If fx-sin2 x -1n- then x-2 is a point of

Given nbsp;that, nbsp;f(x)=sin2 nbsp;x nbsp; 1nfπ2=0fπ2+=→0 n nbsp;and nbsp;fπ2 =→0 nIf nbsp;n nbsp;is nbsp;even, nbsp;fπ2+ nbsp;and nbs ...

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Local Maxima

If fx=sin2 x ...

Question

If $f (x) = {({sin}^{2} x - 1)}^{n}, then x = \frac{π}{2}$ is a point of

local maximum, if n is odd

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local minimum, if n is odd

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local maximum, if n is even

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local minimum, if n is even

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Solution

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f (x) = ({sin}^{2} x - 1)^{n}

x = \frac{π}{2}

f (x) = cos [π^{2}] x + cos [- π^{2}] x

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} - 2 sin x, & - π \leq x \leq \frac{- π}{2} a cos (x) + b, & \frac{- π}{2} < x < 0 - 2 + cos x, & 0 \leq x \leq \frac{π}{2} c sin x, & \frac{π}{2} < x \leq π \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

[- π, π],

f (x)

f (x) = \{\begin{matrix} \frac{1 - \sin x}{π - 2 x}, & x \neq \frac{π}{2} \\ k, & x = \frac{π}{2} \end{matrix}

x = \frac{π}{2},

f (x) = \frac{1 - \tan x}{4 x - π}, x \neq \frac{π}{4}, x \in [0, \frac{π}{2}] .

[0, \frac{π}{2}]

f (\frac{π}{4}) =

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