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Differential Equations question

2019 · 9 Jan · Shift 2 · Q41

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JEE Main
Mathematics
Differential Equations
2019 · 9 Jan · Shift 2 · Q41

Differential Equations question

2019 · 9 Jan · Shift 2 · Q41

JEE MainMathematicsDifferential EquationsMCQ+4 / −1

Let f : [0,1]

\to

R be such that f(xy) = f(x).f(y), for all x, y

\in

[0, 1], and f(0)

e

0. If y = y(x) satiesfies the differential equation,

{{dy} \over {dx}}

= f(x) with y(0) = 1, then y

\left( {{1 \over 4}} \right)

+ y

\left( {{3 \over 4}} \right)

is equal to :

A
3
B
4
C
2
D
5

View written solutionFree

Correct answer: A

Use the functional equation

Given

\quad \text{for all } x,y\in[0,1],$$ and also $f(0)\neq 0$. We will first determine $f$. 2. **Find $f(0)$** Put $x=0$ in the relation: $$f(0\cdot y)=f(0)f(y).$$ Since $0\cdot y=0$, this gives $$f(0)=f(0)f(y).$$ Because $f(0)\neq 0$, we can divide by $f(0)$ to get $$f(y)=1 \quad \text{for all } y\in[0,1].$$ Hence, $$f(x)\equiv 1.$$ 3. **Solve the differential equation** We are given $$\frac{dy}{dx}=f(x)=1,$$ with initial condition $$y(0)=1.$$ Integrating, $$y=x+C.$$ Using $y(0)=1$: $$1=0+C \implies C=1.$$ So, $$y(x)=x+1.$$ 4. **Compute the required value** $$y\left(\frac14\right)=1+\frac14=\frac54,$$ $$y\left(\frac34\right)=1+\frac34=\frac74.$$ Therefore, $$y\left(\frac14\right)+y\left(\frac34\right)=\frac54+\frac74=\frac{12}{4}=3.$$ 5. **Check options** The correct option is: - **A: 3** 6. **Compare with stored answer** Stored correct answer: **A** This matches our derived answer.

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