Question

solve the question!!!​

Answer 1

Answer:

f(x)=x2+2   xϵR

x2≥0

x2+2≥2

∴ Range of f(x)=(2,∞)

Answer 2

Given Question :-

Find the range of real function from Real to Real

$\rm :\longmapsto\:f = \bigg(x, \: \dfrac{ {x}^{2} }{ {x}^{2} + 1} \bigg) : x \: \in \: R$

$\green{\large\underline{\sf{Solution-}}}$

Given function is

$\rm :\longmapsto\:f = \bigg(x, \: \dfrac{ {x}^{2} }{ {x}^{2} + 1} \bigg) : x \: \in \: R$

To find the range of the function f, Let assume that

$\rm :\longmapsto\:y = \dfrac{ {x}^{2} }{ {x}^{2} + 1}$

$\rm :\longmapsto\: {yx}^{2} + y = {x}^{2}$

$\rm :\longmapsto\: {yx}^{2} - {x}^{2} = - y$

$\rm :\longmapsto\: - {x}^{2} (1 - y) = - y$

$\rm :\longmapsto\: {x}^{2} (1 - y) = y$

$\rm :\longmapsto\: {x}^{2} = \dfrac{y}{1 - y}$

$\rm\implies \:x = \sqrt{\dfrac{y}{1 - y} }$

For x to be defined,

$\rm :\longmapsto\:y \geqslant 0 \: \: and \: \: 1 - y > 0$

$\rm :\longmapsto\:y \geqslant 0 \: \: and \: \: - y > - 1$

$\rm :\longmapsto\:y \geqslant 0 \: \: and \: \: y < 1$

$\bf\implies \:y \: \in \: [0, \: 1)$

Hence,

$\bf\implies \:Range \: of \: f \: \in \: [0, \: 1)$

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