JEE PYQ: Differential Equation Question 42
Question 42 - 2019 (09 Jan Shift 2)
Let $f: [0,1] \to \mathbb{R}$ be such that $f(xy) = f(x)\cdot f(y)$, for all $x, y \in [0,1]$, and $f(0) \neq 0$. If $y = y(x)$ satisfies the differential equation $\frac{dy}{dx} = f(x)$ with $y(0) = 1$, then $y\left(\frac{1}{4}\right) + y\left(\frac{3}{4}\right)$ is equal to:
(1) 3
(2) 4
(3) 2
(4) 5
Show Answer
Answer: (1)
Solution
$f(xy) = f(x)f(y)$ with $f(0) \neq 0$. Putting $x = y = 0$: $f(0) = f(0)^2$, so $f(0) = 1$. Putting $y = 0$: $f(0) = f(x)f(0)$, so $f(x) = 1$ for all $x$. Then $\frac{dy}{dx} = 1$, $y = x + 1$. $y\left(\frac{1}{4}\right) + y\left(\frac{3}{4}\right) = \frac{5}{4} + \frac{7}{4} = 3$.
BETA
