NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.3 (Ex 1.3)

Q.1

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\{1,3,4\}\to \{1,2,5\}\mathrm{and}\mathrm{g}:\{1,2,5\}\to \{1,3\}\mathrm{be}\mathrm{given}\\ \mathrm{by}\mathrm{f}=\left\{\right(1,2),(3,5),(4,1\left)\right\}\mathrm{and}\mathrm{g}=\left\{\right(1,3),(2,3),(5,1\left)\right\}.\\ \mathrm{Write}\mathrm{down}\mathrm{go}\mathrm{f}.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{functions}\mathrm{f}:\; \{1,\; 3,\; 4\}\to \{1,\; 2,\; 5\}\mathrm{and}\mathrm{g}:\; \{1,\; 2,\; 5\}\to \{1,\; 3\}\\ \mathrm{are}\mathrm{defined}\mathrm{as}\mathrm{f}=\; \{(1,\; 2),\; (3,\; 5),\; (4,\; 1)\}\mathrm{and}\\ \mathrm{g}=\; \{(1,\; 3),\; (2,\; 3),\; (5,\; 1)\}\mathrm{.}\\ \mathrm{gof}\left(1\right)=\mathrm{g}\left\{\mathrm{f}\left(1\right)\right\}=\mathrm{g}\left(2\right)=3[\mathrm{f}\left(1\right)=2,\mathrm{g}\left(2\right)=3]\\ \mathrm{gof}\left(3\right)=\mathrm{g}\left\{\mathrm{f}\left(3\right)\right\}=\mathrm{g}\left(5\right)=1[\mathrm{f}\left(3\right)=5,\mathrm{g}\left(5\right)=1]\\ \mathrm{gof}\left(4\right)=\mathrm{g}\left\{\mathrm{f}\left(4\right)\right\}=\mathrm{g}\left(1\right)=3[\mathrm{f}\left(4\right)=1,\mathrm{g}\left(1\right)=3]\\ \therefore \mathrm{gof}=\{(1,3),(3,1),(4,3)\}\end{array}$

Q.2 Let f, g and h be functions from R to R. Show that

(f + g)oh = foh + goh

(f. g)oh = (foh) . (goh)

Ans

\begin{array}{l}\text{Let}\left(\left(\text{f}+\text{g}\right)\text{oh}\right)\left(\text{x}\right)=\left(\text{foh}\right)\left(\text{x}\right)+\left(\text{goh}\right)\left(\text{x}\right)\\ =\text{f}\left\{\text{h}\left(\text{x}\right)\right\}+\text{g}\left\{\text{h}\left(\text{x}\right)\right\}\\ =\left(\text{foh}\right)\left(\text{x}\right)+\left(\text{goh}\right)\left(\text{x}\right)\\ \\ =\left\{\left(\text{f}+\text{g}\right)\text{oh}\right\}\left(\text{x}\right)\\ =\left\{\left(\text{foh}\right)+\left(\text{goh}\right)\right\}\left(\text{x}\right)\forall x\in R\\ Hence,\left(f+g\right)oh=\left(\text{foh}\right)+\left(\text{goh}\right).\\ Now,\\ \left\{\left(f.g\right)oh\right\}\left(x\right)=\left(f.g\right)\left(h\left(x\right)\right)\\ =f\left(h\left(x\right)\right).g\left(h\left(x\right)\right)\\ =\left(foh\right)\left(x\right).\left(goh\right)\left(x\right)\\ =\left\{\left(foh\right).\left(goh\right)\right\}\left(x\right)\\ \therefore \left\{\left(f.g\right)oh\right\}\left(x\right)=\left\{\left(foh\right).\left(goh\right)\right\}\left(x\right)\forall x\in R\\ Hence,\left(f.g\right)oh=\left(foh\right).\left(goh\right)\end{array}

