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Question

Mathematics Question on Differentiability

Let f:RRf : R \rightarrow R be a function such that f(x+y)=f(x)+f(y),x,yRf (x + y) = f (x) + f (y), \forall x, y \in R. If f(x)f (x ) is differentiable at x=0x = 0, then

A

f(x)f (x) is differentiable only in a finite interval containing zero

B

f(x)f(x) is continuous xR\forall x \in R

C

f(x)f ( x ) is constant xR\forall x \in R

D

f(x)f (x) is differentiable except at finitely many points

Answer

f(x)f ( x ) is constant xR\forall x \in R

Explanation

Solution

f (x + y) = f (x) + f (y), as f (x) is differentiable at x = 0.
f(0)=k\Rightarrow f \, ' (0) = k \hspace15mm ...(i)
Now, f;(x)=limh0f(x+h)f(x)hf \, ; (x) = lim_{ h \to 0 } \frac{ f (x + h) - f (x)}{ h}
\hspace15mm = limh0f(x)+f(h)f(x)h lim_{ h \to 0 } \frac{ f (x) + f (h) - f (x)}{ h}
\hspace15mm limh0f(h)hlim_{ h \to 0 } \frac{ f (h)}{ h} \hspace15mm [00form]\bigg [ \frac{0}{0} \, form \bigg]
Given, f (x + y) = f (x) + f (y), \forall x, y
\therefore f (0) = f (0) + f (0),
when x = y = 0 \Rightarrow f (0) = 0
Using L'Hospital's rule,
\hspace15mm = limh0f;(h)1lim_{ h \to 0 } \frac{ f \, ; (h)}{ 1} f ' (0) = k
\Rightarrow f ' (x) = k, integrating both sides,
f(x) = k x + C, as f (0) = 0
C=0\Rightarrow C = 0
\therefore f (x) = k x
\therefore f (x) is continuous for all x s R and f ' (x) = k, i.e.
constant for all x \in R
Hence, (b) and (c) are correct.