Question

Let f : R → R be defined by f(x) = 3x² – 5 and g : R → R by $g\left(x\right)=\frac{x}{x^2+1}.$ Then (gof) (x) is
(a) $\frac{3x^2-5}{9x^4-30x^2+26}$

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

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

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

Answer 1

Given: f(x) = 3x² – 5 and $g\left(x\right)=\frac{x}{x^2+1}$

$$\begin{gathered} \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right) \\ =g\left(3x^2-5\right) \\ =\frac{3x^2-5}{{\left(3x^2-5\right)}^2+1} \\ =\frac{3x^2-5}{{\left(3x^2\right)}^2+5^2-2\left(3x^2\right)\left(5\right)+1} \\ =\frac{3x^2-5}{9x^4+25-30x^2+1} \\ =\frac{3x^2-5}{9x^4-30x^2+26} \end{gathered}$$

​Hence, the correct option is (a).

​