Ncert Solutions class 12 maths chapter 1 exercise 1.3

Q.1

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

Ans

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

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

$\begin{array}{l} Let ((f + g) oh) (x) = (foh) (x) + (goh) (x) \\ = f {h (x)} + g {h (x)} \\ = (foh) (x) + (goh) (x) \\ = {(f + g) oh} (x) \\ = {(foh) + (goh)} (x) \forall x \in R \\ H e n c e, (f + g) o h = (foh) + (goh) . \\ N o w, \\ {(f . g) o h} (x) = (f . g) (h (x)) \\ = f (h (x)) . g (h (x)) \\ = (f o h) (x) . (g o h) (x) \\ = {(f o h) . (g o h)} (x) \\ ∴ {(f . g) o h} (x) = {(f o h) . (g o h)} (x) \forall x \in R \\ H e n c e, (f . g) o h = (f o h) . (g o h) \end{array}$

Q.3 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaGqabiaa=zeacaWFPbGaa8NBaiaa=rgacaWFGaGaa83zaiaa=9gacaWFMbGaa8hiaiaa=fgacaWFUbGaa8hzaiaa=bcacaWFMbGaa83Baiaa=DgacaWFSaGaa8hiaiaa=LgacaWFMbaabaGaa8hkaiaa=LgacaWFPaGaaGPaVlaaykW7caWFMbWaaeWaaeaacaWF4baacaGLOaGaayzkaaGaa8xpamaaemaabaGaa8hEaaGaay5bSlaawIa7aiaaykW7caWFHbGaa8NBaiaa=rgacaaMc8Uaa83zamaabmaabaGaa8hEaaGaayjkaiaawMcaaiaa=1dadaabdaqaaiaa=vdacaWF4bGaa8xlaiaa=jdaaiaawEa7caGLiWoaaeaadaqadaqaaiaa=LgacaWFPbaacaGLOaGaayzkaaGaaGPaVlaa=zgadaqadaqaaiaa=HhaaiaawIcacaGLPaaacaWF9aGaa8hoaiaa=HhadaahaaWcbeqaaiaa=ndaaaGccaaMc8UaaGPaVlaa=fgacaWFUbGaa8hzaiaaykW7caaMc8Uaa83zamaabmaabaGaa8hEaaGaayjkaiaawMcaaiaa=1dacaWF4bWaaWbaaSqabeaadaWcaaqaaiaa=fdaaeaacaWFZaaaaaaaaaaa@82A6@

Ans

$\begin{array}{l} (i) f (x) = | x | a n d g (x) = | 5 x - 2 | \\ ∴ (g o f) (x) = g (f (x)) \\ = g (| x |) \\ = | 5 | x | - 2 | \\ (f o g) (x) = f (g (x)) \\ = f (| 5 x - 2 |) \\ = | | 5 x - 2 | | \\ = | 5 x - 2 | \\ (i i) f (x) = 8 x^{3} a n d g (x) = x^{\frac{1}{3}} \\ ∴ (g o f) (x) = g (f (x)) \\ = g (8 x^{3}) \\ = {(8 x^{3})}^{\frac{1}{3}} \\ = 2 x \\ (f o g) (x) = f (g (x)) \\ = f (x^{\frac{1}{3}}) \\ = 8 {(x^{\frac{1}{3}})}^{3} \\ = 8 x \end{array}$

Q.4

$\begin{array}{l} I f f (x) = \frac{4 x + 3}{6 x - 4}, x \neq \frac{2}{3}, s h o w t h a t f o f (x) = x, f o r a l l x \neq \frac{2}{3} . \\ W h a t i s t h e i n v e r s e o f f ? \end{array}$

Ans

$\begin{array}{l} I t is given that f (x) = \frac{4 x + 3}{6 x - 4}, t h e n \\ f o f (x) = f (f (x)) \\ = f (\frac{4 x + 3}{6 x - 4}) \\ = \frac{4 (\frac{4 x + 3}{6 x - 4}) + 3}{6 (\frac{4 x + 3}{6 x - 4}) - 4} \\ = \frac{(\frac{16 x + 12 + 18 x - 12}{6 x - 4})}{(\frac{24 x + 18 - 24 x + 16}{6 x - 4})} \\ = \frac{34 x}{34} \\ f o f (x) = x, for all x \neq \frac{2}{3} . \\ \Rightarrow f o f = I \\ Hence, the given function f is invertible and the inverse of f \\ is f itself . \end{array}$

Q.5

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

Ans

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

Q.6

$\begin{array}{l} S h o w t h a t f : [- 1, 1] \to R, g i v e n b y f (x) = \frac{x}{x + 2} i s o n e - o n e . \\ F i n d t h e i n v e r s e o f t h e f u n c t i o n f : [- 1, 1] \to R a n g e f . \end{array}$

