If g:-2,2→R, where fx=x3+tanx+x2+1P is an odd function, then the value of parametric P, where [.] denotes the greatest integer function, is
-5<P<5
P<5
P>5
none of these
gx=x3+tanx+x2+1P
or g-x=x3+tan-x+-x2+1P
=-x3-tanx+x2+1P
Since g(x) is odd,
gx+g-x=0
∴ x3+tanx+x2+1P+-x3-tanx +x2+1P =0
or 2x2+1P=0 or 0≤x2+1P<1
Now, x∈-2,2
∴ x2+1∈1,5∴ P>5