# Ncert Solutions class 12 maths chapter 1 exercise 1.3

Q.1

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

Ans

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

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{}\text{}\text{}\text{}\text{\hspace{0.17em}}=\text{f}\left\{\text{h}\left(\text{x}\right)\right\}+\text{g}\left\{\text{h}\left(\text{x}\right)\right\}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}=\text{}\left(\text{foh}\right)\left(\text{x}\right)+\left(\text{goh}\right)\left(\text{x}\right)\\ \\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}=\text{}\left\{\left(\text{f}+\text{g}\right)\text{oh}\right\}\left(\text{x}\right)\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}=\text{}\left\{\left(\text{foh}\right)+\left(\text{goh}\right)\right\}\left(\text{x}\right)\text{}\forall x\in R\\ Hence,\text{}\left(f+g\right)oh=\left(\text{foh}\right)+\left(\text{goh}\right).\\ Now,\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left\{\left(f.g\right)oh\right\}\left(x\right)=\left(f.g\right)\left(h\left(x\right)\right)\\ \text{}\text{}\text{}\text{}=f\left(h\left(x\right)\right).g\left(h\left(x\right)\right)\\ \text{}\text{}\text{}\text{}=\left(foh\right)\left(x\right).\left(goh\right)\left(x\right)\\ \text{}\text{}\text{}\text{}=\left\{\left(foh\right).\left(goh\right)\right\}\left(x\right)\\ \therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left\{\left(f.g\right)oh\right\}\left(x\right)=\left\{\left(foh\right).\left(goh\right)\right\}\left(x\right)\text{}\text{}\forall x\in R\\ Hence,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(f.\text{\hspace{0.17em}}g\right)oh=\left(foh\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\left(goh\right)\end{array}$

Ans

$\begin{array}{l}\left(i\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(x\right)=|x|\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\text{\hspace{0.17em}}g\left(x\right)=|5x-2|\\ \therefore \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=g\left(|x|\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|5|x|-2|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(fog\right)\left(x\right)=f\left(g\left(x\right)\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=f\left(|5x-2|\right)\\ \\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=||5x-2||\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|5x-2|\\ \left(ii\right)\text{\hspace{0.17em}}f\left(x\right)=8{x}^{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\text{\hspace{0.17em}}g\left(x\right)={x}^{\frac{1}{3}}\\ \therefore \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=g\left(8{x}^{3}\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left(8{x}^{3}\right)}^{\frac{1}{3}}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2x\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(fog\right)\left(x\right)=f\left(g\left(x\right)\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=f\left({x}^{\frac{1}{3}}\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=8{\left({x}^{\frac{1}{3}}\right)}^{3}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=8x\end{array}$

Q.4

$\begin{array}{l}Iff\left(x\right)=\frac{4x+3}{6x–4},\text{\hspace{0.17em}}x\ne \frac{2}{3},showthatfof\left(x\right)=x,\text{\hspace{0.17em}}forallx\ne \frac{2}{3}.\\ Whatistheinverseoff?\end{array}$

Ans

$\begin{array}{l}It\text{is given that}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(x\right)=\frac{4x+3}{6x–4},\text{\hspace{0.17em}}then\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}fof\left(x\right)=f\left(f\left(x\right)\right)\\ \text{}\text{}\text{}=f\left(\frac{4x+3}{6x–4}\right)\\ \text{}\text{}\text{}=\frac{4\left(\frac{4x+3}{6x–4}\right)+3}{6\left(\frac{4x+3}{6x–4}\right)-4}\\ \text{}\text{}\text{}=\frac{\left(\frac{16x+12+18x-12}{6x-4}\right)}{\left(\frac{24x+18-24x+16}{6x–4}\right)}\\ \text{}\text{}\text{}=\frac{34x}{34}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}fof\left(x\right)=x,\text{for all x}\ne \frac{2}{3}.\\ ⇒\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}fof=I\\ \text{Hence, the given function f is invertible and the inverse of f}\\ \text{is f itself}\text{.}\end{array}$

