Question

Test whether the following function is odd/even/either/neither

Answer 1

\[f(x)=\sin(\log(x-\sqrt{x^2+1}))\]

Answer 2

\[f(-x)=\sin(\log(\sqrt{x^2+1}-x)) \]

Answer 3

log (-A) is not -log A or log A
so, its neither.

Answer 4

its odd...

Answer 5

lets try an simplify
x= tan y

Answer 6

@yrelhan4

Answer 7

rationalize..

Answer 8

kyun?

Answer 9

just rationalize f(-x).. you'll get to know why.

Answer 10

kyun karna hai?kaise pata chalega?

Answer 11

ye rd ka question hai. to mereko pta hai.. :P

Answer 12

you'll be using log (1/A) = - log A property here.

Answer 13

an sin(-t) = - sin t

Answer 14

ye to IIT ques hai..and why cant i see anyone viewing this question?

Answer 15

u cant either?

Answer 16

if I rationalize then how i am I using 1/logA property?

Answer 17

okay now i can see

Answer 18

people viewing

Answer 19

:/

Answer 20

what u got after rationalizing ?

Answer 21

\[\large x-\sqrt{x^2+1} \times \frac{x+\sqrt{x^2+1}}{{x+\sqrt{x^2+1}}}=\frac{1}{\sqrt{x^2+1}}\]
hmm but how am i gonna think of this ?

Answer 22

misprint in = sorry :|

Answer 23

iit ka question hai mere bhai..

Answer 24

x+ also there

Answer 25

still :|

Answer 26

okay i have another doubt

Answer 27

\[\large \frac{1}{x+\sqrt{x^2+1}}\]
=logA? logA to 1/x-sqrt{x^2+1} tha..minus tha :|

Answer 28

nhi.. galat hai method. pta nhi. :/

Answer 29

@hartnn ? :O

Answer 30

what ?

Answer 31

dls's last post. rationalizing wont work.

Answer 32

it does ?

Answer 33

rationalizing changes middle sign :\]

Answer 34

we'll get log(x+ sqrt(x^2+1)) right?
but f(x) is.....
exactly...

Answer 35

dafuq :|

Answer 36

@yrelhan4 RD khol

Answer 37

yr dusre room me padi hai. :/

Answer 38

ja na please :|
mere solution me usse
(a+b)(a-b) ka use kara hai..why :/

Answer 39

\[\LARGE \sin(\log(\frac{1}{x+\sqrt{x^2+1}}))=\sin(\log1-\log(x+\sqrt{x^2+1}\]
..done