Question

let $\text{f(x)}={{2}^{10}}\text{x}+1$ and $\text{g(x)}={{3}^{10}}\text{x}-1$. If $\text{(fog)(x)}=\text{x}$, then x equal to.
$\begin{align}
  & \text{A) }\dfrac{{{2}^{10}}-1}{{{2}^{10}}-{{3}^{-10}}} \\
 & \text{B) }\dfrac{1-{{3}^{-10}}}{{{2}^{10}}-{{3}^{-10}}} \\
 & \text{C) }\dfrac{1-{{2}^{-10}}}{{{3}^{10}}-{{2}^{-10}}} \\
 & \text{D) }\dfrac{{{3}^{10}}-1}{{{3}^{10}}-{{2}^{-10}}} \\
\end{align}$

Answer 1

Hint: In this, we will first find the composition of function f(x) and g(x) and the find the value of x. Composition of two function is an operation that takes between the two function f and g to produce another function h.

Complete step by step answer:
Given that, $\text{f(x)}={{2}^{10}}\text{x}+1$ and $\text{g(x)}={{3}^{10}}\text{x}-1$
Let $\text{h(x) = (fog)(x)}$
$\text{h(x) = fo(g(x))}$
By using the function$\text{g(x)}={{3}^{10}}\text{x}-1$, we get
$\text{h(x) = f(}{{3}^{10}}\text{x}-1\text{)}$
By using the function$\text{f(x)}={{2}^{10}}\text{x}+1$, we get
$\text{h(x) = }{{2}^{10}}({{3}^{10}}\text{x}-1)+1.....(1)$
Since, $\text{(fog)(x)}=\text{x}$
$\Rightarrow \text{h(x)}=\text{x}....\text{(2)}$
From equation (1) and equation (2), we get
${{2}^{10}}({{3}^{10}}\text{x}-1)+1=\text{x}$
${{2}^{10}}{{3}^{10}}\text{x}-{{2}^{10}}+1=\text{x}$
Subtracting both sides by ${{2}^{10}}{{3}^{10}}\text{x}$, we get
${{2}^{10}}{{3}^{10}}\text{x}-{{2}^{10}}+1-{{2}^{10}}{{3}^{10}}\text{x}=\text{x}-{{2}^{10}}{{3}^{10}}\text{x}$
$-{{2}^{10}}+1=\text{x}-{{2}^{10}}{{3}^{10}}\text{x}......\text{(3)}$
Taking x common from right hand side of equation (3), we get
$-{{2}^{10}}+1=(1-{{2}^{10}}{{3}^{10}}\text{)x}....\text{(4)}$
By dividing equation (4) both sides by $1-{{2}^{10}}{{3}^{10}}$, we get
\[\dfrac{-{{2}^{10}}+1}{1-{{2}^{10}}{{3}^{10}}}=\dfrac{(1-{{2}^{10}}{{3}^{10}}\text{)x}}{1-{{2}^{10}}{{3}^{10}}}\text{.}\]
\[\Rightarrow \text{x}=\dfrac{-{{2}^{10}}+1}{1-{{2}^{10}}{{3}^{10}}}....\text{(5)}\]
Taking minus sign common form equation (5) right hand side’s numerator and denominator, we get
\[\Rightarrow \text{x}=\dfrac{-({{2}^{10}}-1)}{-({{2}^{10}}{{3}^{10}}-1)}\]
\[\Rightarrow \text{x}=\dfrac{({{2}^{10}}-1)}{({{2}^{10}}{{3}^{10}}-1)}.....(6)\]
By dividing numerator and denominator of right hand side of equation (6) by $2^{10}$, we get
\[\Rightarrow \text{x}=\dfrac{\dfrac{{{2}^{10}}-1}{{{2}^{10}}}}{\dfrac{{{2}^{10}}{{3}^{10}}-1}{{{2}^{10}}}}\]
\[\Rightarrow \text{x}=\dfrac{\dfrac{{{2}^{10}}}{{{2}^{10}}}-\dfrac{1}{{{2}^{10}}}}{\dfrac{{{2}^{10}}{{3}^{10}}}{{{2}^{10}}}-\dfrac{1}{{{2}^{10}}}}\]
\[\Rightarrow \text{x}=\dfrac{1-\dfrac{1}{{{2}^{10}}}}{{{3}^{10}}-\dfrac{1}{{{2}^{10}}}}\]
\[\Rightarrow \text{x}=\dfrac{1-{{2}^{-10}}}{{{3}^{10}}-{{2}^{-10}}}\]

So, the correct answer is “Option C”.

Note: In this one should know what function is?
Let A and B be two non-empty sets. Then a function from A to B is defined to be a relation which is uniquely related to the element of set A to element of set B.