You are not logged in.
Pages: 1
dear friends,
i need a solution for this sum
the smallest positive root of the equation tanx-x=0 lies in
(a) (0,pi/2) (b) (pi/2,pi)
(c) (pi,3pi/2) (d) (3pi/2,2pi)
please help me.
thanks
bhaskar
Offline
x and tan x coincide at x=0, but then tan x > x for 0<x<π/2. (You can check this with gradients.)
So no solution exists in (0,π/2). Clearly no solution exists in (π/2,π) either, because tan x is negative in that domain.
tan(π) = 0 and tan(3π/2) -> ∞, so within that interval some x must exist such that tanx = x (since tan is continuous within the interval).
c) is the answer.
Why did the vector cross the road?
It wanted to be normal.
Offline
thank you very much
sorry for the delayed reply as i haven't logged in for a while
baskar:)
Offline
Hi,
When plotting the graph of y= tan(x) and y=x the two curves do intersect at some point in the third quardenet i.e between pi and 3pi/2
w.b.w
Mr riad zaidan
Al-quds Open University
Jenin
Palistine
Pages: 1