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TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

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Question : 12 of 160

Marks: +1, -0

The minimum value of

\frac{9\cdot 3^{2x}+6\cdot 3^{x}+4}{9\cdot 3^{2x}-6\cdot 3^{x}+4}

is

$-1$
$\frac{1}{2}$
$\frac{1}{4}$
$\frac{1}{3}$

Solution:

Let

y=\;\frac{9\cdot 3^{2x}+6\cdot 3^{x}+4}{9\cdot 3^{2x}-6\cdot 3^{x}+4}

Put,

3^{x}=t

\therefore

y=\;\frac{9\cdot t^{2}+6t+4}{9t^{2}-6t+4}

\Rightarrow y(9t^{2}-6t+4)=9t^{2}+6t+4

\Rightarrow 9t^{2}(y-1)-6t(y+1)+(4y-4)=0

\Rightarrow \;\; (y-1)t^{2} \;\frac{-6}{9}(y+1)t+\;\frac{4}{9}(y-1)=0

\Rightarrow (y-1)t^{2}-\;\frac{2}{3}(y+1)t+\;\frac{4}{9}(y-1)=0

\because t

is real

\therefore D \geq 0

\Rightarrow \;\frac{4}{9}(y+1)^{2}-4(y-1)\;\frac{4}{9}(y-1) \geq 0

\Rightarrow \;\; \;\frac{4}{9}(y+1)^{2}-\;\frac{16}{9}(y-1)^{2} \geq 0

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

\Rightarrow \;\; y^{2}+1+2y-4y^{2}-4+8y \geq 0

\Rightarrow \;\; -3y^{2}+10y-3 \geq 0

\Rightarrow \;\; 3y^{2}-10y+3 \leq 0

\Rightarrow (y-3)(3y-1) \leq 0 \Rightarrow \;\frac{1}{3} \leq y \leq 3

Hence, minimum value

=\;\frac{1}{3}

.

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