If f ( x ) = 2 x + ∣ x ∣ , g ( x ) = 3 1 ( 2 x − ∣ x ∣ ) and h ( x ) = f ( g ( x )) then sin − 1 ( h ( h ( h ( h … . h . h ( x )))) is n times

Since f [ g ( x )] = x , ∀ x ∈ R ⇒ h ( x ) = x ⇒ sin − 1 [ h ( h ( h … h ( x )))] = sin − 1 x

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MathematicsFunctionsCompositeEasy2 minQB

MathematicsEasysingle choice

If $f (x) = 2 x + ∣ x ∣, g (x) = \frac{1}{3} (2 x - ∣ x ∣)$ and $h (x) = f (g (x))$ then $sin^{- 1} (h (h (h (h \dots . h . h (x))))$ is $n$ times

Answer:

Solution:

Since $f [g (x)] = x, \forall x \in R \Rightarrow h (x) = x$

$\Rightarrow sin^{- 1} [h (h (h \dots h (x)))] = sin^{- 1} x$

Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:Composite

⏱ 2mℹ️ Source: QB

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