First slide

Functions (XII)

Question

$If the function f : R \to R is defined by f (x) = | x | (x - \sin x), then which of the following statements is$ $TRUE ?$

Difficult

A

$f is one-one, but NOT onto$
B

$f is onto, but NOT one-one$
C

$f is BOTH one-one and onto$
D

$f is NEITHER one-one NOR onto$

Solution

$f (x) is a non-periodic, continuous and odd function$
$\begin{array}{l} f (x) = \{\begin{matrix} - x^{2} + x \sin x, x < 0 \\ x^{2} - x \sin x, x \geq 0 \end{matrix} \\ f (- \infty) = \lim_{x \to \infty} (- x^{2}) (1 - \frac{\sin x}{x}) = - \infty \\ f (\infty) = \lim_{x \to \infty} x^{2} (1 - \frac{\sin x}{x}) = \infty \end{array}$
$\begin{array}{l} \Rightarrow Range of f (x) = R \\ \Rightarrow f (x) is an onto function. \end{array}$
$f^{'} (x) = \{\begin{cases} - 2 x + \sin x + xcos x, x < 0 \\ 2 x - \sin x - xcos x, x \geq 0 \end{cases}$
$\begin{array}{l} For (0, \infty) \\ f^{'} (x) = (x - \sin x) + x (1 - \cos x) > 0 (\sin c e \sin x < x f o r x > 0) \end{array}$
$f^{'} (x) > 0$
$f o r x \in (- \infty, 0), f^{'} (x) = (\sin x - x) - x (1 - \cos x) > 0 (\sin c e \sin x > x f o r x < 0)$
$\begin{array}{l} \Rightarrow f^{'} (x) > 0 \\ \Rightarrow f^{'} (x) \geq 0, \forall x \in (- \infty, \infty) \end{array}$
$equality at x = 0$
$\Rightarrow f (x) is one - one function \dots (2)$
$From (1) & (2), f (x) is both one-one & onto.$

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