Q.3

\begin{array}{l}Findgofandfog,if\\ (i)f\left(x\right)=\left|x\right|andg\left(x\right)=\left|5x–2\right|\\ \left(ii\right)f\left(x\right)=8x^{3}andg\left(x\right)=x^{\frac{1}{3}}\end{array}

Ans

\begin{array}{l}\left(i\right)f\left(x\right)=\left|x\right|andg\left(x\right)=\left|5x-2\right|\\ \therefore \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)\\ =g\left(\left|x\right|\right)\\ =\left|5\left|x\right|-2\right|\\ \left(fog\right)\left(x\right)=f\left(g\left(x\right)\right)\\ =f\left(\left|5x-2\right|\right)\\ \\ =\left|\left|5x-2\right|\right|\\ =\left|5x-2\right|\\ \left(ii\right)f\left(x\right)=8x^{3}andg\left(x\right)=x^{\frac{1}{3}}\\ \therefore \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)\\ =g\left(8x^3\right)\\ ={\left(8x^3\right)}^{\frac{1}{3}}\\ =2x\\ \left(fog\right)\left(x\right)=f\left(g\left(x\right)\right)\\ =f\left(x^{\frac{1}{3}}\right)\\ =8{\left(x^{\frac{1}{3}}\right)}^3\\ =8x\end{array}

Q.4

\begin{array}{l}Iff\left(x\right)=\frac{4x+3}{6x–4},x\ne \frac{2}{3},showthatfof\left(x\right)=x,forallx\ne \frac{2}{3}.\\ Whatistheinverseoff?\end{array}

Ans

\begin{array}{l}It\text{is given that}f\left(x\right)=\frac{4x+3}{6x–4},then\\ fof\left(x\right)=f\left(f\left(x\right)\right)\\ =f\left(\frac{4x+3}{6x–4}\right)\\ =\frac{4\left(\frac{4x+3}{6x–4}\right)+3}{6\left(\frac{4x+3}{6x–4}\right)-4}\\ =\frac{\left(\frac{16x+12+18x-12}{6x-4}\right)}{\left(\frac{24x+18-24x+16}{6x–4}\right)}\\ =\frac{34x}{34}\\ fof\left(x\right)=x,\text{for all x}\ne \frac{2}{3}.\\ \Rightarrow fof=I\\ \text{Hence, the given function f is invertible and the inverse of f}\\ \text{is f itself}\text{.}\end{array}

