First slide

Functions (XII)

Question

Which of the following functions is an injective (one-one) function in its respective domain?

Difficult

A

$f (x) = 2 x + \sin 3 x$
B

$f (x) = x \cdot [x], (where [.] denotes the G.I.F)$
C

$f (x) = \frac{2^{x} - 1}{4^{x} + 1}$
D

$f (x) = \frac{2^{x} + 1}{4^{x} - 1}$

Solution

$(1) I f f (x) = 2 x + \sin 3 x \Rightarrow f' (x) = 2 + 3 \cos 3 x = 0 \Rightarrow \cos 3 x = \frac{- 2}{3}, w h i c h i s d e f i n e d$
$a n d f'' (x) = - 9 \sin 3 x = - 3 \sqrt{5} < 0 s o f (x) h a s m a x i m u m v a l u e e x i s t s i n i t s d o m a i n, s o i t i s n o t o n e - o n e$
$(2) f (x) = x . [x] i s n o t o n e - o n e \sin c e f (0.1) = 0.1 \times 0 = 0 a n d f (0.2) = 0$
$(3) f (x) = \frac{2^{x} - 1}{4^{x} + 1}$
$\Rightarrow f (x) i s c o n t i n u o u s f u n c t i o n a n d f (0) = 0 a n d f (x) \to 0 a g a i n w h e n x \to \infty$
$(\sin c e y = \frac{\frac{1}{2^{x}} - \frac{1}{4^{x}}}{1 + \frac{1}{4^{x}}} \to 0 a s x \to \infty)$ $h e n c e f (x) h a s a a t l e a t o n e m a x i m u m e x i s t s, s o i t i s n o t o n e - o n e$
$(4) f (x) = \frac{2^{x} + 1}{4^{x} - 1} = \frac{2^{x} + 1}{(2^{x} - 1) (2^{x} + 1)} = \frac{1}{2^{x} - 1} \Rightarrow f' (x) = - \frac{2^{x} \log 2}{{(2^{x} - 1)}^{2}} < 0 \Rightarrow f (x) i s d e c r e a \sin g f u n c t i o n s o i t i s o n e - o n e$

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