Name: Commutative & Associative Property of Binary Operation
Uploaded: 2019-07-19
Duration: 5 min 47 s
Description: To know more about the properties of Binary operations, watch the video and explore the different problems presented. Commit the concept to your memory with...

Question

Let

f : N \to 5

be a function defined by

f (x) = 9 x^{2} + 6 x - 5

, where

S

is the range of

f

. Show that

f

is invertible. Find the inverse of

f

and hence find

f^{- 1} (43)

and

f^{- 1} (163)

.

Accepted Answer

Given that, $f : R^{+} \to [- 5, \infty)$ is defined by $f (x) = 9 x^{2} + 6 x - 5$

Injective: $\forall x, y \in R^{+}$ such that $f (x) = f (y)$

$\Rightarrow 9 x^{2} + 6 x - 5 = 9 y^{2} + 6 y - 5$

$\Rightarrow 9 (x^{2} - y^{2}) + 6 (x - y) = 0$

$\Rightarrow (x - y) [9 (x + y) + 6] = 0$ $\Rightarrow x - y = 0 [∵ 9 (x + y) + 6 \neq 0 \forall x, y \in R^{+}]$

$\Rightarrow x = y$

Thus, $f^{- 1} (163)$ is injective.

Surjective: Let $y \in [- 5, \infty)$ be any arbitrary element.

Then $f (x) = y \Rightarrow y = 9 x^{2} + 6 x - 5$

$\Rightarrow y = (3 x + 1)^{2} - 6 \Rightarrow y + 6 = (3 x + 1)^{2}$

$\Rightarrow 3 x + 1 = \pm \sqrt{y + 6} \Rightarrow x = \frac{- 1 \pm \sqrt{y + 6}}{3}$

Since $y \in [- 5, \infty) \Rightarrow y \geq - 5 \Rightarrow y + 6 \geq 1$

$\Rightarrow \sqrt{y + 6} \geq 1 \Rightarrow \frac{- 1 + \sqrt{y + 6}}{3} \geq 0$

$\Rightarrow x = \frac{- 1 + \sqrt{y + 6}}{3} \geq 0 l . e . x \in R^{+}$

Thus, $\forall y \in [- 5, \infty) \exists x \in R^{+}$ such that $f (x) = y ∴ f$ is onto.

Hence $f$ is $1 - 1$ and onto, thus $f$ is bijective.

$∴ f^{- 1}$ exists and let $f^{- 1}$ denotes the inverse of $f$ .

Then $(f o f^{- 1}) (x) = x \forall x \in [- 5, \infty)$

$\Rightarrow f [f^{- 1} (x)] = x \forall x \in [- 5, \infty)$

$\Rightarrow 9 {\{f^{- 1} (x)\}}^{2} + 6 \{f^{- 1} (x)\} - 5 = x$

$\forall x \in [- 5, \infty)$

$\Rightarrow {\{3 f^{- 1} (x) + 1\}}^{2} = x + 6 \forall x \in [- 5, \infty)$

$\Rightarrow 3 f^{- 1} (x) + 1 = \sqrt{x + 6} \forall x \in [- 5, \infty)$

$f^{- 1} (x) = \frac{- 1 + \sqrt{x + 6}}{3}, \forall x \in [- 5, \infty)$

$\Rightarrow f^{- 1} (43) = \frac{- 1 + \sqrt{43 + 6}}{3} = \frac{- 1 + \sqrt{49}}{3}$

$= \frac{- 1 + 7}{3} = 2$

$\Rightarrow f^{- 1} (163) = \frac{- 1 + \sqrt{163 + 6}}{3} = \frac{- 1 + \sqrt{169}}{3}$

$= \frac{- 1 + 13}{3} = 4$

Let $f : N \to 5$ be a function defined by $f (x) = 9 x^{2} + 6 x - 5$ , where $S$ is the range of $f$ . Show that $f$ is invertible. Find the inverse of $f$ and hence find $f^{- 1} (43)$ and $f^{- 1} (163)$ .

Important Questions on Relations and Functions

Learn and Prepare for any exam you want !

Let f:N→5 be a function defined by f(x)=9x2+6x-5, where S is the range of f. Show that f is invertible. Find the inverse of f and hence find f-1(43) and f-1(163).

Important Questions on Relations and Functions

Learn and Prepare for any exam you want !

Let $f : N \to 5$ be a function defined by $f (x) = 9 x^{2} + 6 x - 5$ , where $S$ is the range of $f$ . Show that $f$ is invertible. Find the inverse of $f$ and hence find $f^{- 1} (43)$ and $f^{- 1} (163)$ .