Practicing Success
Let $f(x)=x^2$ and $g(x)=2^x$, then the solution set of $fog(x) = gof (x)$ is |
$R$ {0} {0, 2} none of these |
{0, 2} |
$fog(x)=f(g(x))=f(2^x)=(2^x)^2=2^{2x}$ and $gof(x)=g(f(x))=g(x^2)=2^{x^2}$ Thus the solution of $2^{2x}=2^{x^2}$ is given by $x^2= 2x$ which is x = 0, 2. |