JEE Main 2025 23 January Morning Shift
Exam JEE Main
Year 2025
Date 23 January
Shift Morning Shift
Subject Mathematics
Chapter Functions JEE Main 2025 · 23 January · Morning Shift · Mathematics · Functions
Let f(x)= _ e x and g(x)=(x^4-2 x^3+3 x^2-2 x+2)/(2 x^2-2 x+1) ....
Let
f ( x ) = log e x f(x)=\log _{\mathrm{e}} x f ( x ) = log e x and
g ( x ) = x 4 − 2 x 3 + 3 x 2 − 2 x + 2 2 x 2 − 2 x + 1 g(x)=\frac{x^4-2 x^3+3 x^2-2 x+2}{2 x^2-2 x+1} g ( x ) = 2 x 2 − 2 x + 1 x 4 − 2 x 3 + 3 x 2 − 2 x + 2 . Then the domain of
f ∘ g f \circ g f ∘ g is
Options A [ 0 , ∞ ) [0, \infty) [ 0 , ∞ ) B [ 1 , ∞ ) [1, \infty) [ 1 , ∞ ) C ( 0 , ∞ ) (0, \infty) ( 0 , ∞ ) D R \mathbb{R} R Correct answer D. R \mathbb{R} R
Step-by-step solution
f(g(x))=\ln
( x 4 − 2 x 3 + 3 x 2 − 2 x + 2 2 x 2 − 2 x + 1 ) \left(\frac{x^4-2 x^3+3 x^2-2 x+2}{2 x^2-2 x+1}\right) ( 2 x 2 − 2 x + 1 x 4 − 2 x 3 + 3 x 2 − 2 x + 2 )
Since
2 x 2 − 2 x + 1 > 0 ∀ x ∈ R ∵ ( − 2 ) 2 − 4 ( 2 ) < 0 2 x^2-2 x+1>0 \quad \forall x \in \mathbb{R} \because(-2)^2-4(2) < 0 2 x 2 − 2 x + 1 > 0 ∀ x ∈ R ∵ ( − 2 ) 2 − 4 ( 2 ) < 0
Consider
x 4 − 2 x 3 + 3 x 2 − 2 x + 2 = ( x 4 − 2 x 3 + x 2 ) + ( x 2 − 2 x + 1 ) + ( 1 + x 2 ) = x 2 ( x − 1 ) 2 + ( x − 1 ) 2 + ( x 2 + 1 ) > 0 ∀ x ∈ R ⇒ g ( x ) > 0 ∀ x ∈ R ⇒ ln f ( ( x ) ) , f ( x ) > 0 ∀ x ∈ R ⇒ x ∈ R is domain \begin{aligned}
& x^4-2 x^3+3 x^2-2 x+2 \\
& =\left(x^4-2 x^3+x^2\right)+\left(x^2-2 x+1\right)+\left(1+x^2\right) \\
& =x^2(x-1)^2+(x-1)^2+\left(x^2+1\right)>0 \forall x \in \mathbb{R} \\
& \Rightarrow g(x)>0 \forall x \in \mathbb{R} \\
& \Rightarrow \ln f((x)), f(x)>0 \forall x \in \mathbb{R} \\
& \Rightarrow x \in \mathbb{R} \text{ is domain }
\end{aligned} x 4 − 2 x 3 + 3 x 2 − 2 x + 2 = ( x 4 − 2 x 3 + x 2 ) + ( x 2 − 2 x + 1 ) + ( 1 + x 2 ) = x 2 ( x − 1 ) 2 + ( x − 1 ) 2 + ( x 2 + 1 ) > 0∀ x ∈ R ⇒ g ( x ) > 0∀ x ∈ R ⇒ ln f (( x )) , f ( x ) > 0∀ x ∈ R ⇒ x ∈ R is domain
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