If- fx-cos-2-x-cos -2-x- where -x- stands for greatest integer function- then

F(x)=cos[π^2]x+cos[ π^2]x f(x) = cos 9x + cos ( 10)x f(x) = cos 9x + cos 10 x ∴ f ( π/2 )=0 1= 1 f (π) = 1 + 1 = 0 f ( π) = 1 + 1 = 0 f ( π/4 )=1/√(2)+0=1/√(2 ...

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Standard XIII

Mathematics

Domain

If, fx=cos[π2...

Question

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

$f (\frac{π}{2}) = 1$

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$f (π) = 1$

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$f (- π) = 0$

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$f (\frac{π}{4}) = 2$

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Solution

The correct option is C
$f (- π) = 0$

$f (x) = c o s [π^{2}] x + c o s [- π^{2}] x$
f(x) = cos 9x + cos (-10)x
f(x) = cos 9x + cos 10 x
$∴ f (\frac{π}{2}) = 0 - 1 = - 1$
f $(π)$ = -1 + 1 = 0
f $(- π)$ = -1 + 1 = 0
$f (\frac{π}{4}) = \frac{1}{\sqrt{2}} + 0 = \frac{1}{\sqrt{2}}$

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Similar questions

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$f (x) = cos [π^{2}] x + cos [- π^{2}] x$ , where $[x]$ stands for the greatest integer
function, then

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

Q. If

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

, where

[.]

denotes the greatest integer function, then which of the following is not correct

Q. Let

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

where

[x]

is the greatest integer function, then find

f (\frac{π}{2})

and

f (π)

Q. If

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

then

f^{'} (x)

is, here

[π^{2}]

and

[{- π}^{2}]

greatest integer function not greater than its value

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If, f(x)=cos[π2]x+cos[−π2]x, where [x] stands for greatest integer function, then

The correct option is C f(−π)=0 f(x)=cos[π2]x+cos[−π2]x f(x) = cos 9x + cos (-10)x f(x) = cos 9x + cos 10 x ∴ f(π2)=0−1=−1 f (π) = -1 + 1 = 0 f (−π) = -1 + 1 = 0 f(π4)=1√2+0=1√2

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

The correct option is C
$f (- π) = 0$

$f (x) = c o s [π^{2}] x + c o s [- π^{2}] x$
f(x) = cos 9x + cos (-10)x
f(x) = cos 9x + cos 10 x
$∴ f (\frac{π}{2}) = 0 - 1 = - 1$
f $(π)$ = -1 + 1 = 0
f $(- π)$ = -1 + 1 = 0
$f (\frac{π}{4}) = \frac{1}{\sqrt{2}} + 0 = \frac{1}{\sqrt{2}}$