Q.5

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

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{f}: \left\{1, 2, 3, 4\right\}\to \left\{10\right\}\mathrm{defined}\mathrm{as}:\\ \mathrm{ }\mathrm{f}= \left\{\left(1, 10\right), \left(2, 10\right), \left(3, 10\right), \left(4, 10\right)\right\}\\ \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}\left(1\right) =\mathrm{f}\left(2\right)=\mathrm{f}\left(3\right) =\mathrm{f}\left(4\right) = 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{ }\mathrm{g}: \left\{5, 6, 7, 8\right\}\to \left\{1, 2, 3, 4\right\}\mathrm{defined}\mathrm{as}:\\ \mathrm{g}= \left\{\left(5, 4\right), \left(6, 3\right), \left(7, 4\right), \left(8, 2\right)\right\}\\ \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}\left(5\right) =\mathrm{g}\left(7\right) = 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}: \left\{2, 3, 4, 5\right\}\to \left\{7, 9, 11, 13\right\}\mathrm{defined}\mathrm{as}:\\ \mathrm{h}= \left\{\left(2, 7\right), \left(3, 9\right), \left(4, 11\right), \left(5, 13\right)\\ \mathrm{It}\mathrm{is}\mathrm{seen}\mathrm{that}\mathrm{all}\mathrm{distinct}\mathrm{elements}\mathrm{of}\mathrm{the}\mathrm{set}\left\{2, 3, 4, 5\right\}\\ \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}\\ \left\{7, 9, 11, 13\right\},\mathrm{there}\mathrm{exists}\mathrm{an} \mathrm{element}\mathrm{x}\mathrm{in}\mathrm{the}\mathrm{set}\\ \left\{2, 3, 4, 5\right\} \mathrm{such}\mathrm{that}\mathrm{h}\left(\mathrm{x}\right) =\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

$\begin{array}{l}Showthatf:\left[-1,1\right]\to R,givenbyf\left(x\right)=\frac{x}{x+2}is\text{\hspace{0.17em}}one–one.\\ Findtheinverseofthefunctionf:\left[-1,1\right]\to Rangef.\end{array}$

Ans

$f: [−1, 1]→R is given as f x = x x+2 Let f x =f y ⇒ x x+2 = y y+2 ⇒ x y+2 =y x+2 ⇒ xy+2x=xy+2y ⇒ x=y ∴f is one-one function. It is clear that f: [−1, 1]→Range f is onto. ∴f: [−1, 1]→Range f is one-one and onto and therefore, the inverse of the function: f: [−1, 1]→ Range f exists. Let g: Range f→[−1, 1] be the inverse of f. Let y be an arbitrary element of range f. Since f: [−1, 1]→Range f is onto, we have: f x =y for some x∈ −1,1 ⇒ y= x x+2 ⇒xy+2y=x ⇒ 2y=x−xy =x 1−y ⇒ x= 2y 1−y , y≠1 Now, let us define g: Range f→[−1, 1] as g y = 2y 1−y , y≠1. Now, gof x =g f x =g x x+2 = 2 x x+2 1− x x+2 = 2x x+2−x = 2x 2 gof x =x fog y =f g y =f 2y 1−y = 2y 1−y 2y 1−y +2 = 2y 2y+2−2y = 2y 2 fog y =y ∴go f −1 = I −1,1 and fo g −1 = I Range f ∴ f −1 =g ⇒ f −1 y = 2y y−1 , y≠1. ⇒ f −1 x = 2x x−1 , x≠1. 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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

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

Q.8

$\begin{array}{l}Considerf:{R}_{+}\to \left[4,\infty \right)givenbyf\left(x\right)={x}^{2}+4.Showthatfis\\ invertiblewiththeinverse{f}^{–1}ofgivenfby,where{R}^{+}istheset\\ ofallnon–negativereal\text{\hspace{0.17em}}\text{\hspace{0.17em}}numbers.\end{array}$

Ans

$f: R + → [4, ∞ ) is given as f(x) = x 2 + 4. One-one: Let f(x) = f(y) ⇒ x 2 + 4= y 2 + 4 ⇒ x 2 = y 2 ⇒ x=y as x=y∈ R + ∴ f is one−one function. Onto: For y∈ [4,∞), let y= x 2 + 4 x= y−4 ≥0 Therefore, for any y∈R, there exists x= y−4 ≥0, such that f x =f y−4 = y−4 2 +4 =y−4+4 f x =y ∴ f is onto. Thus, f is one-one and onto and therefore, f -1 exists. Let us define g: [4,∞)→ R + by, g y = y−4 Now, gof x =g f x =g x 2 +4 = x 2 +4 −4 gof x =x fog y =f g y =f y−4 = y−4 2 +4 =y−4+4 fog y =y ∴gof= fog= I R + . 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Ans