Q.5

$\begin{array}{l}\mathrm{State}\mathrm{with}\mathrm{reason}\mathrm{whether}\mathrm{following}\mathrm{functions}\hspace{0.17em}\mathrm{have}\mathrm{inverse}\\ \left(\mathrm{i}\right)\mathrm{f}:\{1,2,3,4\}\to \left\{10\right\}\mathrm{with}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{f}=\left\{\right(1,10),(2,10),(3,10),(4,10\left)\right\}\\ \left(\mathrm{ii}\right)\hspace{0.17em}\hspace{0.17em}\mathrm{g}:\{5,6,7,8\}\to \{1,2,3,4\}\mathrm{with}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{g}=\left\{\right(5,4),(6,3),(7,4),(8,2\left)\right\}\\ \left(\mathrm{iii}\right)\mathrm{h}:\{2,3,4,5\}\to \{7,9,11,13\}\mathrm{with}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{h}=\left\{\right(2,7),(3,9),(4,11),(5,13\left)\right\}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{f}:\; \{1,\; 2,\; 3,\; 4\}\to \left\{10\right\}\mathrm{defined}\mathrm{as}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{f}=\; \{(1,\; 10),\; (2,\; 10),\; (3,\; 10),\; (4,\; 10)\}\\ \mathrm{From}\mathrm{the}\mathrm{given}\mathrm{definition}\mathrm{of}\mathrm{f},\mathrm{we}\mathrm{can}\mathrm{see}\mathrm{that}\mathrm{f}\mathrm{is}\mathrm{a}\mathrm{many}\\ \mathrm{one}\mathrm{function}\mathrm{as}:\mathrm{f}(1)\; =\mathrm{f}(2)=\mathrm{f}(3)\; =\mathrm{f}(4)\; =\; 10\\ \therefore \mathrm{f}\mathrm{is}\mathrm{not}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Hence},\mathrm{function}\mathrm{f}\mathrm{does}\mathrm{not}\mathrm{have}\mathrm{an}\mathrm{inverse}\mathrm{.}\\ \left(\mathrm{ii}\right)\mathrm{\hspace{0.17em}}\mathrm{g}:\; \{5,\; 6,\; 7,\; 8\}\to \{1,\; 2,\; 3,\; 4\}\mathrm{defined}\mathrm{as}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{g}=\; \{(5,\; 4),\; (6,\; 3),\; (7,\; 4),\; (8,\; 2)\}\\ \mathrm{From}\mathrm{the}\mathrm{given}\mathrm{definition}\mathrm{of}\mathrm{g},\mathrm{it}\mathrm{is}\mathrm{seen}\mathrm{that}\mathrm{g}\mathrm{is}\mathrm{a}\mathrm{many}\mathrm{one}\\ \mathrm{function}\mathrm{as}:\mathrm{g}(5)\; =\mathrm{g}(7)\; =\; 4\mathrm{.}\\ \therefore \mathrm{g}\mathrm{is}\mathrm{not}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Hence},\mathrm{function}\mathrm{g}\mathrm{does}\mathrm{not}\mathrm{have}\mathrm{an}\mathrm{inverse}\mathrm{.}\\ \left(\mathrm{iii}\right)\mathrm{h}:\; \{2,\; 3,\; 4,\; 5\}\to \{7,\; 9,\; 11,\; 13\}\mathrm{defined}\mathrm{as}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{h}=\; \{(2,\; 7),\; (3,\; 9),\; (4,\; 11),\; (5,\; 13)\\ \mathrm{It}\mathrm{is}\mathrm{seen}\mathrm{that}\mathrm{all}\mathrm{distinct}\mathrm{elements}\mathrm{of}\mathrm{the}\mathrm{set}\{2,\; 3,\; 4,\; 5\}\\ \mathrm{have}\mathrm{distinct}\mathrm{images}\mathrm{under}\mathrm{h}\mathrm{.}\\ \therefore \mathrm{Function}\mathrm{h}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Also},\mathrm{h}\mathrm{is}\mathrm{onto}\mathrm{since}\mathrm{for}\mathrm{every}\mathrm{element}\mathrm{y}\mathrm{of}\mathrm{the}\mathrm{set}\\ \{7,\; 9,\; 11,\; 13\},\mathrm{there}\mathrm{exists}\mathrm{an}\hspace{0.17em}\hspace{0.17em}\mathrm{element}\mathrm{x}\mathrm{in}\mathrm{the}\mathrm{set}\\ \{2,\; 3,\; 4,\; 5\}\hspace{0.17em}\hspace{0.17em}\mathrm{such}\mathrm{that}\mathrm{h}(\mathrm{x})\; =\mathrm{y}\mathrm{.}\\ \mathrm{Thus},\mathrm{h}\mathrm{is}\mathrm{a}\mathrm{one}–\mathrm{one}\mathrm{and}\mathrm{onto}\mathrm{function}.\mathrm{Hence},\mathrm{h}\mathrm{has}\mathrm{an}\mathrm{inverse}\mathrm{.}\end{array}$

Q.6

\begin{array}{l}Showthatf:[-1,1]\to R,givenbyf\left(x\right)=\frac{x}{x+2}isone–one.\\ Findtheinverseofthefunctionf:[-1,1]\to Range\end{array}

