BITSAT 2015 Paper

Let f(x + y) = f(x) + f(y), for x, y ∊ R
Put x = 0 = y
⇒ f(0) = f(0) + f(0)
⇒ f(0) = 0
Now. f ' (0) =

lim

h→0

f(0+h)−f(0)

h

f '' (0) =

lim

h→0

f(h)

h

Now, f (x) =

lim

h→0

f(x+h)−f(x)

h

=

lim

h→0

f(x)+f(h)−f(x)

h

⇒ f ' (x) =

lim

h→0

f(h)

h

= f ' (0)
⇒ f (x) = x f ' (0) + C
But f (0) = 0
∴ C = 0
Hence, f(x) = x f ' (0), for all x ∊ R
Clearly, f(x) is everywhere continuous and differentiable and f ' (x) is constant
for all x ∊ R