Question

Range question

Answer 1

Where is the function?

Answer 2

tan(pi[x^2 - x] / 1 +sin(cos(x)) wher [.] is greatest integer function

Answer 3

\[\tan(\pi \left[x ^{2}-x \right])\div 1+\sin(cosx)\]

Answer 4

\[\left[ . \right]\] is greatest integer function!

Answer 5

\[\tan(\pi \left[x ^{2}-x \right])+\sin(\cos x)\]

First term will always result in \(\tan (n\ \pi)\) n=is an integer
so it'll be zero
DO you get this?

Answer 6

wait.. i think u wrote the quest rong..
numerator is tan(pi[x^2-x])
denominator is 1 + sin(cosx)

Answer 7

Ok, Let's do again

Answer 8

\[\frac {\tan(\pi \left[x ^{2}-x \right])}{1+\sin(\cos x)}\]

Answer 9

yupp..

Answer 10

Tell me what's tan n*pi

Answer 11

0

Answer 12

numerator will always be zero since
\[[x^2-x]\longrightarrow \ integer\]

Answer 13

okk..

Answer 14

so the numerator always becomes zero

Answer 15

Yeah, so what's the range?

Answer 16

just 0?

Answer 17

Yeah:)

Answer 18

but the options to the question are :
a) \[\left( -\infty,\infty \right) - [0, \tan1]\]
b) \[\left( -\infty,\infty \right) - [\tan2,0]\]
c) [tan 2, tan1]
d) \[\left( -\infty,\infty \right)\]

Answer 19

third option is open interval 0 by the way