First slide

Monotonicity

Question

$Let b be a nonzero real number. Suppose f : ℝ \to ℝ is a differentiable function such that f (0) = 1 .$
$If the derivative f^{'} of f satisfies the equation f^{'} (x) = \frac{f (x)}{b^{2} + x^{2}} for all x \in ℝ, then which of the following statements is/are TRUE?$

Difficult

A

$If b > 0, then f is an increasing function$
B

$If b < 0, then f is a decreasing function$
C

$f (x) f (- x) = 1 for all x \in ℝ$
D

$f (x) - f (- x) = 0 for all x \in ℝ$

Solution

$\begin{array}{l} f^{'} (x) = \frac{f (x)}{b^{2} + x^{2}} \\ \int \frac{f^{'} (x)}{f (x)} d x = \int \frac{d x}{x^{2} + b^{2}} \\ \Rightarrow \ln | f (x) | = \frac{1}{b} \tan^{- 1} (\frac{x}{b}) + c \\ Now f (0) = 1 \\ ∴ c = 0 \\ ∴ | f (x) | = e^{\frac{1}{b} \tan^{- 1} (\frac{x}{b})} \\ \Rightarrow f (x) = \pm e^{\frac{1}{b} \tan^{- 1} (\frac{x}{b})} \\ since f (0) = 1 ∴ f (x) = e^{\frac{1}{b} \tan^{- 1} (\frac{x}{b})} \end{array}$
$\begin{array}{l} x \to - x \\ f (- x) = e^{- \frac{1}{b} \tan^{- 1} (\frac{x}{b})} \\ ∴ f (x) \cdot f (- x) = e^{0} = 1 (option C) \end{array}$
$and for b > 0$
$\begin{array}{l} f (x) = e^{\frac{1}{b} \tan^{- 1} (\frac{x}{b})} \\ \Rightarrow f (x) is increasing for all x \in R (option A) \end{array} (\sin c e f^{'} (x) > 0 f o r a l l x \in R, b \in R)$

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