JEE Advance - Mathematics (2020 - Paper 2 Offline - No. 9)
Let f : R $$ \to $$ R and g : R $$ \to $$ R be functions
satisfying f(x + y) = f(x) + f(y) + f(x)f(y)
and f(x) = xg(x) for all x, y$$ \in $$R.
If $$\mathop {\lim }\limits_{x \to 0} g(x) = 1$$, then which of the following statements is/are TRUE?
satisfying f(x + y) = f(x) + f(y) + f(x)f(y)
and f(x) = xg(x) for all x, y$$ \in $$R.
If $$\mathop {\lim }\limits_{x \to 0} g(x) = 1$$, then which of the following statements is/are TRUE?
f is differentiable at every x$$ \in $$R
If g(0) = 1, then g is differentiable at every x$$ \in $$R
The derivative f'(1) is equal to 1
The derivative f'(0) is equal to 1
Explanation
The given function f : R $$ \to $$ R is satisfying as
$$f(x + y) = f(x) + f(y) + f(x)f(y)$$
So, $$f'(x) = \mathop {\lim }\limits_{h \to 0} {{f(x + h) - f(x)} \over h}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{f(x) + f(h) + f(x)f(h) - f(x)} \over h}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{f(h)} \over h}(1 + f(x))$$
$$ \because $$ $$f(x) = xg(x) \Rightarrow g(x) = {{f(x)} \over x}$$
$$ \therefore $$ $$\mathop {\lim }\limits_{x \to 0} {{f(x)} \over x} = \mathop {\lim }\limits_{x \to 0} g(x) = 1$$ (given)
$$ \therefore $$ $$f'(x) = 1 + f(x)) \Rightarrow {{f'(x)} \over {1 + f(x)}} = 1$$
$$ \Rightarrow {\log _e}(1 + f(x)) = x + c$$
$$ \Rightarrow 1 + f(x) = {e^{x + c}}$$
$$ \Rightarrow f(x) = {e^{x + c}} - 1$$
$$ \because $$ $$f(0) = 0 \Rightarrow c = 0$$
Therefore, $$f(x) = {e^x} - 1$$ is differentiable at every $$x \in R$$.
And $$f'(x) = {e^x} \Rightarrow f'(0) = 1$$
Now, $$g(x) = f{{(x)} \over x} = {{{e^x} - 1} \over x}$$ and if g(0) = 1
LHD (at x = 0) of
$$g(x) = \mathop {\lim }\limits_{h \to 0} {{g(0 - h) - g(0)} \over { - h}}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{{{{e^{ - h}} - 1} \over { - h}} - 1} \over { - h}}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{{e^{ - h}} - 1 + h} \over {{h^2}}} = {1 \over 2}$$
and, RHD (at x = 0) of
$$g(x) = \mathop {\lim }\limits_{h \to 0} {{g(0 + h) - g(0)} \over h}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{{e^h} - 1 - h} \over {{h^2}}} = {1 \over 2}$$
So, if g(0) = 1, then g is differentiable at every x$$ \in $$R.
$$f(x + y) = f(x) + f(y) + f(x)f(y)$$
So, $$f'(x) = \mathop {\lim }\limits_{h \to 0} {{f(x + h) - f(x)} \over h}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{f(x) + f(h) + f(x)f(h) - f(x)} \over h}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{f(h)} \over h}(1 + f(x))$$
$$ \because $$ $$f(x) = xg(x) \Rightarrow g(x) = {{f(x)} \over x}$$
$$ \therefore $$ $$\mathop {\lim }\limits_{x \to 0} {{f(x)} \over x} = \mathop {\lim }\limits_{x \to 0} g(x) = 1$$ (given)
$$ \therefore $$ $$f'(x) = 1 + f(x)) \Rightarrow {{f'(x)} \over {1 + f(x)}} = 1$$
$$ \Rightarrow {\log _e}(1 + f(x)) = x + c$$
$$ \Rightarrow 1 + f(x) = {e^{x + c}}$$
$$ \Rightarrow f(x) = {e^{x + c}} - 1$$
$$ \because $$ $$f(0) = 0 \Rightarrow c = 0$$
Therefore, $$f(x) = {e^x} - 1$$ is differentiable at every $$x \in R$$.
And $$f'(x) = {e^x} \Rightarrow f'(0) = 1$$
Now, $$g(x) = f{{(x)} \over x} = {{{e^x} - 1} \over x}$$ and if g(0) = 1
LHD (at x = 0) of
$$g(x) = \mathop {\lim }\limits_{h \to 0} {{g(0 - h) - g(0)} \over { - h}}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{{{{e^{ - h}} - 1} \over { - h}} - 1} \over { - h}}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{{e^{ - h}} - 1 + h} \over {{h^2}}} = {1 \over 2}$$
and, RHD (at x = 0) of
$$g(x) = \mathop {\lim }\limits_{h \to 0} {{g(0 + h) - g(0)} \over h}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{{e^h} - 1 - h} \over {{h^2}}} = {1 \over 2}$$
So, if g(0) = 1, then g is differentiable at every x$$ \in $$R.
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