Answer:
[tex]\text{Solution:}\\\text{f}(x)=3x-1\\\text{g}(x)=x^2+4\\\text{Now,}\\\text{fg}(x)=\text{f\{g}(x)\}=\text{f}(x^2+4)=3(x^2+4)-1=3x^2+12-1=3x^2+11\\\text{gf}(x)=\text{g\{f}(x)\}=\text{g}(3x-1)=(3x-1)^2+4=9x^2-6x+1+4=9x^2-6x+5\\[/tex]
[tex]\text{Also we are given:}\\\text{fg}(x)=2\text{gf}(x)\\\text{or,\ }3x^2+11=2(9x^2-6x+4)\\\text{or,\ }3x^2+11=18x^2-12x+10\\\text{or,\ }15x^2-12x-1=0\\\text{proved}[/tex]
