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# 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

Last updated date: 20th Jun 2024
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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.