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Question: let \(\text{f(x)}={{2}^{10}}\text{x}+1\) and \(\text{g(x)}={{3}^{10}}\text{x}-1\). If \(\text{(fog)(...

let f(x)=210x+1\text{f(x)}={{2}^{10}}\text{x}+1 and g(x)=310x−1\text{g(x)}={{3}^{10}}\text{x}-1. If (fog)(x)=x\text{(fog)(x)}=\text{x}, then x equal to.
A) 210−1210−3−10 B) 1−3−10210−3−10 C) 1−2−10310−2−10 D) 310−1310−2−10 \begin{aligned} & \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{aligned}

Explanation

Solution

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, f(x)=210x+1\text{f(x)}={{2}^{10}}\text{x}+1 and g(x)=310x−1\text{g(x)}={{3}^{10}}\text{x}-1
Let h(x) = (fog)(x)\text{h(x) = (fog)(x)}
h(x) = fo(g(x))\text{h(x) = fo(g(x))}
By using the functiong(x)=310x−1\text{g(x)}={{3}^{10}}\text{x}-1, we get
h(x) = f(310x−1)\text{h(x) = f(}{{3}^{10}}\text{x}-1\text{)}
By using the functionf(x)=210x+1\text{f(x)}={{2}^{10}}\text{x}+1, we get
h(x) = 210(310x−1)+1.....(1)\text{h(x) = }{{2}^{10}}({{3}^{10}}\text{x}-1)+1.....(1)
Since, (fog)(x)=x\text{(fog)(x)}=\text{x}
⇒h(x)=x....(2)\Rightarrow \text{h(x)}=\text{x}....\text{(2)}
From equation (1) and equation (2), we get
210(310x−1)+1=x{{2}^{10}}({{3}^{10}}\text{x}-1)+1=\text{x}
210310x−210+1=x{{2}^{10}}{{3}^{10}}\text{x}-{{2}^{10}}+1=\text{x}
Subtracting both sides by 210310x{{2}^{10}}{{3}^{10}}\text{x}, we get
210310x−210+1−210310x=x−210310x{{2}^{10}}{{3}^{10}}\text{x}-{{2}^{10}}+1-{{2}^{10}}{{3}^{10}}\text{x}=\text{x}-{{2}^{10}}{{3}^{10}}\text{x}
−210+1=x−210310x......(3)-{{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
−210+1=(1−210310)x....(4)-{{2}^{10}}+1=(1-{{2}^{10}}{{3}^{10}}\text{)x}....\text{(4)}
By dividing equation (4) both sides by 1−2103101-{{2}^{10}}{{3}^{10}}, we get
−210+11−210310=(1−210310)x1−210310.\dfrac{-{{2}^{10}}+1}{1-{{2}^{10}}{{3}^{10}}}=\dfrac{(1-{{2}^{10}}{{3}^{10}}\text{)x}}{1-{{2}^{10}}{{3}^{10}}}\text{.}
⇒x=−210+11−210310....(5)\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
⇒x=−(210−1)−(210310−1)\Rightarrow \text{x}=\dfrac{-({{2}^{10}}-1)}{-({{2}^{10}}{{3}^{10}}-1)}
⇒x=(210−1)(210310−1).....(6)\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 2102^{10}, we get
⇒x=210−1210210310−1210\Rightarrow \text{x}=\dfrac{\dfrac{{{2}^{10}}-1}{{{2}^{10}}}}{\dfrac{{{2}^{10}}{{3}^{10}}-1}{{{2}^{10}}}}
⇒x=210210−1210210310210−1210\Rightarrow \text{x}=\dfrac{\dfrac{{{2}^{10}}}{{{2}^{10}}}-\dfrac{1}{{{2}^{10}}}}{\dfrac{{{2}^{10}}{{3}^{10}}}{{{2}^{10}}}-\dfrac{1}{{{2}^{10}}}}
⇒x=1−1210310−1210\Rightarrow \text{x}=\dfrac{1-\dfrac{1}{{{2}^{10}}}}{{{3}^{10}}-\dfrac{1}{{{2}^{10}}}}
⇒x=1−2−10310−2−10\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.