IIT JEE , AIEEE Forum - Mathematics , Trignometry

vahiniprasad

m,n are any 2 +ve values of x for wich 2cosx,|cosx| and 1-3cos^2x are in g.p.the max value of |m-n|

/3

/4

/2

Decoder

first..apply the condition for..g.p..solving tht u get..
.cosx= 2 - 6cos^2x..
now we know m-n is rootD/a..(diff. of roots)...
calculating..roots...1 N other is not possible..so next value of m-n..can have max as infinity...so strange..

sprinkle

Vahiniprasad!

Suppose cos x = y

=> y

[-1, 1]

now 2y, |y| and 1-3y^2 are in G.P.

=> |y|^2 = 2y(1-3y^2)
=> y^2 = 2y - 6y^3
=> 6y^3 + y^2 - 2y = 0
solving we get, y = 0, 1/2, -2/3

=> cos x = 0, 1/2, -2/3 (though cos x can't be 0 as then 0, 0, 1 are not in G.P.)
=> x = (2k pi + pi/2), (2k pi - pi/2), (2k pi + pi/3), (2k pi - pi/3),
(2k pi + (sqrt(3) - 1) pi), (2k pi - (sqrt(3) - 1) pi) where k is an integer

so answer is none of ur values given. rather it can be infinity.

Please buzz me if any doubts!

Ask & Discuss Questions with Community & Experts