Ans

\begin{array}{l}\text{f: [}-\text{1, 1]}\to \text{R is given as f}\left(x\right)=\frac{x}{x+2}\\ Let\text{f}\left(x\right)=f\left(y\right)\\ \Rightarrow \frac{x}{x+2}=\frac{y}{y+2}\\ \Rightarrow x\left(y+2\right)=y\left(x+2\right)\\ \Rightarrow xy+2x=xy+2y\\ \Rightarrow x=y\\ \therefore f\text{is one-one function}\text{.}\\ \text{It is clear that f: [}-\text{1, 1]}\to \text{Range f is onto}\text{.}\\ \therefore \text{f: [}-\text{1, 1]}\to \text{Range f is one-one and onto and therefore,}\\ \text{the inverse of the function:}\\ \text{f: [}-\text{1, 1]}\to \text{Range f exists}\text{.}\\ \text{Let g: Range f}\to \text{[}-\text{1, 1] be the inverse of f}\text{.}\\ \text{Let y be an arbitrary element of range f}\text{.}\\ \text{Since f: [}-\text{1, 1]}\to \text{Range f is onto, we have:}\\ \text{f}\left(x\right)=yfor\text{some x}\in \left[-1,1\right]\\ \Rightarrow y=\frac{x}{x+2}\\ \Rightarrow xy+2y=x\\ \Rightarrow 2y=x-xy\\ =x\left(1-y\right)\\ \Rightarrow x=\frac{2y}{1-y},y\ne 1\\ \text{Now, let us define g: Range f}\to \text{[}-\text{1, 1] as}\\ g\left(y\right)=\frac{2y}{1-y},y\ne 1.\\ Now,\\ \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)\\ =g\left(\frac{x}{x+2}\right)\\ =\frac{2\left(\frac{x}{x+2}\right)}{1-\left(\frac{x}{x+2}\right)}\\ =\frac{2x}{x+2-x}\\ =\frac{2x}{2}\\ \left(gof\right)\left(x\right)=x\\ \left(fog\right)\left(y\right)=f\left(g\left(y\right)\right)\\ =f\left(\frac{2y}{1-y}\right)\\ =\frac{\left(\frac{2y}{1-y}\right)}{\frac{2y}{1-y}+2}\\ =\frac{2y}{2y+2-2y}\\ =\frac{2y}{2}\\ \left(fog\right)\left(y\right)=y\\ \therefore go{f}^{-1}=I_{\left[-1,1\right]}andfo{g}^{-1}=I_{Rangef}\\ \therefore f^{-1}=g\\ \Rightarrow f^{-1}\left(y\right)=\frac{2y}{y-1},y\ne 1.\\ \Rightarrow f^{-1}\left(x\right)=\frac{2x}{x-1},x\ne 1.\end{array}

Q.7

$\begin{array}{l}\mathrm{Consider}\mathrm{f}:\mathrm{R}\to \mathrm{Rgiven}\mathrm{by}\mathrm{f}\left(\mathrm{x}\right)=4\mathrm{x}+3.\mathrm{Show}\mathrm{that}\mathrm{f}\mathrm{is}\\ \mathrm{invertible}.\mathrm{Find}\mathrm{the}\mathrm{inverse}\mathrm{off}.\end{array}$

