Let f : [ 0 , 3 ] → [ 0 , 3 π + lo g e 2 ] defined by f ( x ) = lo g e x 2 + 1 + tan − 1 x then f ( x ) is

f ′ ( x ) = x 2 + 1 x + 1 ⇒ f ( x ) is increasin g in [ 0 , 3 ]

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MathematicsFunctionsTypes of Function (Mapping)Medium2 minQB

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Let $f : [0, 3] \to [0, \frac{π}{3} + lo g_{e} 2]$ defined by $f (x) = lo g_{e} x^{2} + 1 + tan^{- 1} x$ then $f (x)$ is

Answer:

Solution:

$f^{'} (x) = \frac{x + 1}{x ^{2} + 1} \Rightarrow f (x)$ is increasin $g$ in $[0, 3]$

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

⏱ 2mℹ️ Source: QB

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