Mathematics - Continuity - Differentiability Question with Solution | TestHub

(i) Since $f(x)$ is continuous at $x=0$, ${\lim }_{x\to a}f(x)=f(0)$ ---(1)

Let $a\in R$, then ${\lim }_{x\to a}f(x)={\lim }_{h\to 0}f(a+h)$

$$\begin{array}{rl}& =\underset{h\to 0}{\lim }f(a)+f(h)\\ & =f(a)+\underset{x\to 0}{\lim }f(h)\\ & =f(a)+f(0)=f(a+0)\\ & =f(a)\end{array}$$

$\Rightarrow f$ is continuous $\forall x\in R$, as 'a' is arbitrary.

(ii) Because $f(x+y)=f(x)+f(y)\Rightarrow f(0)=0$. $f(1)$ - --(1)

For any positive integer 'n', $f(n)=f(1+1+\dots +1)=n\cdot f(1)$ ---(2)

For any negative integer $m$, we have $0=f(0)=f[m+(-m)]=f(m)+f(-m)$

$\Rightarrow f(m)=-f(-m)=-(-m)\cdot f(1)$

$$=m\cdot f(1)---(3)$$

(iii) Let $p/q$ be any rational number where 'q' is a positive integer and p is any positive, negative, or zero integer. Then $f\left(q\cdot \frac{p}{q}\right)=f\left(\frac{p}{q}+\frac{p}{q}+\dots (qtimes)\right)$

$$\begin{array}{rl}& =f\left(\frac{p}{q}\right)+f\left(\frac{p}{q}\right)+\dots +f\left(\frac{p}{q}\right)\\ & =q\cdot f\left(\frac{p}{q}\right)\\ & \Rightarrow f(p)=q\cdot f(p/q)----\left(4\right)\\ & \text{ But }f(p)=p\cdot f(1)\text{ from (2) and (3)}\\ & \therefore p\cdot f(1)=q\cdot f(p/q)\\ & \Rightarrow f(p/q)=\frac{p}{q}\cdot f(1)------\left(5\right)\end{array}$$