$f: R + → [−5, ∞ ) is given as f(x) = 9x 2 + 6x−5. Let y be an arbitrary element of [−5,∞). Let y = 9x 2 + 6x−5 ⇒ 9x 2 + 6x− 5+y =0 ⇒ x= −6± 6 2 −4×9×− 5+y 2×9 = −6± 36+36 5+y 18 = −6±6 1+5+y 18 x= −1± 6+y 3 = 6+y −1 3 âˆµy≥−6⇒y+6≥0 ∴f is onto, thereby range f = [−5, ∞). Let us define g: [−5, ∞)→ R + as g y = y+6 −1 3 Now, gof x =g f x =g 9x 2 + 6x−5 = 9x 2 + 6x−5 +6 −1 3 = 9x 2 + 6x+1 −1 3 = 3x+1 2 −1 3 = 3x+1−1 3 =x And, fog y =f g y =f y+6 −1 3 =9 y+6 −1 3 2 + 6 y+6 −1 3 −5 =9 y+6−2 y+6 +1 9 +2 y+6 −2−5 =y+6−2 y+6 +1+2 y+6 −2−5 fog y =y ∴ gof= I R and fog= I −5,∞ Hence, f is invertible and the inverse of f is given by f −1 y =g y = y+6 −1 3 . 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Q.10

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

Ans

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

Q.11

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

Ans

$\begin{array}{l}\mathrm{Function}\mathrm{f}: \left\{1, 2, 3\right\}\to \left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\}\mathrm{is}\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{If}\mathrm{we}\mathrm{define}\mathrm{g}: \left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\}\to \left\{1, 2, 3\right\}\mathrm{as}\mathrm{g}\left(\mathrm{a}\right) = 1,\mathrm{g}\left(\mathrm{b}\right) = 2,\\ \mathrm{g}\left(\mathrm{c}\right) = 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}=\left\{1,2,3\right\}\mathrm{and}\mathrm{Y}=\left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\}.\\ \mathrm{Thus},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{exists}\mathrm{and}{\mathrm{f}}^{–1}=\mathrm{g}\mathrm{.}\\ \therefore {\mathrm{f}}^{-1}:\left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\}\to \left\{1,2,3\right\}\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{}⇒\mathrm{The}\mathrm{inverse}\mathrm{of}\mathrm{g}\\ \mathrm{We}\mathrm{define},\mathrm{h}:\left\{1,2,3\right\}\to \left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\} \mathrm{as}\\ \mathrm{h}\left(1\right) =\mathrm{a},\mathrm{h}\left(2\right) =\mathrm{b},\mathrm{h}\left(3\right) =\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{ }\mathrm{and} \mathrm{hog}={\mathrm{I}}_{\mathrm{Y}},\mathrm{ }\mathrm{where}\mathrm{ }\mathrm{X}=\left\{1,2,3\right\}\mathrm{and}=\left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\}.\\ \mathrm{Thus},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{g}\mathrm{exists}\mathrm{and}\mathrm{ }{\mathrm{g}}^{-1}=\mathrm{h}⇒{\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}$

