Let fx - cos -2- x - cos -2- x- where - - represents greatest integer function- then f -2 - -1 and f- -

The correct option is C 0 [ We know that,; π =3 ...

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Continuity in an Interval

Let fx = co...

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Let $f (x) = cos [π^{2}] x + cos [- π^{2}] x, w h e r e$ [ .] $r e p r e s e n t s g r e a t e s t i n t e g e r f u n c t i o n, t h e n f (\frac{π}{2}) = - 1 a n d f (- π) =$ ?

-1

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π

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Similar questions

If, $f (x) = c o s [π^{2}] x + c o s [- π^{2}] x,$ where [x] stands for greatest integer function, then

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

f (\frac{- π}{4})

y = \sqrt{\frac{1 - cos 2 x}{1 + cos 2 x}}, x \in (0, \frac{π}{2}) \cup (\frac{π}{2}, π)

\frac{d y}{d x}

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

[x]

f (\frac{π}{2})

f (π)

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)

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