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

29/06/202414/03/2023 by Team Gmath

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 $</span>f(x) = \sqrt{16-x^2}$

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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