Let f ( x ) be a non negative differentiable function such that f 1 ( x ) ≤ f ( x ) ∀ x ≥ 0 and f ( 0 ) = 0 thena) f ( 1 ) + f ( 2 ) = 0

f 1 ( x ) ≤ f ( x ) , let f 1 ( x ) − f ( x ) = K ⇒ f ( x ) = − K + Ke x = K ( e z − 1 ) ≥ 0∀ x ≥ 0 ⇒ K = 0 ∴ f is a constant function but f ( 0 ) = 0 ∴ f ≡ 0

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Mathematics - Continuity - Differentiability Question with Solution | TestHub

MathematicsContinuity - DifferentiabilityDifferentiabilityMedium2 minQB

MathematicsMediummultiple choice

Let $f (x)$ be a non negative differentiable function such that $f^{1} (x) \leq f (x) \forall x \geq 0$ and $f (0) = 0$ thena) $f (1) + f (2) = 0$

Options:(select one or more)

Answer:

A, B, C

Solution:

$f^{1} (x) \leq f (x)$ , let $f^{1} (x) - f (x) = K \Rightarrow f (x) = - K + Ke^{x} = K (e^{z} - 1) \geq 0\forall x \geq 0 \Rightarrow K = 0$ ∴ f is a constant function but $f (0) = 0$ $∴ f \equiv 0$

Stream:JEESubject:MathematicsTopic:Continuity - DifferentiabilitySubtopic:Differentiability

⏱ 2mℹ️ Source: QB

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