JEE MAIN - Mathematics (2020 - 2nd September Evening Slot - No. 7)
Let f : (–1,
$$\infty $$)
$$ \to $$ R be defined by f(0) = 1 and
f(x) = $${1 \over x}{\log _e}\left( {1 + x} \right)$$, x $$ \ne $$ 0. Then the function f :
f(x) = $${1 \over x}{\log _e}\left( {1 + x} \right)$$, x $$ \ne $$ 0. Then the function f :
decreases in (–1, $$\infty $$)
decreases in (–1, 0) and increases in (0, $$\infty $$)
increases in (–1, $$\infty $$)
increases in (–1, 0) and decreases in (0, $$\infty $$)
Explanation
$$f(x) = {1 \over x}{\log _e}\left( {1 + x} \right)$$
$$ \Rightarrow f'(x) = {{x{1 \over {1 + x}} - 1{{\log }_e}\left( {1 + x} \right)} \over {{x^2}}}$$
$$ \Rightarrow f'(x) = {{x - \left( {1 + x} \right){{\log }_e}\left( {1 + x} \right)} \over {{x^2}\left( {1 + x} \right)}}$$
Let $$g(x) = x - \left( {1 + x} \right){\log _e}\left( {1 + x} \right)$$
$$ \Rightarrow g'(x) = 1 - \left( {1 + x} \right){1 \over {1 + x}} - \left( 1 \right) \times {\log _e}\left( {1 + x} \right)$$
$$ = 1 - 1 - {\log _e}\left( {1 + x} \right)$$
$$ = - {\log _e}\left( {1 + x} \right)$$
For $$x \in \left( { - 1,0} \right),g'(x) > 0$$
and for $$x \in \left( {0,\infty } \right),g'(x) < 0$$
Also, $$g(0) = 0 - \left( {1 + 0} \right){\log _e}\left( {1 + 0} \right) = 0$$
$$ \therefore g'(x) < 0\,\,\forall\,\, x \in \left( { - 1,\infty } \right)$$
$$ \Rightarrow f'(x) < 0\,\,\forall\,\, x \in \left( { - 1,\infty } \right)$$
$$ \Rightarrow $$ f(x) is a decreasing function for all $$x \in \left( { - 1,\infty } \right)$$
$$ \Rightarrow f'(x) = {{x{1 \over {1 + x}} - 1{{\log }_e}\left( {1 + x} \right)} \over {{x^2}}}$$
$$ \Rightarrow f'(x) = {{x - \left( {1 + x} \right){{\log }_e}\left( {1 + x} \right)} \over {{x^2}\left( {1 + x} \right)}}$$
Let $$g(x) = x - \left( {1 + x} \right){\log _e}\left( {1 + x} \right)$$
$$ \Rightarrow g'(x) = 1 - \left( {1 + x} \right){1 \over {1 + x}} - \left( 1 \right) \times {\log _e}\left( {1 + x} \right)$$
$$ = 1 - 1 - {\log _e}\left( {1 + x} \right)$$
$$ = - {\log _e}\left( {1 + x} \right)$$
For $$x \in \left( { - 1,0} \right),g'(x) > 0$$
and for $$x \in \left( {0,\infty } \right),g'(x) < 0$$
Also, $$g(0) = 0 - \left( {1 + 0} \right){\log _e}\left( {1 + 0} \right) = 0$$
$$ \therefore g'(x) < 0\,\,\forall\,\, x \in \left( { - 1,\infty } \right)$$
$$ \Rightarrow f'(x) < 0\,\,\forall\,\, x \in \left( { - 1,\infty } \right)$$
$$ \Rightarrow $$ f(x) is a decreasing function for all $$x \in \left( { - 1,\infty } \right)$$
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