Let f: R arrow R be a function such that f(x+y)=f(x)+f(y), x, y ∈ ℝ If f(x) is differentiable at x=0 , then which one of the following is incorrect?

Correct answer: D. f(x) is differentiable only in a finite interval containing zera. Let f(x+y)=f(x)+f(y), x, y ∈ R Put x=0=y f(0)=f(0)+f(0) f(0)=0 Now, f^ (0)= _ h arrow 0 (f(0+h)-f(0))/(h)f^ (0)= _ h arrow 0 (f(h))/(h) Now, f(x)= _ h arrow 0 (f(x+h)-f(x))/(h)= _ h arrow 0 (f(x)+f(h)-f(x))/(h) array l ^ (x)= _ h arrow 0 f(h) h = ^ (0) f(x)=x f(0)+C array But f(0)=0 C=0 Hence, f(x)=x f(0), x ∈ R Clearly, f(x) is everywhere continuous and differentiable and ^ (x) is constant. x ∈ R

BITSAT 2015

Exam: BITSAT
Year: 2015
Subject: Mathematics
Chapter: Continuity and Differentiability

BITSAT 2015 · Mathematics · Continuity and Differentiability

Let f: R arrow R be a function such that f(x+y)=f(x)+f(y), x, y ∈ ℝ...

Let

f: R \rightarrow R

be a function such that

f(x+y)=f(x)+f(y), \forall x, y \in \mathbb{R}

\mathrm{f}(\mathrm{x})

is differentiable at

\mathrm{x}=0

, then which one of the following is incorrect?

Options

A $\mathrm{f}(\mathrm{x})$ is continuous, $\forall \mathrm{x} \in \mathrm{R}$
B $\mathrmf^{\prime}(\mathrm{x})$ is constant, $\forall \mathrm{x} \in \mathrm{R}$
C $\mathrm{f}(\mathrm{x})$ is differentiable, $\forall \mathrm{x} \in \mathrm{R}$
D $\mathrm{f}(\mathrm{x})$ is differentiable only in a finite interval containing zera.

Correct answer

D. $\mathrm{f}(\mathrm{x})$ is differentiable only in a finite interval containing zera.

Step-by-step solution

Let

f(x+y)=f(x)+f(y), \forall x, y \in R

Put

x=0=y\Rightarrow f(0)=f(0)+f(0)\Rightarrow f(0)=0

Now,

f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}f^{\prime \prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)}{h}

Now,

f(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{f(x)+f(h)-f(x)}{h}

\begin{array}{l} \Rightarrow \mathrmf^{\prime}(\mathrm{x})=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(\mathrm{h})}{\mathrm{h}}=\mathrmf^{\prime}(0) \\ \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{x} \mathrm{f}(0)+\mathrm{C} \end{array}

But

f(0)=0

\therefore

\mathrm{C}=0 Hence,

f(x)=x f(0), \forall x \in R

Clearly, \mathrm{f}(\mathrm{x})

is everywhere continuous and differentiable and

\mathrmf^{\prime}(\mathrm{x})

is constant.

\forall \mathrm{x} \in \mathrm{R}$

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