Let f : (0, \infty) → R be given by f(x) = \int_{\frac{1}{x}}^{x} e^{- \textrm{ } \left(\right. t + \frac{1}{t} \left.\right)} \frac{dt}{t}. Then

Engineering

Mathematics

Leibnitz Rule Derivative of Anti Derivatives

Question

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

Then

f (2^x) is an odd function of x on R.

f(x) is monotonically increasing on [1, $\infty$ ).

f(x) + $f (\frac{1}{x})$ = 0, for all x $\in$ (0, $\infty$ ).

f(x) is monotonically decreasing on (0, 1).

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Replacing $x \to \frac{1}{x}$

$f (\frac{1}{x}) = \int_{x}^{\frac{1}{x}} e^{- (t + \frac{1}{t})} \frac{dt}{t} = - \int_{\frac{1}{x}}^{x} e^{- (t + \frac{1}{t})} \frac{dt}{t}$

$∴ f (x) = - f (\frac{1}{x})$

$f (x) + f (\frac{1}{x}) = 0 . \Rightarrow (C)$

Applying Leibnitz rule

$f' (x) = (\frac{2}{x}) e^{- (x + \frac{1}{x})}$

$∴$ f '(x) > 0

Thus, f(x) is monotonically increasing ⇒ (A)

$f (2^{x}) = \int_{2^{- x}}^{2^{x}} e^{- (t + \frac{1}{t})} \frac{dt}{t}$

$f (2^{- x}) = \int_{2^{x}}^{2^{- x}} e^{- (t + \frac{1}{t})} \frac{dt}{t} = - \int_{2^{- x}}^{2^{x}} e^{- (t + \frac{1}{t})} \frac{dt}{t}$

f (2^–x) = – f(2^x) ⇒ (D)