Ans

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{f}(\mathrm{x})\; =\; 4\mathrm{x}+\; 3\\ \mathrm{One}–\mathrm{one}:\\ \mathrm{Let}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\mathrm{.}\\ \Rightarrow 4\mathrm{x}+\; 3=4\mathrm{y}+\; 3\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}4\mathrm{x}=4\mathrm{y}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=\mathrm{y}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{a}\mathrm{one}–\mathrm{one}\mathrm{function}\mathrm{.}\\ \mathrm{Onto}:\\ \mathrm{For}\mathrm{y}\in \mathrm{R},\\ \mathrm{let}\mathrm{y}=\; 4\mathrm{x}+\; 3\mathrm{.}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}=\frac{\mathrm{y}-3}{4}\in \mathrm{R}\\ \\ \mathrm{Therefore},\mathrm{for}\mathrm{any}\mathrm{y}\in \mathrm{R},\mathrm{there}\mathrm{exists},{\mathrm{f}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Let}\mathrm{us}\mathrm{define}\mathrm{g}:\mathrm{R}\to \mathrm{R}\mathrm{by}\mathrm{g}\left(\mathrm{y}\right)=\frac{\mathrm{y}-3}{4}\\ \mathrm{Now},\mathrm{\hspace{0.17em}}\left(\mathrm{gof}\right)\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{g}\left(4\mathrm{x}+3\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{4\mathrm{x}+3-3}{4}\\ \left(\mathrm{gof}\right)\left(\mathrm{x}\right)=\mathrm{x}\\ \mathrm{\hspace{0.17em}}\left(\mathrm{fog}\right)\left(\mathrm{y}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{y}\right)\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{f}\left(\frac{\mathrm{y}-3}{4}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=4\left(\frac{\mathrm{y}-3}{4}\right)+3\\ \mathrm{\hspace{0.17em}}\left(\mathrm{fog}\right)\left(\mathrm{y}\right)=\mathrm{y}\\ \therefore \mathrm{gof}=\mathrm{fog}={\mathrm{I}}_{\mathrm{R}}\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{invertible}\mathrm{and}\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{is}\mathrm{given}\mathrm{by}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{f}}^{-1}\left(\mathrm{y}\right)=\mathrm{g}\left(\mathrm{y}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{\mathrm{y}-3}{4}.\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\frac{\mathrm{x}-3}{4}\end{array}$

Q.8

\begin{array}{l}Considerf:R_{+}\to [4,\infty )givenbyf(x)=x^2+4.Showthatfis\\ invertiblewiththeinverse{f}^{–1}ofgivenfby,where{R}^{+}istheset\\ ofallnon–negativerealnumbers.\end{array}

Ans

\begin{array}{l}{\text{f: R}}_{\text{+}}\to \text{[4,}\infty {\text{) is given as f(x) = x}}^{\text{2}}\text{+ 4}\text{.}\\ \text{One-one:}\\ \text{Let f(x) = f(y)}\\ \Rightarrow {\text{x}}^{\text{2}}\text{+ 4}={\text{y}}^{\text{2}}\text{+ 4}\\ \Rightarrow {\text{x}}^{\text{2}}=y^2\\ \Rightarrow x=y\left[asx=y\in R_{+}\right]\\ \therefore fisone-one\text{function}.\\ Onto:\\ \text{For y}\in \text{[4,}\infty \text{),}\\ \text{let y}={\text{x}}^{\text{2}}\text{+ 4}\\ \text{x}=\sqrt{y-4}\ge 0\\ \text{Therefore, for any y}\in \text{R, there exists}\text{x}=\sqrt{y-4}\ge 0,\text{such that}\\ \text{f}\left(x\right)=f\left(\sqrt{y-4}\right)\\ ={\left(\sqrt{y-4}\right)}^2+4\\ =y-4+4\\ \text{f}\left(x\right)=y\\ \therefore f\text{is onto}\text{.}\\ {\text{Thus, f is one-one and onto and therefore, f}}^{\text{-1}}\text{exists}\text{.}\\ \text{Let us define g: [4,}\infty \text{)}\to {\text{R}}_{{}^{\text{+}}}\text{by,}\\ \text{g}\left(y\right)=\sqrt{y-4}\\ Now,\left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)\\ =g\left(x^2+4\right)\\ =\sqrt{\left(x^2+4\right)-4}\\ \left(gof\right)\left(x\right)=x\\ \left(fog\right)\left(y\right)=f\left(g\left(y\right)\right)\\ =f\left(\sqrt{y-4}\right)\\ ={\left(\sqrt{y-4}\right)}^2+4\\ =y-4+4\\ \left(fog\right)\left(y\right)=y\\ \therefore gof=fog=I_{R^{+}}.\\ \text{Hence, f is invertible and the inverse of f is given by}\\ f^{-1}\left(y\right)=g\left(y\right)=\sqrt{y-4}\\ \Rightarrow f^{-1}\left(x\right)=\sqrt{x-4}\end{array}

