Given f ( x ) = { x 2 e 2 ( x − 1 ) a cos ( 2 x − 2 ) + b x 2 0 ≤ x ≤ 1 1 < x ≤ 2 is differentiable at x = 1 provided

f ( 1 + 0 ) = f ( 1 − 0 ) ⇒ a + b = 1 f 1 ( 1 − 0 ) = f 1 ( 1 + 0 ) ⇒ 4 = 2 b ⇒ b = 2 , a = − 1

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MathematicsDifferentiationMiscellaneous/MixedMedium2 minQB

MathematicsMediumsingle choice

Given $f (x) = {x^{2} e^{2 (x - 1)} a cos (2 x - 2) + b x^{2} 0 \leq x \leq 1 1 < x \leq 2$ is differentiable at $x = 1$ provided

Answer:

Solution:

$f (1 + 0) = f (1 - 0) \Rightarrow a + b = 1$

$f^{1} (1 - 0) = f^{1} (1 + 0) \Rightarrow 4 = 2 b \Rightarrow b = 2, a = - 1$

Stream:JEESubject:MathematicsTopic:DifferentiationSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: QB

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