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) = $\lim\limits_{h\rightarrow0}$ $\frac{f(0+h)-f(0)}{h}$ f '' (0) = $\lim\limits_{h\rightarrow0}$ $\frac{f(h)}{h}$ Now, f (x) = $\lim\limits_{h\rightarrow0}$ $\frac{f(x+h)-f(x)}{h}$ = $\lim\limits_{h\rightarrow0}$ $\frac{f(x) + f(h) - f(x)}{h}$ ⇒ f ' (x) = $\lim\limits_{h\rightarrow0}$ $\frac{f(h)}{h}$ = 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