If f : R → R and g : R → R defined by $f(x) = 3x-1$ and $g(x) = { x }^{ 2 } +1$, then the value of x for which f(g(x)) = 29 is

$\pm 4$ $\pm 5$ $\pm 3$ $\pm 2$

$\pm 3$

$f(x) = { x }^{ 3 }$, $g(x) =x^2+1$ $f(g(x))=f(x^2+1)=3(x^2+1)-1=3x^2+2$ so $f(g(x))=29⇒3x^2+2=29$ $⇒3x^2=27$ $x^2=9$ $x=\pm 3$