IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

priyesh

f(x) = x - sinx

so

f ' (x) = 1 - cosx

now for x b/w (0,pi/2)

f ' (x) is >0

=> f(x) is increasing

so minimum value of f(x) is lim x>>>>0 f(x) = 0 - 0 = 0

so f(x) is > 0 in the given interval

similarly g(x) = x - tanx

g ' (x) = 1 - sec^2 x = 1 - 1/cos^2 x

now when x lies b/w (0,pi/2)

cos^2 x lies b/w (0,1)

so g ' (x) < 0 in the given interval

=>so maximum value of g(x) is limx>>>>0 g(x) = 0 - 0 = 0

so g(x) is < 0 in the given interval

hence f(x) g(x) < 0 in the given interval