Let f be a function defined by f(x)=2x2−log⁡|x|x≠0 then

f′(x)=4x−1x=4x2−1x=(2x−1)(2x+1)xx2 2x – 12x + 1xf(x)x 1/2++++Thus, f increases on [−1/2,0]∪[1/2,∞].

Let f be a function defined by f(x)=2x2−log⁡|x|x≠0 then

f′(x)=4x−1x=4x2−1x=(2x−1)(2x+1)xx2 2x – 12x + 1xf(x)x 1/2++++Thus, f increases on [−1/2,0]∪[1/2,∞].

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Let $f$ be a function defined by $f (x) = 2 x^{2} - \log | x |$
$x \neq 0$ then

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$f$ decreases on $[1 / 2, \infty]$

$f$ increase on $[- 1 / 2, 0] \cup [1 / 2, \infty)$

$f$ increases on $n (- \infty, - 1 / 2)$

$f$ decreases on $(- \infty, 0)$

answer is A.

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Detailed Solution

$f^{'} (x) = 4 x - \frac{1}{x} = \frac{4 x^{2} - 1}{x} = \frac{(2 x - 1) (2 x + 1) x}{x^{2}}$

	$2 x - 1$	$2 x + 1$	$x$	$f (x)$
$x < - 1 / 2$	-	-	-	-
$- 1 / 2 < x < 0$	-	+	-	+
$0 < x < 1 / 2$	-	+	+	-
$x > 1 / 2$	+	+	+	+

Thus, $f$ increases on $[- 1 / 2, 0] \cup [1 / 2, \infty]$ .

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Let $f$ be a function defined by $f (x) = 2 x^{2} - \log | x |$
$x \neq 0$ then