Let f be a differentiable bijection with f ( 0 ) = 1 , g = f − 1 , and f ′ ( x ) − ∫ 0 x f ( t ) d t = ∫ 1 f ( x ) g ( t ) d t for x > 0 . Which statements are true?

The inverse-area identity gives ∫ 0 x f + ∫ 1 f ( x ) g = x f ( x ) . Therefore f ′ ( x ) = x f ( x ) , so f ( x ) = e x 2 /2 from f ( 0 ) = 1 . Hence f ( 1 ) = e and g ( e ) = 2 .

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MathematicsFunctionsInverseHard2 minQB

MathematicsHardmultiple choice

Let $f$ be a differentiable bijection with $f (0) = 1, g = f^{- 1}$ , and $f^{'} (x) - \int_{0}^{x} f (t) d t = \int_{1}^{f (x)} g (t) d t$ for $x > 0$ . Which statements are true?

Options:(select one or more)

Answer:

A, C

Solution:

The inverse-area identity gives $\int_{0}^{x} f + \int_{1}^{f (x)} g = x f (x)$ . Therefore $f^{'} (x) = x f (x)$ , so $f (x) = e^{x^{2} /2}$ from $f (0) = 1$ . Hence $f (1) = e$ and $g (e) = 2$ .

Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:Inverse

⏱ 2mℹ️ Source: QB

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