If `g[f(x)]=|sinx|, f[g(x)]=(sinsqrtx)^(2)` then

If g(f(x))=|sinx|a n df(g(x))=(sinsqrt(x))^2 , then f(x)=sin^2x ,g(x)=sqrt(x) f(x)=sinx ,g(x)=|x| f(x=x^2,g(x)=sinsqrt(x) fa n dg cannot be determined

If g(f(x))=|sinx|a n df(g(x))=(sinsqrt(x))^2 , then (a).f(x)=sin^2x ,g(x)=sqrt(x) (b).f(x)=sinx ,g(x)=|x| (c)f(x=x^2,g(x)=sinsqrt(x) (d).f\ a n d \g cannot be determined

If g(f(x))=|sinx|a n df(g(x))=(sinsqrt(x))^2 , then (a) f(x)=sin^2x ,g(x)=sqrt(x) (b) f(x)=sinx ,g(x)=|x| (c) f(x) =x^2,g(x)=sinsqrt(x) (d) f and g cannot be determined

If g(f(x))= |sinx | and f(g(x))=(sin sqrt(x))^(2) , then

For x in RR, f(x)=|log2-sinx| and g(x)=f(f(x)) , then

If f(x) = 2 sinx, g(x) = cos^(2) x , then the value of (f+g)((pi)/(3))

If f(x) = 2 sinx, g(x) = cos^(2) x , then the value of (f+g)((pi)/(3))

If f(x)=sinx+cosx, g(x)= x^(2)-1 , then g{f(x)} is invertible in the domain

If f(x)=sinx+cosx, g(x)=x^(2)-1, then g(f(x)) is invertible in the domain

Let f(x)=2x-sinx and g(x) = 3^(sqrtx) . Then