JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

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Analyzing +4 | -1

For

\frac{{\left| {x - 1} \right|}}{{x + 2}} < 1,\;x

lies in the interval

View answer and explanation

Correct option: A

\left( { - \infty ,\; - 2} \right) \cup \left( { - \frac{1}{2},\;\infty } \right)

Option A

\left( { - \infty ,\; - 2} \right) \cup \left( { - \frac{1}{2},\;\infty } \right)

is correct because it represents the solution set of the inequality considering both cases when

x < 1

and

x \ge 1

More Questions from QUADRATIC EQUATIONS

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

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a

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

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

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