Ans

$\begin{array}{l} f: [- 1, 1] \to R is given as f (x) = \frac{x}{x + 2} \\ L e t f (x) = f (y) \\ \Rightarrow \frac{x}{x + 2} = \frac{y}{y + 2} \\ \Rightarrow x (y + 2) = y (x + 2) \\ \Rightarrow x y + 2 x = x y + 2 y \\ \Rightarrow x = y \\ ∴ f is one-one function . \\ It is clear that f: [- 1, 1] \to Range f is onto . \\ ∴ f: [- 1, 1] \to Range f is one-one and onto and therefore, \\ the inverse of the function: \\ f: [- 1, 1] \to Range f exists . \\ Let g: Range f \to [- 1, 1] be the inverse of f . \\ Let y be an arbitrary element of range f . \\ Since f: [- 1, 1] \to Range f is onto, we have: \\ f (x) = y f o r some x \in [- 1, 1] \\ \Rightarrow y = \frac{x}{x + 2} \\ \Rightarrow x y + 2 y = x \\ \Rightarrow 2 y = x - x y \\ = x (1 - y) \\ \Rightarrow x = \frac{2 y}{1 - y}, y \neq 1 \\ Now, let us define g: Range f \to [- 1, 1] as \\ g (y) = \frac{2 y}{1 - y}, y \neq 1. \\ N o w, \\ (g o f) (x) = g (f (x)) \\ = g (\frac{x}{x + 2}) \\ = \frac{2 (\frac{x}{x + 2})}{1 - (\frac{x}{x + 2})} \\ = \frac{2 x}{x + 2 - x} \\ = \frac{2 x}{2} \\ (g o f) (x) = x \\ (f o g) (y) = f (g (y)) \\ = f (\frac{2 y}{1 - y}) \\ = \frac{(\frac{2 y}{1 - y})}{\frac{2 y}{1 - y} + 2} \\ = \frac{2 y}{2 y + 2 - 2 y} \\ = \frac{2 y}{2} \\ (f o g) (y) = y \\ ∴ g o f^{- 1} = I_{[- 1, 1]} a n d f o g^{- 1} = I_{R a n g e f} \\ ∴ f^{- 1} = g \\ \Rightarrow f^{- 1} (y) = \frac{2 y}{y - 1}, y \neq 1. \\ \Rightarrow f^{- 1} (x) = \frac{2 x}{x - 1}, x \neq 1. \end{array}$

Q.7

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

Ans

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

Q.8

$\begin{array}{l} C o n s i d e r f : R_{+} \to [4, \infty) g i v e n b y f (x) = x^{2} + 4 . S h o w t h a t f i s \\ i n v e r t i b l e w i t h t h e i n v e r s e f^{- 1} o f g i v e n f b y, w h e r e R^{+} i s t h e s e t \\ o f a l l n o n - n e g a t i v e r e a l n u m b e r s . \end{array}$

Ans

$\begin{array}{l} {f: R}_{+} \to [4, \infty {) is given as f(x) = x}^{2} + 4 . \\ One-one: \\ Let f(x) = f(y) \\ \Rightarrow x^{2} + 4 = y^{2} + 4 \\ \Rightarrow x^{2} = y^{2} \\ \Rightarrow x = y [a s x = y \in R_{+}] \\ ∴ f i s o n e - o n e function . \\ O n t o : \\ For y \in [4, \infty), \\ let y = x^{2} + 4 \\ x = \sqrt{y - 4} \geq 0 \\ Therefore, for any y \in R, there exists x = \sqrt{y - 4} \geq 0, such that \\ f (x) = f (\sqrt{y - 4}) \\ = {(\sqrt{y - 4})}^{2} + 4 \\ = y - 4 + 4 \\ f (x) = y \\ ∴ f is onto . \\ {Thus, f is one-one and onto and therefore, f}^{-1} exists . \\ Let us define g: [4, \infty) \to R_{^{+}} by, \\ g (y) = \sqrt{y - 4} \\ N o w, (g o f) (x) = g (f (x)) \\ = g (x^{2} + 4) \\ = \sqrt{(x^{2} + 4) - 4} \\ (g o f) (x) = x \\ (f o g) (y) = f (g (y)) \\ = f (\sqrt{y - 4}) \\ = {(\sqrt{y - 4})}^{2} + 4 \\ = y - 4 + 4 \\ (f o g) (y) = y \\ ∴ g o f = f o g = I_{R^{+}} . \\ Hence, f is invertible and the inverse of f is given by \\ f^{- 1} (y) = g (y) = \sqrt{y - 4} \\ \Rightarrow f^{- 1} (x) = \sqrt{x - 4} \end{array}$

