JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

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If for

x \in R,\frac{1}{3} < \frac{{{x^2} - 2x + 4}}{{{x^2} + 2x + 4}} < 3,

then

\frac{{{{9.3}^{2x}} - {{6.3}^x} + 4}}{{{{9.3}^{2x}} + {{6.3}^x} + 4}}\;

lies between

View answer and explanation

Correct option: B

\frac{1}{3}

and 3

Option B (

\frac{1}{3}

and 3) is correct because the given expression can be transformed into the form of the inequality, and thus lies between the same bounds.

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{x^2} + 4ax + 2 > 0

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a

f\left( x \right) = a{x^2} + bx + c\;

\;f\left( { - 1} \right) < 1,f\left( 1 \right) > - 1,f\left( 3 \right) < - 4\;

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9{x^2} - 18\left| x \right| + 5 = 0

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