Q.12

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\mathrm{X}\to \mathrm{Y}\mathrm{be}\mathrm{aninvertible}\mathrm{function}.\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{inverse}\\ \mathrm{of}{\mathrm{f}}^{–1}\mathrm{is}\mathrm{f},\mathrm{i}.\mathrm{e}.,{\left({\mathrm{f}}^{–1}\right)}^{–1}=\mathrm{f}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\mathrm{X}\to \mathrm{Y}\mathrm{be}\mathrm{an}\mathrm{invertible}\mathrm{function}\mathrm{.}\\ \mathrm{Then},\mathrm{there}\mathrm{exists}\mathrm{a}\mathrm{function}\mathrm{g}:\mathrm{Y}\to \mathrm{X}\mathrm{such}\mathrm{that}\mathrm{gof}={\mathrm{I}}_{\mathrm{X}}\\ \mathrm{and}\mathrm{fog}={\mathrm{I}}_{\mathrm{Y}}\mathrm{.}\\ \mathrm{Here},{\mathrm{f}}^{-1}=\mathrm{g}\mathrm{.}\\ \mathrm{Now},\mathrm{gof}={\mathrm{I}}_{\mathrm{X}}\mathrm{and}\mathrm{fog}={\mathrm{I}}_{\mathrm{Y}}\\ ⇒{\mathrm{f}}^{–1}\mathrm{of}={\mathrm{I}}_{\mathrm{X}}\mathrm{ }\mathrm{and}{\mathrm{fof}}^{–1}={\mathrm{I}}_{\mathrm{Y}}\\ \mathrm{Hence},{\mathrm{f}}^{-\mathrm{1}}:\mathrm{Y}\to \mathrm{X}\mathrm{is}\mathrm{invertible}\mathrm{and}\mathrm{f}\mathrm{is}\mathrm{the}\mathrm{inverse}\mathrm{of}{\mathrm{f}}^{-\mathrm{1}}\\ \mathrm{i}.\mathrm{e}., \left({\mathrm{f}}^{-\mathrm{1}}{\mathrm{\right)}}^{-1}=\mathrm{f}\mathrm{.}\end{array}$

Q.13

$\begin{array}{l}\mathrm{If}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{be}\mathrm{given}\mathrm{by},\mathrm{f}\left(\mathrm{x}\right)={\left(3-{\mathrm{x}}^{3}\right)}^{\frac{1}{3}}\mathrm{ }\mathrm{then}\mathrm{f}\mathrm{of}\left(\mathrm{x}\right)\mathrm{is}\\ \left(\mathrm{A}\right){\mathrm{x}}^{\frac{1}{3}} \left(\mathrm{B}\right){\mathrm{x}}^{3}\left(\mathrm{C}\right)\mathrm{x}\left(\mathrm{D}\right)\left(3-{\mathrm{x}}^{3}\right).\end{array}$

Ans

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{given}\mathrm{as} \mathrm{f}\left(\mathrm{x}\right)={\left(3-{\mathrm{x}}^{3}\right)}^{\frac{1}{3}}\\ \therefore \mathrm{fof}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\\ \mathrm{ }=\mathrm{f}\left\{{\left(3-{\mathrm{x}}^{3}\right)}^{\frac{1}{3}}\right\}\\ \mathrm{ }={\left[3-{\left\{{\left(3-{\mathrm{x}}^{3}\right)}^{\frac{1}{3}}\right\}}^{3}\right]}^{\frac{1}{3}}\\ \mathrm{ }={\left\{3-\left(3-{\mathrm{x}}^{3}\right)\right\}}^{\frac{1}{3}}\\ \mathrm{ }={\left(3-3+{\mathrm{x}}^{3}\right)}^{\frac{1}{3}}\\ \mathrm{ }\mathrm{fof}\left(\mathrm{x}\right)=\mathrm{x}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\left(\mathrm{C}\right)\mathrm{.}\\ \end{array}$

Q.14

$\begin{array}{l}\mathrm{Let} \mathrm{f}:\mathrm{R}-\left\{-\frac{4}{3}\right\}\to \mathrm{R}\mathrm{be}\mathrm{a}\mathrm{function}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)=\frac{4\mathrm{x}}{3\mathrm{x}+4}.\\ \mathrm{The}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{is}\mathrm{map}\mathrm{g}:\mathrm{R}\mathrm{ange}\mathrm{f}\to \mathrm{R}-\left\{-\frac{4}{3}\right\}\mathrm{given}\mathrm{by}\\ \left(\mathrm{A}\right)\mathrm{ }\mathrm{g}\left(\mathrm{y}\right)=\frac{3\mathrm{y}}{3-4\mathrm{y}}\left(\mathrm{B}\right)\mathrm{ }\mathrm{g}\left(\mathrm{y}\right)=\frac{4\mathrm{y}}{4-3\mathrm{y}}\\ \left(\mathrm{C}\right)\mathrm{ }\mathrm{g}\left(\mathrm{y}\right)=\frac{4\mathrm{y}}{3-4\mathrm{y}}\left(\mathrm{D}\right)\mathrm{g}\left(\mathrm{y}\right)=\frac{3\mathrm{y}}{4-3\mathrm{y}}\end{array}$

