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
Answer & explanation
Correct answer: option 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 $f(g(x))=29⇒3x^2+2=29$
$⇒3x^2=27$
$x^2=9$
$x=\pm 3$