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JEE Main 2019 · 9 Jan · Shift II

JEE Main 2019 Mathematics question on Differential Equations

Let f:[0,1]→Rf:[0,1] \to R be such that f(xy)=f(x)⋅f(y)f(xy) = f(x) \cdot f(y), for all x,y∈[0,1]x, y \in [0,1] and f(0)≠0f(0) \neq 0. If y=y(x)y = y(x) satisfies the differential equation, dydx=f(x)\frac{dy}{dx} = f(x) with y(0)=1y(0) = 1, then y(14)+y(34)y\left(\frac{1}{4}\right) + y\left(\frac{3}{4}\right) is equal to
  1. A.
    5
  2. B.
    3
  3. C.
    2
  4. D.
    4
Show answer and solution

Correct option: B

Solution

From f(xy) = f(x)·f(y), set x = y = 0: f(0) = f(0)², so f(0) = 1 (since f(0) ≠ 0). Set y = 0: f(0) = f(x)·f(0), giving f(x) = 1 for all x. Thus dy/dx = 1 with y(0) = 1, so y(x) = x + 1. Therefore y(1/4) + y(3/4) = (5/4) + (7/4) = 3.

[JEE Main 2019 · 9 Jan · Shift II]

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