# Let f : R → R be defined by

Question:

Let $f: R \rightarrow R$ be defined by $f(x)=3 x^{2}-5$ and $g: R \rightarrow R$ by $g(x)=\frac{x}{x^{2}+1} .$ Then $(g$ of $)(x)$ is

(a) $\frac{3 x^{2}-5}{9 x^{4}-30 x^{2}+26}$

(b) $\frac{3 x^{2}-5}{9 x^{4}-6 x^{2}+26}$

(c) $\frac{3 x^{2}}{x^{4}+2 x^{2}-4}$

(d) $\frac{3 x^{2}}{9 x^{4}+30 x^{2}-2}$

Solution:

Given: $f(x)=3 x^{2}-5$ and $g(x)=\frac{x}{x^{2}+1}$

$(g o f)(x)=g(f(x))$

$=g\left(3 x^{2}-5\right)$

$=\frac{3 x^{2}-5}{\left(3 x^{2}-5\right)^{2}+1}$

$=\frac{3 x^{2}-5}{\left(3 x^{2}\right)^{2}+5^{2}-2\left(3 x^{2}\right)(5)+1}$

$=\frac{3 x^{2}-5}{9 x^{4}+25-30 x^{2}+1}$

$=\frac{3 x^{2}-5}{9 x^{4}-30 x^{2}+26}$

​Hence, the correct option is (a).