JEE Main 2025 Jan 23 Shift 1, Mathematics Q6: Invertible And Composite Functions

JEE Main2025Jan 23, Shift 1Mathematics

Let $f (x) = lo g_{e} x$ and $g (x) = \frac{x ^{4} - 2 x ^{3} + 3 x ^{2} - 2 x + 2}{2 x ^{2} - 2 x + 1}$ .

Then the domain of $f \circ g$ is

A
$R$
B
$(0, \infty)$
C
$[0, \infty)$
D
$[1, \infty)$

Solution

The composition $f (g (x)) = ln (g (x))$ is defined whenever $g (x) > 0$ .

Denominator: $2 x^{2} - 2 x + 1 = 2 (x - \frac{1}{2})^{2} + \frac{1}{2} > 0$ for all real $x$ .

Numerator: rewrite as a sum of squares. Notice $x^{4} - 2 x^{3} + 3 x^{2} - 2 x + 2 = (x^{2} - x)^{2} + 2 (x^{2} - x) + 2 = (x^{2} - x + 1)^{2} + 1$ after grouping. Since this is at least 1, the numerator is strictly positive for every real $x$ .

Hence $g (x) > 0$ for all $x \in R$ , so the domain of $f \circ g$ is $R$ .

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JEE Main 2025 Jan 23 Shift 1, Mathematics Q6: Invertible And Composite Functions

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