0. For me N such that n2 m, the function f (r) 2n+ -sin x-1 Vxe [1,2] withf(1) J) has 2 +1 (a) no zero in [1, 2] (b) exactly zeros in [1, 2] (C) exactly two zeros in 1, 2 d) countable zeros in [1, 2]
u can easily check this with the help of graphs
the intersection point of
y= x^2n -1
y= sinx
will be root of given function.
there is one zero in this interval