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# If $f:R\to R,g:R\to R$ are defined by $f\left( x \right)=5x-3,\text{ }g\left( x \right)={{x}^{2}}+3$, then $\left( go{{f}^{-1}} \right)\left( 3 \right)=$ (a) $\dfrac{25}{3}$(b) $\dfrac{111}{25}$(c) $\dfrac{9}{25}$(d) $\dfrac{25}{111}$  Hint: First of all find ${{f}^{-1}}\left( x \right)$ by finding x in terms of f(x) and replace x and f(x) by ${{f}^{-1}}\left( x \right)$ and x respectively. Substitute x = 3 to find ${{f}^{-1}}\left( 3 \right)$. Now to find $g{{f}^{-1}}\left( 3 \right)$, substitute $x={{f}^{-1}}\left( 3 \right)$ in g(x).

We are given a function $f\left( x \right)=5x-3\text{ and }g\left( x \right)={{x}^{2}}+3$. We have to find $go{{f}^{-1}}\left( 3 \right)$ or $g{{f}^{-1}}\left( 3 \right)$.
First of all, let us consider the function f(x).
$f\left( x \right)=5x-3....\left( i \right)$
First of all, we will express x in terms of f(x).
By adding 3 to both sides of the equation (i), we will get,
$f\left( x \right)+3=5x$
By dividing both the sides of the above equation by 5, we will get
$\dfrac{f\left( x \right)+3}{5}=x$
Or, $x=\dfrac{f\left( x \right)+3}{5}$
Now, to find ${{f}^{-1}}\left( x \right)$, we will replace x by ${{f}^{-1}}\left( x \right)$ and f(x) by x, we will get,
${{f}^{-1}}\left( x \right)=\dfrac{x+3}{5}$
By substituting x = 3, we will get,
${{f}^{-1}}\left( 3 \right)=\dfrac{3+3}{5}=\dfrac{6}{5}$
Now we know that,
$g\left( x \right)={{x}^{2}}+3$
By substituting $x={{f}^{-1}}\left( 3 \right)$ in the above equation, we will get,
$g\left( {{f}^{-1}}\left( 3 \right) \right)={{\left[ {{f}^{-1}}\left( 3 \right) \right]}^{2}}+3$
Now, we know that ${{f}^{-1}}\left( 3 \right)=\dfrac{6}{5}$, so by substituting the value of ${{f}^{-1}}\left( 3 \right)$ in RHS of the above equation, we will get
$g\left( {{f}^{-1}}\left( 3 \right) \right)={{\left( \dfrac{6}{5} \right)}^{2}}+3$
By simplifying the above equation, we will get,
$g{{f}^{-1}}\left( 3 \right)=\dfrac{36}{25}+3$
$\Rightarrow g{{f}^{-1}}\left( 3 \right)=\dfrac{36+75}{25}=\dfrac{111}{25}$
Therefore, we get $g{{f}^{-1}}\left( 3 \right)=\dfrac{111}{25}\text{ or }\left( go{{f}^{-1}} \right)\left( 3 \right)=\dfrac{111}{25}$.
Hence, option (b) is the right answer.

Note: Many students confuse between gf(x) and f(g(x)) and many even consider them the same function. But fg(x) and gf(x) are not the same. When we substitute x = g(x) in f(x), then we get f(g(x)) whereas when we substitute x = f(x) in g(x), then we get gf(x). Also it must be noted that if fg(x) = gf(x), then f(x) and g(x) are inverse of each other. Also, in this question some students first calculate $g{{f}^{-1}}\left( x \right)$ and then substitute x = 3, this is correct but is time consuming, so it is better to find ${{f}^{-1}}\left( x \right)$ and substitute x = 3 there only and then proceed to find $g{{f}^{-1}}\left( 3 \right)$ directly.

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