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Maths-

General

Easy

Question

Exhaustive set of value of x satisfying log_|x|(x² + x + 1) ≥0 is –

(–1, 0)
$(- \infty, 1) \cup (1, \infty)$
$(- \infty, \infty) - {- 1, 0, 1}$
$(- \infty, - 1) \cup (- 1, 0) \cup (1, \infty)$

hint Hint:

In this question, we are required to find the exhaustive set of value of x satisfying $log subscript open vertical bar x close vertical bar end subscript$ (x2 + x + 1) ≥0 is

The correct answer is: $left parenthesis negative straight infinity comma negative 1 right parenthesis union left parenthesis negative 1 comma 0 right parenthesis union left parenthesis 1 comma straight infinity right parenthesis$

Step by step solution:
$log subscript open vertical bar x close vertical bar end subscript left parenthesis x squared plus x plus 1 right parenthesis space greater or equal than 0$
$rightwards double arrow log subscript x left parenthesis x squared plus x plus 1 right parenthesis greater or equal than 0$ [ we can remove the mod from the base since bases of log cannot be negative]
CASE 1 : when base, i.e., x $element of space left parenthesis 0 comma 1 right parenthesis$
$rightwards double arrow x squared plus x plus 1 less or equal than 1$
$rightwards double arrow x left parenthesis x plus 1 right parenthesis less or equal than 0$
$rightwards double arrow x space element of space left parenthesis negative 1 comma 0 right parenthesis$ ...(i)
CASE 2: when base, i.e., $x greater than 1$
$rightwards double arrow x squared plus x plus 1 greater or equal than 1$
$rightwards double arrow x left parenthesis x plus 1 right parenthesis greater or equal than 0$
$rightwards double arrow x element of left parenthesis negative infinity comma negative 1 right parenthesis$ $union left parenthesis 0 comma infinity right parenthesis$ ...(ii)
Thus, from eq (i) and eq (ii), we get,
$x element of left parenthesis negative infinity comma negative 1 right parenthesis$ $union left parenthesis negative 1 comma 0 right parenthesis$ $union left parenthesis 0 comma infinity right parenthesis$
Hence, option(c) is the correct option.

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