Let f : 0 , ∞ → R be given by f x = ∫ 1 x x e - t + 1 t d t t , then

f ′ x = 2 e - x + 1 x x Which is increasing in 1 , ∞ Also, f x + f 1 x = 0 g x = f 2 x = ∫ 2 - x 2 x e - t + 1 t t d t g - x = ∫ 2 x 2 - x e - t + 1 t t d t = - g x Hence, an odd function

Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMonotonicity-Increasing-DecreasingMedium2 minPYQ_2014

MathematicsMediummultiple choice

Let $f : (0, \infty) \to R$ be given by $f (x) = \int_{\frac{1}{x}}^{x} e^{- (t + \frac{1}{t})} \frac{d t}{t},$ then

Options:(select one or more)

Answer:

A, C, D

Solution:

$f^{'} (x) = \frac{2 e^{- (x + \frac{1}{x})}}{x}$
Which is increasing in $[1, \infty)$
Also, $f (x) + f (\frac{1}{x}) = 0$
$g (x) = f (2^{x}) = \int_{2^{- x}}^{2^{x}} \frac{e^{- (t + \frac{1}{t})}}{t} d t$
$g (- x) = \int_{2^{x}}^{2^{- x}} \frac{e^{- (t + \frac{1}{t})}}{t} d t = - g (x)$
Hence, an odd function

Stream:JEE_ADVSubject:MathematicsTopic:Application of DerivativeSubtopic:Monotonicity-Increasing-Decreasing

⏱ 2mℹ️ Source: PYQ_2014

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