Let a ∈ R − { 0 } . then

f ′ ( x ) − x 2 + 1 2 x f ( x ) = − x 2 ( x 2 + 1 ) ⇒ I . F = e ∫ x 2 + 1 − 2 x d x = x 2 + 1 1 ∴ solution is f ( x ) × x 2 + 1 1 = ∫ x 2 − ( x 2 + 1 ) × x 2 + 1 1 d x = x 1 + c As f ( 1 ) = 2 ⇒ c = 0 ⇒ f ( x ) = x + x 1 g ′ ( x ) = x 3 ( 1 + x 4 ) 2 ( x + 2 ) ( x − 2 ) ( x + 1 ) ( x − 1 ) ( x 2 + 1 )

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Mathematics - Functions Question with Solution | TestHub

MathematicsFunctionsFunctional EquationMedium2 minQB

MathematicsMediumsingle choice

Passage / Comprehension

Let $f (x)$ be a differentiable function for all $x \in R ∖ {0}$ that satisfies the equation $(1, 2)$ is on the graph of $y = f (x)$ .

Let $a \in R - {0}$ . then

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Answer:

Solution:

$f^{'} (x) - \frac{2 x}{x ^{2} + 1} f (x) = - \frac{( x ^{2} + 1 )}{x ^{2}} \Rightarrow I . F = e^{\int \frac{- 2 x}{x ^{2} + 1} d x} = \frac{1}{x ^{2} + 1}$ ∴ solution is $f (x) \times \frac{1}{x ^{2} + 1} = \int \frac{- ( x ^{2} + 1 )}{x ^{2}} \times \frac{1}{x ^{2} + 1} d x = \frac{1}{x} + c$ As $f (1) = 2 \Rightarrow c = 0 \Rightarrow f (x) = x + \frac{1}{x}$ $g^{'} (x) = \frac{2 ( x + 2 ) ( x - 2 ) ( x + 1 ) ( x - 1 ) ( x ^{2} + 1 )}{x ^{3} ( 1 + x ^{4} )}$

Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:Functional Equation

⏱ 2mℹ️ Source: QB

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