First slide

Functions (XII)

Question

Which of the following functions is one-one?

Moderate

A

$(a) f : R \to R defined as f (x) = e^{sgn x} + e^{x^{2}}$
B

$(b) f : [- 1, \infty) \to (0, \infty) defined by f (x) = e^{x^{2} + | x |}$
C

$(c) f : [3, 4] \to [4, 6] defined by f (x) = | x - 1 | + | x - 2 | +$ $| x - 3 | + | x - 4 |$
D

$(d) f (x) = \sqrt{\ln (\cos (\sin x))}$

Solution

$(a) f (x) = e^{sgn x} + e^{x^{2}}$
$\begin{aligned} when x = 0 \\ when x > 0 \end{aligned}$ $\begin{aligned} then f (0) = 2 \\ then f (x) = e + e^{x^{2}} \end{aligned}$
$when x < 0$ $then f (x) = \frac{1}{e} + e^{x^{2}}$
$(b) We have f (x) = e^{x^{2} + | x |}, x \in [- 1, \infty)$
$clearly f (- 1) = e^{2} = f (1)$
$also x^{2} + | x | \geq 0 \forall x \in [- 1, \infty]$
$\Rightarrow R_{f} = [1, \infty)$
$∴ f (x) is many-one into function.$
$\begin{array}{l} (c) f (x) =∣ x - 1 | + | x - 2 | + | x - 3 | + | x - 4 ∣, x \in [3, 4] \\ ∴ f (x) = (x - 1) + (x - 2) + (x - 3) - (x - 4) \\ = (3 x - 6) - (x - 4) \end{array}$
$f (x) = 2 x - 2, which is increasing function$
$R_{f} = [4, 6]$
$Clearly, f (x) is one-one onto function.$
$(d) f (x) = \sqrt{\ln (\cos (\sin x))}$
$For domain, \ln (\cos (\sin x)) \geq 0$
$\begin{array}{ll} \Rightarrow (\cos (\sin x)) \geq 1 \\ \Rightarrow \cos (\sin x) = 1 \\ ∴ \sin x = 0 \\ \Rightarrow x = n π, n \in I \\ R_{f} = {0} \end{array}$
$since f (x) = 0$
$Thus, f (x) is many-one function.$

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