Given - fx-4x3-6x2cos2a-3x sin 2a-sin 6a-ln 2a-a2 then

The correct option is C f'(1/2)>0 f'(x)=12x^2 12xcos2a+3sin2a.sin6a+0 f'(1/2)=12/4 12/2cos2a+3sin2a.sin6a =3 6cos2a+1.5(cos4a cos8a) So, it ...

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Algebra of Derivatives

Given : fx=...

Question

Given : $f (x) = 4 x^{3} - 6 x^{2} cos 2 a + 3 x sin 2 a . sin 6 a + \sqrt{ln (2 a - a^{2})}$ then

f (x)

is not defined at

x = \frac{1}{2}

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f^{'} (\frac{1}{2}) < 0

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f^{'} (x)

is not defined at

x = \frac{1}{2}

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f^{'} (\frac{1}{2}) > 0

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Similar questions

f (x) = 4 x^{3} - 6 x^{2} cos 2 a + 3 x \cdot sin 2 a \cdot sin 6 a + \sqrt{log (2 a - a^{2})}

f^{'} (\frac{1}{2}) > 0

lim x \to 0 \frac{\sqrt{1 - cos 2 x}}{x}

f (x) = \frac{\sqrt{1 - cos 2 x}}{x}

x = 0

f

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} 2 x + 3 & if - 3 \leq x < - 2 x + 1 & if - 2 \leq x < 0 - x + 2 & if 0 \leq x \leq 1 \end{matrix}

x = - 2

x = 0.

f : {1, 2, 3, . . .} \to {0, \pm 1, \pm 2, . . .}

y = f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{x}{2} if x is even - \frac{(x - 1)}{2}, if x is odd \end{matrix}

f^{- 1} (100)

f [- 1, \frac{- 1}{2}] \to [- 1, 1]

f (x) = 4 x^{3} - 3 x

f^{- 1} (x) =

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