Let f ( x ) = 2 x 2 − ln ∣ x ∣ , x  = 0 , then f ( x ) is

∵ f ( x ) = 2 x 2 − ln ∣ x ∣ ∴ f ′ ( x ) = 4 x − x 1 = x ( 2 x + 1 ) ( 2 x − 1 ) For increasing, f ′ ( x ) > 0 ∴ x ∈ ( − 2 1 , 0 ) ∪ ( 2 1 , ∞ ) And for decreasing, f ′ ( x ) < 0 ∴ x ∈ ( − ∞ , − 2 1 ) ∪ ( 0 , 2 1 )

Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMonotonicity-Increasing-DecreasingMedium2 minQB

MathematicsMediummultiple choice

Let $f (x) = 2 x^{2} - ln ∣ x ∣, x \neq = 0$ , then $f (x)$ is

Options:(select one or more)

Answer:

A, D

Solution:

$∵ f (x) = 2 x^{2} - ln ∣ x ∣$ $∴ f^{'} (x) = 4 x - \frac{1}{x} = \frac{( 2 x + 1 ) ( 2 x - 1 )}{x}$ For increasing, $f^{'} (x) > 0$ $∴ x \in (- \frac{1}{2}, 0) \cup (\frac{1}{2}, \infty)$ And for decreasing, $f^{'} (x) < 0$ $∴ x \in (- \infty, - \frac{1}{2}) \cup (0, \frac{1}{2})$

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Monotonicity-Increasing-Decreasing

⏱ 2mℹ️ Source: QB

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