The Permutational Composition

Using the given equation : $\displaystyle \prod_{r=1}^n (fog(r))^2=^{n+3}P_n$

$(fog(n))^2=\dfrac{\displaystyle \prod_{r=1}^n (fog(r))^2}{\displaystyle \prod_{r=1}^{n-1} (fog(r))^2}=\dfrac{^{n+3}P_n}{^{(n-1)+3}P_{n-1}}$

Putting $n=4$ , we get :

$(fog(4))^2=\dfrac{^7P_4}{^6P_3}=7$

From the given definition of function : $f(x)=\dfrac{\sqrt{3x^2+1}}{x}$

$fog(4)=\dfrac{\sqrt{3(g(4))^2+1}}{g(4)}$

On squaring both sides, we will get :

$(fog(4))^2=\dfrac{3(g(4))^2+1}{(g(4))^2}$

$\Rightarrow 7=\dfrac{3(g(4))^2+1}{(g(4))^2}$

$\Rightarrow (g(4))^2=\frac{1}{4}$

$\Rightarrow g(4)=\dfrac{1}{2} \ \text{ or } \ g(4)=-\dfrac{1}{2}$

• $g(4)=-\dfrac{1}{2}$ is neglected because $g$ is defined as a positive function.

Hence $g(4)=\dfrac{1}{2} \Rightarrow 4g(4)=\boxed{2}$

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