
If \[f\left( x \right) = \dfrac{{\left( {x - 1} \right)}}{4} + \dfrac{{{{\left( {x - 1} \right)}^3}}}{{12}} + \dfrac{{{{\left( {x - 1} \right)}^5}}}{{20}} + \dfrac{{{{\left( {x - 1} \right)}^7}}}{{28}} + ...\] where \[0 < x < 4\], then find the value of \[f'\left( x \right)\].
A. \[\dfrac{1}{{4x\left( {2 - x} \right)}}\]
B. \[\dfrac{1}{{4{{\left( {x - 2} \right)}^2}}}\]
C. \[\dfrac{1}{{\left( {2 - x} \right)}}\]
D. \[\dfrac{1}{{\left( {2 + x} \right)}}\]
Answer
162.6k+ views
Hint In the given question the sum of infinite terms is given . We will find the derivative of the equation \[f\left( x \right)\] with respect to \[x\]. The infinite terms of the derivative equation \[f'\left( x \right)\] are in geometric progression. So, by using the formula of sum of infinite terms in geometric progression, we will find the value of \[f'\left( x \right)\].
Formula used
Power rule of derivative: \[\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}\]
The sum of infinite terms in GP: \[{S_\infty } = \dfrac{a}{{\left( {1 - r} \right)}}\]
where \[a\] is the first term and \[r\] is the common ratio, where \[r \ne 0\] and \[r < 1\] .
Complete step by step solution:
The given equation is \[f\left( x \right) = \dfrac{{\left( {x - 1} \right)}}{4} + \dfrac{{{{\left( {x - 1} \right)}^3}}}{{12}} + \dfrac{{{{\left( {x - 1} \right)}^5}}}{{20}} + \dfrac{{{{\left( {x - 1} \right)}^7}}}{{28}} + ...\] where \[0 < x < 4\].
Let’s calculate the derivative of the above equation with respect to \[x\].
\[f'\left( x \right) = \dfrac{1}{4} + \dfrac{{3{{\left( {x - 1} \right)}^2}}}{{12}} + \dfrac{{5{{\left( {x - 1} \right)}^4}}}{{20}} + \dfrac{{7{{\left( {x - 1} \right)}^6}}}{{28}} + ...\]
Simplify the above equation
\(f'\left( x \right) = \dfrac{1}{4} + \dfrac{{{{\left( {x - 1} \right)}^2}}}{4} + \dfrac{{{{\left( {x - 1} \right)}^4}}}{4} + \dfrac{{{{\left( {x - 1} \right)}^6}}}{4} + ...\)
Simplify further the above equation
\[ \Rightarrow \] \[f'\left( x \right) = \dfrac{1}{4}\left[ {1 + {{\left( {x - 1} \right)}^2} + {{\left( {x - 1} \right)}^4} + {{\left( {x - 1} \right)}^6} + ...} \right]\]
The terms present in the square bracket are in geometric progression where the first term is \(a = 1\) and the common ratio is \[r = \frac{{\frac{{{{\left( {x - 1} \right)}^2}}}{4}}}{{\frac{1}{4}}} = {\left( {x - 1} \right)^2}\].
Apply the formula of the sum of infinite terms in geometric progression in the square bracket.
\[f'\left( x \right) = \dfrac{1}{4}\left[ {\dfrac{1}{{\left( {1 - {{\left( {x - 1} \right)}^2}} \right)}}} \right]\]
Simplify the above equation
\[f'\left( x \right) = \dfrac{1}{4}\left[ {\dfrac{1}{{1 - \left( {{x^2} - 2x + 1} \right)}}} \right]\]
\[ \Rightarrow \]\[f'\left( x \right) = \dfrac{1}{4}\left[ {\dfrac{1}{{2x - {x^2}}}} \right]\]
\[ \Rightarrow \]\[f'\left( x \right) = \dfrac{1}{{4x\left( {2 - x} \right)}}\]
Hence the correct option is option A.
