 2 are lies in [-4, 4]

f(x) = \sqrt{16-x^2}

f(-2)= \sqrt{16-(-2)^2}

=\sqrt{16-4} = \sqrt{12}

f(-2)= \sqrt{16-(2)^2}

=\sqrt{16-4} = \sqrt{12}

Therefore,

f(-2)=f(2)=\sqrt{12}

But -2\neq 2

Thus the given function is not one-one

Onto :- Since f(x) = \sqrt{16-x^2}

Let y =\sqrt{16-x^2}

\Rightarrow y^2 = 16 - x^2

\Rightarrow x^2=16-y^2

\Rightarrow x =\sqrt{16-y^2}

Let y = 4\in[0,4 ]\Rightarrow x = \sqrt{16-16}=0\in[-4,4]

Again let y = 0\in [0, 4] \Rightarrow x =\sqrt{16-0}=\pm4\in[-4,4]

Thus, every value of co-domain have pre-image in domain so the given function is onto

Since, f(x) = \sqrt{16-x^2}

f(a)=\sqrt{16-a^2}

\Rightarrow \sqrt{7}=\sqrt{16-a^2}

Sqaring both side

\Rightarrow 7 = 16-a^2

\Rightarrow a^2 = 16-7=9

\Rightarrow a = \pm 3

Possible value of a = -3, 3

 

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