
If \[f(x) = xx - 1\], then\[\left( {fofo...of} \right)\left( x \right)19\] times is equal to:
A. \[xx - 1\]
B. \[\left( {xx - 1} \right)19\]
C. \[19xx - 1\]
D. \[x\]
Answer
162.9k+ views
Hint: If you can locate the inverse function of \[f\], you can solve for the missing value \[\left( x \right)\]. Take \[f\] 's derivative with respect to \[x\] and utilize substitution along with that knowledge to do this. You can accomplish this by determining the value of fo (or any other variable), adding it to the equation, and then changing back to x to obtain \[xofo...ofx = \dfrac{x}{{x - 1}}\].
Complete step by step solution: The given equation is \[f(x) = \dfrac{x}{{x - 1}}\]
In order to solve for \[x\] in this equation, we first need to simplify the negative divisor.
That is \[(fof)(x) = f\{ f(x)\} = f(\dfrac{x}{{x - 1}})\]
\[\begin{array}{l}\dfrac{{\dfrac{x}{{x - 1}}}}{{\dfrac{x}{{x - 1}} - 1}} = \dfrac{{\dfrac{x}{{x - 1}}}}{{\dfrac{{x - x + 1}}{{x - 1}}}}\\ = > \dfrac{{\dfrac{x}{{x - 1}}}}{{\dfrac{1}{{x - 1}}}} = x\end{array}\]
So, \[(fofof)(x) = f(fof)(x) = f(x) = \dfrac{x}{{x - 1}}\]
\[(fofof...of)(x) = f(fof)(x) = f(x) = \dfrac{x}{{x - 1}}\]
If \[f\left( x \right)\]is a function that takes on one or more real numbers as input, and \[x\] occurs in the domain of \[f\left( x \right)\], then \[\left( {fofo...of} \right)\left( x \right)\]will always be equal to: \[\dfrac{x}{{x - 1}}\]. This property can be verified using integration by parts.
So, A is the answer.
That is , the \[(fofof...of)\]part is of \[19\] times which also gives the same value as it does not change much.
If \[f(x) = \dfrac{x}{{x - 1}}\], then \[(fofof...of)(x)\] \[19\]times is equal to: \[\dfrac{x}{{x - 1}}\]. That's because when we multiply two negative numbers together, the result will be a positive number. In this case, \[x\] gets multiplied by \[ - 1\]and that equals \[1\].
So, Option ‘A’ is correct.
Note: "If \[f\left( x \right) = \dfrac{x}{{x - 1}}\], then \[\left( {fofo...of} \right)\left( x \right)19\]times is equal to \[\dfrac{x}{{x - 1}}\]"
The equation's parentheses inform us that we are multiplying \[19\] terms together. Therefore, all we have to do to put the result of each phrase in parenthesis is place parentheses beneath the multiplication symbol and separate them.
The outcome of \[19\] times adding \[f\left( x \right)\]to itself is a real number in the range of \[ - 1\] and \[1\]. The distributive property can be used to determine this number.
Complete step by step solution: The given equation is \[f(x) = \dfrac{x}{{x - 1}}\]
In order to solve for \[x\] in this equation, we first need to simplify the negative divisor.
That is \[(fof)(x) = f\{ f(x)\} = f(\dfrac{x}{{x - 1}})\]
\[\begin{array}{l}\dfrac{{\dfrac{x}{{x - 1}}}}{{\dfrac{x}{{x - 1}} - 1}} = \dfrac{{\dfrac{x}{{x - 1}}}}{{\dfrac{{x - x + 1}}{{x - 1}}}}\\ = > \dfrac{{\dfrac{x}{{x - 1}}}}{{\dfrac{1}{{x - 1}}}} = x\end{array}\]
So, \[(fofof)(x) = f(fof)(x) = f(x) = \dfrac{x}{{x - 1}}\]
\[(fofof...of)(x) = f(fof)(x) = f(x) = \dfrac{x}{{x - 1}}\]
If \[f\left( x \right)\]is a function that takes on one or more real numbers as input, and \[x\] occurs in the domain of \[f\left( x \right)\], then \[\left( {fofo...of} \right)\left( x \right)\]will always be equal to: \[\dfrac{x}{{x - 1}}\]. This property can be verified using integration by parts.
So, A is the answer.
That is , the \[(fofof...of)\]part is of \[19\] times which also gives the same value as it does not change much.
If \[f(x) = \dfrac{x}{{x - 1}}\], then \[(fofof...of)(x)\] \[19\]times is equal to: \[\dfrac{x}{{x - 1}}\]. That's because when we multiply two negative numbers together, the result will be a positive number. In this case, \[x\] gets multiplied by \[ - 1\]and that equals \[1\].
So, Option ‘A’ is correct.
Note: "If \[f\left( x \right) = \dfrac{x}{{x - 1}}\], then \[\left( {fofo...of} \right)\left( x \right)19\]times is equal to \[\dfrac{x}{{x - 1}}\]"
The equation's parentheses inform us that we are multiplying \[19\] terms together. Therefore, all we have to do to put the result of each phrase in parenthesis is place parentheses beneath the multiplication symbol and separate them.
The outcome of \[19\] times adding \[f\left( x \right)\]to itself is a real number in the range of \[ - 1\] and \[1\]. The distributive property can be used to determine this number.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Instantaneous Velocity - Formula based Examples for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

JEE Advanced 2025 Notes