Q.9 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaGqabiaa=neacaWFVbGaa8NBaiaa=nhacaWFPbGaa8hzaiaa=vgacaWFYbGaa8hiaiaa=zgacaWF6aGaa8hiaiaa=jfadaWgaaWcbaGaa83kaaqabaGccqGHsgIRcaWFGaGaa83waiaa=1cacaWF1aGaa8hlaiaa=bcacqGHEisPcaWFPaGaa8hiaiaa=DgacaWFPbGaa8NDaiaa=vgacaWFUbGaa8hiaiaa=jgacaWF5bGaa8hiaiaa=zgacaWFOaGaa8hEaiaa=LcacaWFGaGaeyypa0Jaa8hiaiaa=LdacaWF4bWaaWbaaSqabeaacaWFYaaaaOGaa8hiaiaa=TcacaWFGaGaa8Nnaiaa=HhacaWFTaGaa8xnaiaa=5cacaWFGaaabaGaa83uaiaa=HgacaWFVbGaa83Daiaa=bcacaWF0bGaa8hAaiaa=fgacaWF0bGaa8hiaiaa=zgacaWFGaGaa8xAaiaa=nhacaWFGaGaa8xAaiaa=5gacaWF2bGaa8xzaiaa=jhacaWF0bGaa8xAaiaa=jgacaWFSbGaa8xzaiaa=bcacaWF3bGaa8xAaiaa=rhacaWFObGaa8hiaiaa=zgadaahaaWcbeqaaiabgkHiTiaa=fdaaaGcdaqadaqaaiaa=LhaaiaawIcacaGLPaaacqGH9aqpdaqadaqaamaalaaabaWaaeWaaeaadaGcaaqaaiaa=LhacaWFRaGaa8NnaaWcbeaaaOGaayjkaiaawMcaaiabgkHiTiaa=fdaaeaacaWFZaaaaaGaayjkaiaawMcaaiaa=5caaaaa@8D83@

Ans

$\begin{array}{l} {f: R}_{+} \to [- 5, \infty {) is given as f(x) = 9x}^{2} + 6x - 5 . \\ Let y be an arbitrary element of [- 5, \infty) . \\ Let y = {9x}^{2} + 6x - 5 \\ \Rightarrow {9x}^{2} + 6x - (5+y) = 0 \\ \Rightarrow x = \frac{- 6 \pm \sqrt{6^{2} - 4 \times 9 \times - (5 + y)}}{2 \times 9} \\ = \frac{- 6 \pm \sqrt{36 + 36 (5 + y)}}{18} \\ = \frac{- 6 \pm 6 \sqrt{1 + 5 + y}}{18} \\ x = \frac{- 1 \pm \sqrt{6 + y}}{3} \\ = \frac{\sqrt{6 + y} - 1}{3} [âˆµ y \geq - 6 \Rightarrow y + 6 \geq 0] \\ ∴ f is onto, thereby range f = [- 5, \infty) . \\ Let us define g: [- 5, \infty) \to R_{+} as \\ g (y) = \frac{\sqrt{(y + 6)} - 1}{3} \\ N o w, (g o f) (x) = g (f (x)) \\ = g ({9x}^{2} + 6x - 5) \\ = \frac{\sqrt{({9x}^{2} + 6x - 5) + 6} - 1}{3} \\ = \frac{\sqrt{({9x}^{2} + 6x + 1)} - 1}{3} \\ = \frac{\sqrt{{(3 x + 1)}^{2}} - 1}{3} \\ = \frac{3 x + 1 - 1}{3} = x \\ A n d, \\ (f o g) (y) = f (g (y)) \\ = f (\frac{\sqrt{(y + 6)} - 1}{3}) \\ = 9 {(\frac{\sqrt{(y + 6)} - 1}{3})}^{2} + 6 (\frac{\sqrt{(y + 6)} - 1}{3}) - 5 \\ = 9 (\frac{y + 6 - 2 \sqrt{y + 6} + 1}{9}) + 2 \sqrt{(y + 6)} - 2 - 5 \\ = y + 6 - 2 \sqrt{y + 6} + 1 + 2 \sqrt{(y + 6)} - 2 - 5 \\ (f o g) (y) = y \\ ∴ g o f = I_{R} a n d f o g = I_{[- 5, \infty]} \\ Hence, f is invertible and the inverse of f is given by \\ f^{- 1} (y) = g (y) = \frac{\sqrt{y + 6} - 1}{3} . \end{array}$

Q.10

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

Ans

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

Q.11

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

Ans

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

Q.12

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

Ans

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

Q.13

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

Ans

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

Q.14

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

Ans

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

Ncert Solutions class 12 maths chapter 1 exercise 1.3

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