If the function f defined by f ( x ) = x 2 l o g ( 1 + x ) 1 + x − x 1 if x  = 0 is continuous at x = 0 , then 6 ( f ( 0 )) =

f ( 0 ) = = x → 0 Lim x 2 l n ( 1 + x ) 1 + x − x = = x → 0 Lim x 2 ( 1 + x ) l n ( 1 + x ) − x = Lim x → 0 2 x 1 + ln ( 1 + x ) − 1 = 2 1 ∴ 6 f ( 0 ) = 3

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Mathematics - Continuity - Differentiability Question with Solution | TestHub

MathematicsContinuity - DifferentiabilityContinuity- MiscellaneousEasy2 minQB

MathematicsEasyinteger

If the function $f$ defined by $f (x) = \frac{l o g ( 1 + x ) ^{1 + x}}{x ^{2}} - \frac{1}{x}$ if $x \neq = 0$ is continuous at $x = 0$ , then $6 (f (0)) =$

Answer:

Solution:

$f (0) = =_{x \to 0}^{Lim} \frac{l n ( 1 + x ) ^{1 + x} - x}{x ^{2}} = =_{x \to 0}^{Lim} \frac{( 1 + x ) l n ( 1 + x ) - x}{x ^{2}}$

$= Lim_{x \to 0} \frac{1 + ln ( 1 + x ) - 1}{2 x} = \frac{1}{2} ∴ 6 f (0) = 3$

Stream:JEESubject:MathematicsTopic:Continuity - DifferentiabilitySubtopic:Continuity- Miscellaneous

⏱ 2mℹ️ Source: QB

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