Question

Function \[f(x)= \frac{\sin^4x+\cos^4x}{x+x^tanx} \] is
A.Even
B.Odd
C.Periodic with period pi
D.Periodic with period 2pi

Answer 1

rewrite the function

Answer 2

done

Answer 3

seems periodic with 2pi .. but i wouldn't exactly call that

Answer 4

can you explain?

Answer 5

definitely not even and odd,
since f(-x) != -f(x)
and f(-x) != f(x) either

Answer 6

\[(\sin ^4 (-x) + \cos ^4 (-x)) /[ -x +(-x) \tan(-x)]\]
numerator is equal to
\[(sin4(x)+cos4(x))\]

denominator, we know that tan (-x) = -tan(x) so
-x + (-x) (- tan(x))=
-x + tan(x) =
so this will mean that the function is not odd or even

Answer 7

is this the function

\[f(x)= \frac{\sin^4x+\cos^4x}{x+x^{\tan(x)}}\]?

Answer 8

the sin and cosines have period of 360, they are squares ... so 180 would be fine ... but there's tan involved ... so period would be no way 180

Answer 9

If it is ... the answer is E: none of the above

Answer 10

now if you plot the graph ... tan(x) would be inf at 90, so at odd multiple of 90's , graph would return to 0 .. at odd multiple of ... odd multiples of 90's

Answer 11

oh man ... that would make period pi

Answer 12

Graph the function
((sin(x))^4 + (cos(x))^4 )/ (x + x*tan(x))
in the
http://www.coolmath.com/graphit/
I guess your function is wrong. check again

Answer 13

looks like i have to check again the definition of periodicity ....
http://en.wikipedia.org/wiki/Periodic_function

can positive side of this graph be called periodic
http://www4c.wolframalpha.com/Calculate/MSP/MSP94931a0iagba3h605ia400001b76666b87b9h4aa?MSPStoreType=image/gif&s=45&w=320&h=127&cdf=RangeControl

Answer 14

if you can call it periodic then your answer would be C

Answer 15

Oh sorry i lost connection in between

Function is \[\huge f(x)= \frac{\sin^4x+\cos^4x}{x+x^2tanx}\]
I'm so sorry!

Answer 16

@experimentX @Zarkon @alireza.safdari

Answer 17

i will give you the similar answer, if you can call it periodic, then the answer would be D .. ie period 2pi
http://www.wolframalpha.com/input/?i=y+%3D+%28%28sinx%29%5E4%2B%28cos%29%5E4%29%2F%28x%2Bx%5E2*%28tanx%29%29

Answer 18

I'm confused :S

Answer 19

It is an odd function
continue from last post
denominator, we know that tan (-x) = -tan(x) so
-x + (-x) (- tan(x))=
-x + ((-x)^2) *(-tan(x)) =
-x - (x^2)(tanx)

then we factor -1 (x + (x^2)(tanx))
now we know that
f(x) = -f(-x)

Answer 20

Diyadiya did you understand?

Answer 21

Oo.. he's right

Answer 22

I'm going through it
just a min

Answer 23

odd function

Answer 24

Ok yeah
i got it :) Thank you so much :)