Q.9

\begin{array}{l}Considerf:R_{+}\to [–5,\infty )givenbyf(x)=9x^2+6x–5.\\ Showthatfisinvertiblewith{f}^{-1}\left(y\right)=\left(\frac{\left(\sqrt{y+6}\right)-1}{3}\right)\end{array}

Ans

\begin{array}{l}{\text{f: R}}_{\text{+}}\to \text{[}-\text{5,}\infty {\text{) is given as f(x) = 9x}}^{\text{2}}\text{+ 6x}-\text{5}\text{.}\\ \text{Let y be an arbitrary element of [}-\text{5,}\infty \text{)}\text{.}\\ \text{Let}\text{y}={\text{9x}}^{\text{2}}\text{+ 6x}-\text{5}\\ \Rightarrow {\text{9x}}^{\text{2}}\text{+ 6x}-\left(\text{5+y}\right)=\text{0}\\ \Rightarrow x=\frac{-6\pm \sqrt{6^2-4\times 9\times -\left(5+y\right)}}{2\times 9}\\ =\frac{-6\pm \sqrt{36+36\left(5+y\right)}}{18}\\ =\frac{-6\pm 6\sqrt{1+5+y}}{18}\\ x=\frac{-1\pm \sqrt{6+y}}{3}\\ =\frac{\sqrt{6+y}-1}{3}\left[âˆ\mu y\ge -6\Rightarrow y+6\ge 0\right]\\ \therefore \text{f is onto, thereby range f = [}-\text{5,}\infty \text{)}\text{.}\\ \text{Let us define g: [}-\text{5,}\infty \text{)}\to {\text{R}}_{\text{+}}\text{as}\\ g\left(y\right)=\frac{\sqrt{\left(y+6\right)}-1}{3}\\ Now,\left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)\\ =g\left({\text{9x}}^{\text{2}}\text{+ 6x}-\text{5}\right)\\ =\frac{\sqrt{\left({\text{9x}}^{\text{2}}\text{+ 6x}-\text{5}\right)+6}-1}{3}\\ =\frac{\sqrt{\left({\text{9x}}^{\text{2}}\text{+ 6x}+1\right)}-1}{3}\\ =\frac{\sqrt{{\left(3x+1\right)}^2}-1}{3}\\ =\frac{3x+1-1}{3}=x\\ And,\\ \left(fog\right)\left(y\right)=f\left(g\left(y\right)\right)\\ =f\left(\frac{\sqrt{\left(y+6\right)}-1}{3}\right)\\ =\text{9}{\left(\frac{\sqrt{\left(y+6\right)}-1}{3}\right)}^{\text{2}}\text{+ 6}\left(\frac{\sqrt{\left(y+6\right)}-1}{3}\right)-\text{5}\\ =9\left(\frac{y+6-2\sqrt{y+6}+1}{9}\right)+2\sqrt{\left(y+6\right)}-2-5\\ =y+6-2\sqrt{y+6}+1+2\sqrt{\left(y+6\right)}-2-5\\ \left(fog\right)\left(y\right)=y\\ \therefore gof=I_{R}andfog=I_{\left[-5,\infty \right]}\\ \text{Hence, f is invertible and the inverse of f is given by}\\ f^{-1}\left(y\right)=g\left(y\right)=\frac{\sqrt{y+6}-1}{3}.\end{array}

