# 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) \[\dfrac{25}{111}\]

Last updated date: 19th Mar 2023

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Answer

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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.

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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