Ans

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{ }\mathrm{f}:\mathrm{R}-\left\{-\frac{4}{3}\right\}\to \mathrm{R}\mathrm{ }\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)=\frac{4\mathrm{x}}{3\mathrm{x}+4}.\\ \mathrm{Let}\mathrm{ }\mathrm{y}\mathrm{be}\mathrm{an}\mathrm{arbitrary}\mathrm{element}\mathrm{of}\mathrm{Range}\mathrm{f}\mathrm{.}\\ \mathrm{Then},\mathrm{there}\mathrm{exists}\mathrm{x}\in \mathrm{R}-\left\{-\frac{4}{3}\right\}\mathrm{ }\mathrm{such}\mathrm{ }\mathrm{that}\mathrm{ }\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right).\\ ⇒ \mathrm{ }\mathrm{y}=\frac{4\mathrm{x}}{3\mathrm{x}+4}\\ ⇒3\mathrm{xy}+4\mathrm{y}=4\mathrm{x}\\ ⇒-3\mathrm{xy}+4\mathrm{x}=4\mathrm{y}\\ ⇒\mathrm{ }\mathrm{x}\left(4-3\mathrm{y}\right)=4\mathrm{y}\\ ⇒ \mathrm{ }\mathrm{x}=\frac{4\mathrm{y}}{4-3\mathrm{y}}\\ \mathrm{Let}\mathrm{us}\mathrm{define}\mathrm{g}:\mathrm{Range} \mathrm{f}\to \mathrm{R}-\left\{-\frac{4}{3}\right\} \mathrm{as}\mathrm{ }\mathrm{g}\left(\mathrm{y}\right)=\frac{4\mathrm{y}}{4-3\mathrm{y}}\mathrm{ }\\ \mathrm{Now},\mathrm{ }\left(\mathrm{gof}\right)\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\\ \\ \mathrm{ }=\mathrm{g}\left(\frac{4\mathrm{x}}{3\mathrm{x}+4}\right)\\ \mathrm{ }=\frac{4\left(\frac{4\mathrm{x}}{3\mathrm{x}+4}\right)}{4-3\left(\frac{4\mathrm{x}}{3\mathrm{x}+4}\right)}\\ \mathrm{ }=\frac{16\mathrm{x}}{12\mathrm{x}+16-12\mathrm{x}}\\ \mathrm{ }=\mathrm{x}\\ \mathrm{And},\mathrm{ }\left(\mathrm{fog}\right)\left(\mathrm{y}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{y}\right)\right)\\ \mathrm{ }=\mathrm{f}\left(\frac{4\mathrm{y}}{4-3\mathrm{y}}\right)\\ \mathrm{ }=\frac{4\left(\frac{4\mathrm{y}}{4-3\mathrm{y}}\right)}{3\left(\frac{4\mathrm{y}}{4-3\mathrm{y}}\right)+4}\\ \mathrm{ }=\frac{16\mathrm{y}}{12\mathrm{y}+16-12\mathrm{y}}\\ \mathrm{ }=\frac{16\mathrm{y}}{16}\\ \mathrm{ }=\mathrm{y}\\ \therefore \mathrm{gof}={\mathrm{I}}_{\mathrm{R}-\left\{-\frac{4}{3}\right\}} \mathrm{and} \mathrm{fog}={\mathrm{I}}_{\mathrm{Range}\mathrm{ }\mathrm{f}}\\ \mathrm{Thus},\mathrm{g}\mathrm{is}\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{i}.\mathrm{e}.,{\mathrm{f}}^{–1}=\mathrm{g}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{is}\mathrm{the}\mathrm{map}\mathrm{g}:\mathrm{Range}, \mathrm{R}\to \mathrm{R}-\left\{-\frac{4}{3}\right\},\\ \mathrm{which}\mathrm{is}\mathrm{given}\mathrm{by} \mathrm{g}\left(\mathrm{y}\right)=\frac{4\mathrm{y}}{4-3\mathrm{y}}.\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{B}\mathrm{.}\end{array}$