JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

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Analyzing +4 | -1

0 < a < 1,

then the solution set of the inequation

\frac{{1 + {{\left( {{{\log }_a}x} \right)}^2}}}{{1 + \left( {{{\log }_a}x} \right)}} > 1,

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Correct option: C

\left( {1,\;1/a} \right) \cup \left( {0,a} \right)

Option C (

\left( {1,\;1/a} \right) \cup \left( {0,a} \right)

) is correct because solving the logarithmic inequality results in two intervals for x: (1, 1/a) and (0, a).

More Questions from QUADRATIC EQUATIONS

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