Verified Solution • 2025–26 Session moderate

QJEEM20Ma3993: JEE Main 2020 Mathematics - Class 12 JEEM20Ma3993 Solution: Let f : (–1, ) R be defined by f(0) = 1 and f(x) = , x 0. Then…

Let f : (–1, ∞\infty ) →\to R be defined by f(0) = 1 and
f(x) = 1xlog⁡e(1+x){1 \over x}{\log _e}\left( {1 + x} \right) , x ≠\ne 0. Then the function f :

Multiple Choice Options

  • A decreases in (–1, ∞\infty )
  • B decreases in (–1, 0) and increases in (0, ∞\infty )
  • C increases in (–1, ∞\infty )
  • D increases in (–1, 0) and decreases in (0, ∞\infty )

Step-by-Step Verified Solution

f(x)=1xlog⁡e(1+x)f(x) = {1 \over x}{\log _e}\left( {1 + x} \right)

⇒f′(x)=x11+x−1log⁡e(1+x)x2\Rightarrow f'(x) = {{x{1 \over {1 + x}} - 1{{\log }_e}\left( {1 + x} \right)} \over {{x^2}}}

⇒f′(x)=x−(1+x)log⁡e(1+x)x2(1+x)\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−(1+x)log⁡e(1+x)g(x) = x - \left( {1 + x} \right){\log _e}\left( {1 + x} \right)

⇒g′(x)=1−(1+x)11+x−(1)×log⁡e(1+x)\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(1+x)= 1 - 1 - {\log _e}\left( {1 + x} \right)

=−log⁡e(1+x)= - {\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−(1+0)log⁡e(1+0)=0g(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∈(−1,∞)x \in \left( { - 1,\infty } \right)

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