Note: Students are often confused with the formula \[\dfrac{d}{{dx}}{\left( {ax + b} \right)^n} = n{\left( {ax + b} \right)^{n - 1}}\] and \[\dfrac{d}{{dx}}{\left( {ax + b} \right)^n} = n{\left( {ax + b} \right)^{n - 1}} \times a\] . But the correct formula is \[\dfrac{d}{{dx}}{\left( {ax + b} \right)^n} = n{\left( {ax + b} \right)^{n - 1}} \times a\]. Because we also have to calculate the derivative of the terms present in the bracket.
Formula used
Power rule of derivative: \[\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}\]
The sum of infinite terms in GP: \[{S_\infty } = \dfrac{a}{{\left( {1 - r} \right)}}\]
where \[a\] is the first term and \[r\] is the common ratio, where \[r \ne 0\] and \[r < 1\] .
Complete step by step solution:
The given equation is \[f\left( x \right) = \dfrac{{\left( {x - 1} \right)}}{4} + \dfrac{{{{\left( {x - 1} \right)}^3}}}{{12}} + \dfrac{{{{\left( {x - 1} \right)}^5}}}{{20}} + \dfrac{{{{\left( {x - 1} \right)}^7}}}{{28}} + ...\] where \[0 < x < 4\].
Let’s calculate the derivative of the above equation with respect to \[x\].
\[f'\left( x \right) = \dfrac{1}{4} + \dfrac{{3{{\left( {x - 1} \right)}^2}}}{{12}} + \dfrac{{5{{\left( {x - 1} \right)}^4}}}{{20}} + \dfrac{{7{{\left( {x - 1} \right)}^6}}}{{28}} + ...\]
Simplify the above equation
\(f'\left( x \right) = \dfrac{1}{4} + \dfrac{{{{\left( {x - 1} \right)}^2}}}{4} + \dfrac{{{{\left( {x - 1} \right)}^4}}}{4} + \dfrac{{{{\left( {x - 1} \right)}^6}}}{4} + ...\)
Simplify further the above equation
\[ \Rightarrow \] \[f'\left( x \right) = \dfrac{1}{4}\left[ {1 + {{\left( {x - 1} \right)}^2} + {{\left( {x - 1} \right)}^4} + {{\left( {x - 1} \right)}^6} + ...} \right]\]
The terms present in the square bracket are in geometric progression where the first term is \(a = 1\) and the common ratio is \[r = \frac{{\frac{{{{\left( {x - 1} \right)}^2}}}{4}}}{{\frac{1}{4}}} = {\left( {x - 1} \right)^2}\].
Apply the formula of the sum of infinite terms in geometric progression in the square bracket.
\[f'\left( x \right) = \dfrac{1}{4}\left[ {\dfrac{1}{{\left( {1 - {{\left( {x - 1} \right)}^2}} \right)}}} \right]\]
Simplify the above equation
\[f'\left( x \right) = \dfrac{1}{4}\left[ {\dfrac{1}{{1 - \left( {{x^2} - 2x + 1} \right)}}} \right]\]
\[ \Rightarrow \]\[f'\left( x \right) = \dfrac{1}{4}\left[ {\dfrac{1}{{2x - {x^2}}}} \right]\]
\[ \Rightarrow \]\[f'\left( x \right) = \dfrac{1}{{4x\left( {2 - x} \right)}}\]
Hence the correct option is option A.
Note: Students are often confused with the formula \[\dfrac{d}{{dx}}{\left( {ax + b} \right)^n} = n{\left( {ax + b} \right)^{n - 1}}\] and \[\dfrac{d}{{dx}}{\left( {ax + b} \right)^n} = n{\left( {ax + b} \right)^{n - 1}} \times a\] . But the correct formula is \[\dfrac{d}{{dx}}{\left( {ax + b} \right)^n} = n{\left( {ax + b} \right)^{n - 1}} \times a\]. Because we also have to calculate the derivative of the terms present in the bracket.
Recently Updated Pages
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

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals 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

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

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

NEET 2025 – Every New Update You Need to Know
