The range of the real valued function f(x)={\;{x²+2 x+8}{x²+2 x+4}} is

Correct Answer is : (1, 73](1, {\;7/3}](1,37]

Correct Answer is : (1, 73](1, {\;7/3}](1,37]

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AP EAPCET 04-Jul-2022 Shift 1 Paper

Section: Mathematics

Question : 1 of 160

Marks: +1, -0

f(x)=\sqrt{\;\frac{x^{2}+2 x+8}{x^{2}+2 x+4}}

Solution: 👈: Video Solution

f(x)=\sqrt{\;\frac{x^2+2 x+8}{x^2+2 x+4}}

Let

y=\;\frac{x^2+2 x+8}{x^2+2 x+4}

[ \because f(x)=\sqrt{y}]

\;\Rightarrow \;\; y x^2+2 x y+4 y=x^2+2 x+8

\;\Rightarrow \;\; (y-1) x^2+2 x(y-1)+4 y-8=0

For

x

to be real, discriminant of equation must be greater than equals to zero.

\;\{ [2(y-1)]^2-4(y-1)(4 y-8) \} \ge 0

\Rightarrow 4(y-1)^2-4(y-1) 4(y-2) \ge 0

\Rightarrow\;(y-1)^2-(y-1) 4(y-2) \ge 0

\Rightarrow\;(y-1)[y-1-4 y+8] \ge 0

\Rightarrow\;(y-1)(-3 y+7) \ge 0

\Rightarrow\;y\;=1,\;\frac{7}{3}

\therefore \;\; (y-1)(-3 y+7) \ge 0 \Rightarrow y \in (1,\;\frac{7}{3}]

[ \because .

y=1, x^2+2 x+4 \neq x^2+2 x+8]

\; \because \;\; y \in (1,\;\frac{7}{3}]

\therefore\; f(x) \in (1, \sqrt{\;\frac{7}{3}}] .

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