IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

neeraj_agarwal_1990

Question:

The number of roots of the equation

is:

Options:

A.

0

B.

1

C.

2

D.	more than two

Answer

Answer:

C

pottermania1990

for my comfort im changing the domain from [0-4pie]

cosx shud always b positive...i.e in 1st and 4th quadrant, also sin x shud be positive..so the ans .shud lie in the 1st quadrant..when we draw the graph of x^2 and cosx,they both intersect at two points..and hence the answer....

neeraj_agarwal_1990

but doesn't the value of sin-1sinx change with domain?

dinesh_ddt

graph of sin-1sin x 's domain is whole real numbers
from this itself can say the solutions will be infinite
but here the range in only -2pi to 2pi so no of solns is 2

arpan1

rewrite the ques as

sin (

cos x) = sin x

then

cos x = x

or, cos x = x²

obviously it has 2 solutions

rate me

neeraj_agarwal_1990

but sin-1sinx is not always x..?

arpan1

see thats why i converted the inverse function into general trigonometric function so that u do not have any restrictions on its domain

hope u got it

rate me

dinesh_ddt

see this ull get it

srini

Hey,

in [0.pi/2], sin^-1(sinx)=x =>\cosx\=x^2;

[pi/2,pi] pi/2 <x<pi

=>pi-x => |cosx|=(pi-x)^2

Its this way?

arpan1

@ dinesh

which graph hav u drawn

dinesh_ddt

its d graph of sin-1(sin x)

Ask & Discuss Questions with Community & Experts