If for-x - R- 1-3 lt- x-2 - 2x - 4-x-2 - 2x - 4 lt- 3- -then-9-3-2x - 6-3-x - 4-9-3-2x - 6-3-x - 4 lies between1-3-and-3None-of-these1-2- and-20-and-2

The correct option is A 1/3 and 39.3^2x 6.3^x + 4/9.3^2x + 6.3^x + 4 = (3^2(x + 1)) 2 (3^(x +1)) + 4/(3^2(x + 1)) + 2 (3^(x +1)) + 4 = t^2 2t + 4/t^2 + 2 ...

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Quantitative Aptitude

Equations

If for x ϵ R,...

Question

If for $x ϵ R,$
$\frac{1}{3} < \frac{x^{2} - 2 x + 4}{x^{2} + 2 x + 4} < 3, t h e n \frac{{9.3}^{2 x} - {6.3}^{x} + 4}{{9.3}^{2 x} + {6.3}^{x} + 4}$ lies between

\frac{1}{2} a n d 2

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\frac{1}{3} a n d 3

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0 a n d 2

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N o n e o f t h e s e

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Solution

The correct option is B $\frac{1}{3} a n d 3$
$\frac{{9.3}^{2 x} - {6.3}^{x} + 4}{{9.3}^{2 x} + {6.3}^{x} + 4} = \frac{(3^{2 (x + 1)}) - 2 (3^{(x + 1)}) + 4}{(3^{2 (x + 1)}) + 2 (3^{(x + 1)}) + 4}$
$= \frac{t^{2} - 2 t + 4}{t^{2} + 2 t + 4} (w h e r e t = 3^{x + 1}) . . . (1)$
$S i n c e, \frac{1}{3} < \frac{x^{2} - 2 x + 4}{x^{2} + 2 x + 4} < 3$
$∴$ From (1), the given expression lies between $\frac{1}{3}$ and 3.

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