JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Differential Equations

Let

f : [0, 1] \to R

be such that

f (x y) = f (x) f (y)

for all

x, y \in [0, 1]

, and

f (0) \neq = 0

. If

y = y (x)

satisfies the differential equation,

\frac{d y}{d x} = f (x)

with

y (0) = 1

, then

y (\frac{1}{4}) + y (\frac{3}{4})

is equal to

Answer: (B)

Solution

The functional equation

f (x y) = f (x) f (y)

with

f (0) \neq = 0

pins the function down. Putting

x = y = 0

gives

f (0) = f (0)^{2}

, so

f (0) = 1

; putting

y = 0

gives

f (0) = f (x) f (0)

, forcing

f (x) = 1

for every

x

. The differential equation is then

\frac{d y}{d x} = 1

with

y (0) = 1

, whose solution is

y = x + 1

. Hence

y (\frac{1}{4}) + y (\frac{3}{4}) = \frac{5}{4} + \frac{7}{4} = 3

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