
Let a function \[f:[0,1]\to R\] such that \[f(xy)=f(x).f(y)\] for all \[x,y\in [0,1]\] , and \[f(0)\ne 0\]. If \[y=y(x)\] satisfies the deferential equation ,\[\dfrac{dy}{dx}=f(x)\] with \[y(0)=1\], then \[y\left( \dfrac{1}{4} \right)+y\left( \dfrac{3}{4} \right)\] is equal to
(A) 4
(B) 3
(C) 5
(D) 2
Answer
154.5k+ views
Hint: We start solving this problem by substituting the suitable values in the given functional equation \[f(xy)=f(x).f(y)\]. Then we make use of the given result \[f(0)\ne 0\]. From the given differential equation \[\dfrac{dy}{dx}=f(x)\], we can find y which includes an integral constant. By using the result \[y(0)=1\] we will get the constant value so we get the function y and by substituting the values we find \[y\left( \dfrac{1}{4} \right)+y\left( \dfrac{3}{4} \right)\] value.
Complete step-by-step solution:
Let us consider the given functional equation, \[f(xy)=f(x).f(y)\] for all\[x,y\in [0,1]\]
Now, let us put \[x=0\], \[y=0\] and substitute in the given functional equation \[f(xy)=f(x).f(y)\].
\[\Rightarrow f(0\times 0)=f(0).f(0)\]
\[\Rightarrow f(0)-f(0).f(0)=0\]
\[\Rightarrow f(0)\left( 1-f(0) \right)=0\]
As we were given \[f(0)\ne 0\]
\[\Rightarrow \]\[f(0)=1\]
Let us take the differential equation given in the question \[\dfrac{dy}{dx}=f(x)\] and integrate it on both sides at x=0, we get,
\[\int{\dfrac{dy}{dx}}=\int{f(0)}\]
\[\Rightarrow \]\[\int{dy}=\int{dx}\]
\[\Rightarrow \]\[y=x+c\]
We were given \[y(0)=1\]
So, we get,
\[\Rightarrow \]\[1=0+c\]
\[\Rightarrow \]\[c=1\]
\[\Rightarrow \]\[y=x+1\]
So, our required equation is \[y=x+1\].
Now substituting the values \[x=\dfrac{1}{4}\], \[x=\dfrac{3}{4}\] and adding the values we will get \[y\left( \dfrac{1}{4} \right)+y\left( \dfrac{3}{4} \right)\]
\[y\left( \dfrac{1}{4} \right)+y\left( \dfrac{3}{4} \right)=\dfrac{1}{4}+1+\dfrac{3}{4}+1\]
\[\Rightarrow \]\[y\left( \dfrac{1}{4} \right)+y\left( \dfrac{3}{4} \right)=\dfrac{1+3}{4}+1++1\]
\[\Rightarrow \]\[y\left( \dfrac{1}{4} \right)+y\left( \dfrac{3}{4} \right)=1+1+1\]
\[\Rightarrow \]\[y\left( \dfrac{1}{4} \right)+y\left( \dfrac{3}{4} \right)=3\]
Note: The possibilities for making mistakes in this type of problem are one may make mistakes while substituting the values of x and y in the given functional, one must check the domain and then have to select the values only from the domain. One must not forget to write the constant term while integrating. While integrating the given differential equation you have to substitute \[x=0\] in the functional equation and then we have to integrate, also after integrating you may forget to put the integral constant which may lead to wrong answers.
Complete step-by-step solution:
Let us consider the given functional equation, \[f(xy)=f(x).f(y)\] for all\[x,y\in [0,1]\]
Now, let us put \[x=0\], \[y=0\] and substitute in the given functional equation \[f(xy)=f(x).f(y)\].
\[\Rightarrow f(0\times 0)=f(0).f(0)\]
\[\Rightarrow f(0)-f(0).f(0)=0\]
\[\Rightarrow f(0)\left( 1-f(0) \right)=0\]
As we were given \[f(0)\ne 0\]
\[\Rightarrow \]\[f(0)=1\]
Let us take the differential equation given in the question \[\dfrac{dy}{dx}=f(x)\] and integrate it on both sides at x=0, we get,
\[\int{\dfrac{dy}{dx}}=\int{f(0)}\]
\[\Rightarrow \]\[\int{dy}=\int{dx}\]
\[\Rightarrow \]\[y=x+c\]
We were given \[y(0)=1\]
So, we get,
\[\Rightarrow \]\[1=0+c\]
\[\Rightarrow \]\[c=1\]
\[\Rightarrow \]\[y=x+1\]
So, our required equation is \[y=x+1\].
Now substituting the values \[x=\dfrac{1}{4}\], \[x=\dfrac{3}{4}\] and adding the values we will get \[y\left( \dfrac{1}{4} \right)+y\left( \dfrac{3}{4} \right)\]
\[y\left( \dfrac{1}{4} \right)+y\left( \dfrac{3}{4} \right)=\dfrac{1}{4}+1+\dfrac{3}{4}+1\]
\[\Rightarrow \]\[y\left( \dfrac{1}{4} \right)+y\left( \dfrac{3}{4} \right)=\dfrac{1+3}{4}+1++1\]
\[\Rightarrow \]\[y\left( \dfrac{1}{4} \right)+y\left( \dfrac{3}{4} \right)=1+1+1\]
\[\Rightarrow \]\[y\left( \dfrac{1}{4} \right)+y\left( \dfrac{3}{4} \right)=3\]
Note: The possibilities for making mistakes in this type of problem are one may make mistakes while substituting the values of x and y in the given functional, one must check the domain and then have to select the values only from the domain. One must not forget to write the constant term while integrating. While integrating the given differential equation you have to substitute \[x=0\] in the functional equation and then we have to integrate, also after integrating you may forget to put the integral constant which may lead to wrong answers.
Recently Updated Pages
JEE Main 2022 (June 29th Shift 2) Maths Question Paper with Answer Key

JEE Main 2023 (January 25th Shift 1) Maths Question Paper with Answer Key

JEE Main 2022 (July 29th Shift 1) Maths Question Paper with Answer Key

JEE Main 2022 (July 26th Shift 2) Chemistry Question Paper with Answer Key

JEE Main 2022 (June 26th Shift 2) Maths Question Paper with Answer Key

JEE Main 2022 (June 29th Shift 1) Physics Question Paper with Answer Key

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

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Displacement-Time Graph and Velocity-Time Graph for JEE

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

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

Electrical Field of Charged Spherical Shell - JEE

Ideal and Non-Ideal Solutions Raoult's Law - JEE

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
