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QJEEM22Mabc50: JEE Main 2022 Mathematics - Class 11 JEEM22Mabc50 Solution: Let and be two functions defined by and . Then, for which of…

Let f:R→Rf:R \to R and g:R→Rg:R \to R be two functions defined by f(x)=log⁡e(x2+1)−e−x+1f(x) = {\log _e}({x^2} + 1) - {e^{ - x}} + 1 and g(x)=1−2e2xexg(x) = {{1 - 2{e^{2x}}} \over {{e^x}}} . Then, for which of the following range of α\alpha , the inequality f\left( {g\left( {{{{{(\alpha - 1)}^2}} \over 3}} \right)} \right) > f\left( {g\left( {\alpha -{5 \over 3}} \right)} \right) holds ?

Multiple Choice Options

  • A (2, 3)
  • B ( −- 2, −- 1)
  • C (1, 2)
  • D ( −- 1, 1)

Step-by-Step Verified Solution

f(x)=log⁡e(x2+1)−e−x+1f(x) = {\log _e}({x^2} + 1) - {e^{ - x}} + 1

f′(x)=2xx2+1+e−xf'(x) = {{2x} \over {{x^2} + 1}} + {e^{ - x}}

= {2 \over {x + {1 \over x}}} + {e^{ - x}} > 0\,\,\forall x \in R

g(x)=e−x−2exg(x) = {e^{ - x}} - 2{e^x}

g'(x) - - {e^{ - x}} - 2{e^x} < 0\,\,\,\,\forall x \in R

⇒\Rightarrow f(x) is increasing and g(x) is decreasing function.

f\left( {g\left( {{{{{(\alpha - 1)}^2}} \over 3}} \right)} \right) > f\left( {g\left( {\alpha - {5 \over 3}} \right)} \right)

\Rightarrow {{{{(\alpha - 1)}^2}} \over 3} < \alpha - {5 \over 3}

= {\alpha ^2} - 5\alpha + 6 < 0

= (\alpha - 2)(\alpha - 3) < 0

=α∈(2, 3)= \alpha \in (2,\,3)

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