Let f x -2 x 2-ln - x - y -0- thenf x isA- Monotonically increasing in -1-2- 0 -1-2- -B- Monotonically decreasing in -1-2- 0 -1-2- -C- Monotonically increasing in -1-2 -0- 1-2D- Monotonically decreasing in -1-2 -0- 1-2

The correct options are A Monotonically increasing in ( 1/2,0 )∪ ( 1/2,∞ ) D Monotonically decreasing in ( ∞, 1/2 )∪ ( 0,1/2 )f'(x)=4x 1/x=(2x+1)(2x 1)/x f ...

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Byju's Answer

Standard XII

Mathematics

Monotonicity

Let f x =2 x ...

Question

Let $f (x) = 2 x^{2} - l n | x |, x \neq 0, t h e n f (x)$ is

Monotonically increasing in

(- \frac{1}{2}, 0) \cup (\frac{1}{2}, \infty)

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Monotonically decreasing in

(- \frac{1}{2}, 0) \cup (\frac{1}{2}, \infty)

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Monotonically increasing in

(- \infty, - \frac{1}{2}) \cup (0, \frac{1}{2})

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Monotonically decreasing in

(- \infty, - \frac{1}{2}) \cup (0, \frac{1}{2})

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Solution

The correct options are
A Monotonically increasing in $(- \frac{1}{2}, 0) \cup (\frac{1}{2}, \infty)$
D Monotonically decreasing in $(- \infty, - \frac{1}{2}) \cup (0, \frac{1}{2})$
$f^{'} (x) = 4 x - \frac{1}{x} = \frac{(2 x + 1) (2 x - 1)}{x}$

f is increasing $\Rightarrow f^{1} (x) > 0$
f is decreasing $\Rightarrow f^{1} (x) > 0$

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The correct options are A Monotonically increasing in (−12,0)∪(12,∞) D Monotonically decreasing in (−∞,−12)∪(0,12)f′(x)=4x−1x=(2x+1)(2x−1)x f is increasing ⇒f1(x)>0 f is decreasing ⇒f1(x)>0

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