Q.10 $\text{Let f :X→Y be an invertible function.Show that f has unique inverse.}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\mathrm{X}\to \mathrm{Y}\mathrm{be}\mathrm{an}\mathrm{invertible}\mathrm{function}\mathrm{.}\\ \mathrm{Also},\mathrm{suppose}\mathrm{f}\mathrm{has}\mathrm{two}\mathrm{inverses}(\mathrm{say}{\mathrm{g}}_{\mathrm{1}}\mathrm{and}{\mathrm{g}}_{\mathrm{2}}\mathrm{)}\\ \mathrm{Then},\mathrm{for}\mathrm{all}\mathrm{y}\in \mathrm{Y},\mathrm{we}\mathrm{have}:\\ {\mathrm{fog}}_{\mathrm{1}}\left(\mathrm{y}\right)={\mathrm{I}}_{\mathrm{Y}}\left(\mathrm{y}\right)={\mathrm{fog}}_2\left(\mathrm{y}\right)\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{f}\left({\mathrm{g}}_1\left(\mathrm{y}\right)\right)=\mathrm{f}\left({\mathrm{g}}_2\left(\mathrm{y}\right)\right)\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{g}}_1\left(\mathrm{y}\right)={\mathrm{g}}_2\left(\mathrm{y}\right)\left[\mathrm{f}\mathrm{is}\mathrm{invertible}\Rightarrow \mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\right]\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{g}}_1={\mathrm{g}}_2\left[\mathrm{g}\mathrm{is}\mathrm{one}–\mathrm{one}\right]\\ \mathrm{Hence},\mathrm{f}\mathrm{has}\mathrm{a}\mathrm{unique}\mathrm{inverse}\mathrm{.}\end{array}$

