Let f : R → R be a function such that f x = x 2 + 2 x + 1 x 2 + 1 . Then

Given, f x = x 2 + 2 x + 1 x 2 + 1 f ' ( x ) = x 2 + 1 ( 2 x + 2 ) - ( x 2 + 2 x + 1 ) 2 x x 2 + 1 2 ⇒ f ' ( x ) = 2 x 3 + 2 x 2 + 2 x + 2 - 2 x 3 - 4 x 2 - 2 x x 2 + 1 2 ⇒ f ' ( x ) = 2 ( 1 - x 2 ) x 2 + 1 2 = - 2 ( x + 1 ) ( x - 1 ) x 2 + 1 2 Clearly f ( x ) is one-one in - ∞ , - 1 and also in ( 1 , ∞ ) but f ( x ) is not one-one in - ∞ , ∞

Mathematics - Functions Question with Solution | TestHub

MathematicsFunctionsTypes of Function (Mapping)Medium2 minPYQ_2023

MathematicsMediumsingle choice

Let $f : R \to R$ be a function such that $f (x) = \frac{x^{2} + 2 x + 1}{x^{2} + 1}$ . Then

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Answer:

Solution:

Given,

$f (x) = \frac{x^{2} + 2 x + 1}{x^{2} + 1}$

$f^{'} (x) = \frac{(x^{2} + 1) (2 x + 2) - (x^{2} + 2 x + 1) 2 x}{{(x^{2} + 1)}^{2}}$

$\Rightarrow f^{'} (x) = \frac{2 x^{3} + 2 x^{2} + 2 x + 2 - 2 x^{3} - 4 x^{2} - 2 x}{{(x^{2} + 1)}^{2}}$

$\Rightarrow f^{'} (x) = \frac{2 (1 - x^{2})}{{(x^{2} + 1)}^{2}} = - \frac{2 (x + 1) (x - 1)}{{(x^{2} + 1)}^{2}}$

Clearly $f (x)$ is one-one in $(- \infty, - 1)$ and also in $(1, \infty)$ but $f (x)$ is not one-one in $(- \infty, \infty)$

Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:Types of Function (Mapping)

⏱ 2mℹ️ Source: PYQ_2023

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