Let $f: R \rightarrow R$ be a function given by $f(x + y) = f(x)f(y)\quad {\rm{ for}}\,\,{\rm{ all}}\,\,{\rm{ }}x,y \in R$. If $f(x) = 1 + xg(x) + {x^2}g(x)\phi (x)\,\,\,\,{\rm{such}}\,\,\,{\rm{that}}\,\,\,\,\mathop {\lim }\limits_{x \to 0} g(x) = a\,\,\,{\rm{and}}\,\,\mathop {\lim }\limits_{x \to 0} \phi (x) = b{\rm{,}}\,\,\,{\rm{then}}\,\,\,f'(x)\,\,\,{\rm{is}}\,\,\,{\rm{equal}}\,\,\,{\rm{to}}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

Let

f: R \rightarrow R

be a function given by

f(x + y) = f(x)f(y)\quad \mathrm{ for}\,\,\mathrm{ all}\,\,\mathrm{ }x,y \in R

.
If

f(x) = 1 + xg(x) + {x^2}g(x)\phi (x)\,\,\,\,\mathrm{such}\,\,\,\mathrm{that}\,\,\,\,\lim\limits_{x \to 0} g(x) = a\,\,\,\mathrm{and}\,\,\lim\limits_{x \to 0} \phi (x) = b\mathrm{,}\,\,\,\mathrm{then}\,\,\,f'(x)\,\,\,\mathrm{is}\,\,\,\mathrm{equal}\,\,\,\mathrm{to}

View answer and explanation

Correct option: B

a f (x)

Option B is correct because using the given condition

f(x) = 1 + xg(x) + {x^2}g(x)\phi (x)

and the functional equation, we find

f'(x) = a f (x)