Q.11

$\begin{array}{l}\mathrm{Consider}\mathrm{f}:\{1,2,3\}\to \{\mathrm{a},\mathrm{b},\mathrm{c}\}\mathrm{given}\mathrm{by}\mathrm{f}\left(1\right)=\mathrm{a},\mathrm{f}\left(2\right)=\mathrm{b}\\ \mathrm{and}\mathrm{f}\left(3\right)=\mathrm{c}.\mathrm{Find}{\mathrm{f}}^{–1}\mathrm{and}\mathrm{show}\mathrm{that}{\left({\mathrm{f}}^{–1}\right)}^{–1}=\mathrm{f}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Function}\mathrm{f}:\; \{1,\; 2,\; 3\}\to \{\mathrm{a},\mathrm{b},\mathrm{c}\}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{f}(1)\; =\mathrm{a},\mathrm{f}(2)\; =\mathrm{b},\mathrm{and}\mathrm{f}(3)\; =\mathrm{c}\\ \mathrm{If}\mathrm{we}\mathrm{define}\mathrm{g}:\; \{\mathrm{a},\mathrm{b},\mathrm{c}\}\to \{1,\; 2,\; 3\}\mathrm{as}\mathrm{g}(\mathrm{a})\; =\; 1,\mathrm{g}(\mathrm{b})\; =\; 2,\\ \mathrm{g}(\mathrm{c})\; =\; 3,\mathrm{then}\mathrm{we}\mathrm{have}:\\ \left(\mathrm{fog}\right)\left(\mathrm{a}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{a}\right)\right)=\mathrm{f}\left(1\right)=\mathrm{a}\\ \left(\mathrm{fog}\right)\left(\mathrm{b}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{b}\right)\right)=\mathrm{f}\left(2\right)=\mathrm{b}\\ \left(\mathrm{fog}\right)\left(\mathrm{c}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{c}\right)\right)=\mathrm{f}\left(3\right)=\mathrm{c}\\ \mathrm{and}\\ \left(\mathrm{gof}\right)\left(1\right)=\mathrm{g}\left(\mathrm{f}\left(1\right)\right)=\mathrm{g}\left(\mathrm{a}\right)=1\\ \left(\mathrm{gof}\right)\left(2\right)=\mathrm{g}\left(\mathrm{f}\left(2\right)\right)=\mathrm{g}\left(\mathrm{b}\right)=2\\ \left(\mathrm{gof}\right)\left(3\right)=\mathrm{g}\left(\mathrm{f}\left(3\right)\right)=\mathrm{g}\left(\mathrm{c}\right)=3\\ \therefore \mathrm{gof}={\mathrm{I}}_{\mathrm{X}}\mathrm{and}\mathrm{fog}={\mathrm{I}}_{\mathrm{Y}},\mathrm{where}\mathrm{X}=\{1,2,3\}\mathrm{and}\mathrm{Y}=\{\mathrm{a},\mathrm{b},\mathrm{c}\}.\\ \mathrm{Thus},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{exists}\mathrm{and}{\mathrm{f}}^{–1}=\mathrm{g}\mathrm{.}\\ \therefore {\mathrm{f}}^{-1}:\{\mathrm{a},\mathrm{b},\mathrm{c}\}\to \{1,2,3\}\mathrm{is}\mathrm{given}\mathrm{by},\\ {\mathrm{f}}^{-\mathrm{1}}\left(\mathrm{a}\right)=1,{\mathrm{f}}^{-\mathrm{1}}\left(\mathrm{b}\right)=2,{\mathrm{f}}^{-\mathrm{1}}\left(\mathrm{c}\right)=3\\ \mathrm{The}\mathrm{inverse}\mathrm{of}{\mathrm{f}}^{–1}\mathrm{}\Rightarrow \mathrm{The}\mathrm{inverse}\mathrm{of}\mathrm{g}\\ \mathrm{We}\mathrm{define},\mathrm{h}:\{1,2,3\}\to \{\mathrm{a},\mathrm{b},\mathrm{c}\}\hspace{0.17em}\hspace{0.17em}\mathrm{as}\\ \mathrm{h}(1)\; =\mathrm{a},\mathrm{h}(2)\; =\mathrm{b},\mathrm{h}(3)\; =\mathrm{c},\mathrm{then}\mathrm{we}\mathrm{have}:\\ \left(\mathrm{goh}\right)\left(1\right)=\mathrm{g}\left(\mathrm{h}\left(1\right)\right)=\mathrm{g}\left(\mathrm{a}\right)=1\\ \left(\mathrm{goh}\right)\left(2\right)=\mathrm{g}\left(\mathrm{h}\left(2\right)\right)=\mathrm{g}\left(\mathrm{b}\right)=2\\ \left(\mathrm{goh}\right)\left(3\right)=\mathrm{g}\left(\mathrm{h}\left(3\right)\right)=\mathrm{g}\left(\mathrm{c}\right)=3\\ \mathrm{and}\\ \left(\mathrm{hog}\right)\left(\mathrm{a}\right)=\mathrm{h}\left(\mathrm{g}\left(\mathrm{a}\right)\right)=\mathrm{h}\left(1\right)=\mathrm{a}\\ \left(\mathrm{hog}\right)\left(\mathrm{b}\right)=\mathrm{h}\left(\mathrm{g}\left(\mathrm{b}\right)\right)=\mathrm{h}\left(2\right)=\mathrm{b}\\ \left(\mathrm{hog}\right)\left(\mathrm{c}\right)=\mathrm{h}\left(\mathrm{g}\left(\mathrm{c}\right)\right)=\mathrm{h}\left(3\right)=\mathrm{c}\\ \therefore \mathrm{goh}={\mathrm{I}}_{\mathrm{X}}\mathrm{\hspace{0.17em}}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{hog}={\mathrm{I}}_{\mathrm{Y}},\mathrm{\hspace{0.17em}}\mathrm{where}\mathrm{\hspace{0.17em}}\mathrm{X}=\{1,2,3\}\mathrm{and}=\{\mathrm{a},\mathrm{b},\mathrm{c}\}.\\ \mathrm{Thus},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{g}\mathrm{exists}\mathrm{and}\mathrm{\hspace{0.17em}}{\mathrm{g}}^{-1}=\mathrm{h}\Rightarrow {\left({\mathrm{f}}^{-\mathrm{1}}\right)}^{-\mathrm{1}}=\mathrm{h}\mathrm{.}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{noted}\mathrm{that}\mathrm{h}=\mathrm{f}\mathrm{.}\\ \mathrm{Hence},{\left({\mathrm{f}}^{–1}\right)}^{–1}=\mathrm{f}\mathrm{.}\end{array}$