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TG EAMCET 3 May 2025 Shift 2 Paper

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Question : 2 of 160

Marks: +1, -0

If the range of the real valued function

f(x)=\;\frac{x^2+x+k}{x^2-x+k}

\left[\;\frac{1}{3}, 3\right]

k=

$\; -2$
$\; -1$
$\; 1$
$\; 2$

Solution:

\;f(x)=\;\frac{x^2+x+k}{x^2-x+k}

\; \because \; \text{ Range } \; = \left[\;\frac{1}{3}, 3\right]

For range let

f(x)=y

y\;=\;\frac{x^2+x+k}{x^2-x+k}

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

\therefore \; D \geq 0

\Rightarrow \;\; (y+1)^2-4 k(y-1)(y-1) \geq 0

\Rightarrow \;\; (y+1)^2-(2 \sqrt{k}(y-1))^2 \geq 0

\Rightarrow \;\; (y+1+2 \sqrt{k}(y-1))

\; \;\; (y+1-2 \sqrt{k}(y-1) \geq 0

\Rightarrow \;\; (y(2 \sqrt{k}+1)+1-2 \sqrt{k})\;

\; \;\; (y(2 \sqrt{k}-1)-2 \sqrt{k}-1) \leq 0 \;\;\;\cdots (i)

\because \; y \in \left[\;\frac{1}{3}, 3\right]

\; (3 y-1)(y-3) \leq 0 \;\;\;\cdots (ii)

Comparing Eqs. (i) and (ii),

2 \sqrt{k}+1=3 \Rightarrow 2 \sqrt{k}=2 \Rightarrow k=1

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