Ncert Solutions class 12 maths chapter 1 exercise 1.2

Q.1

$\begin{array}{l} Show that the function f : R_{*} \to R_{*} definedby f (x) = \frac{1}{x} is \\ one - one and onto, where R_{*} is these t of all non - zero \\ realnumbers . Is there sulttrue, if the domain R_{*} is \\ replaced by Nwithco - domain being same as R_{*} ? \end{array}$

Ans

$\begin{array}{l} Let x, y \in R_{*} such f (x) = f (y) \\ one - one : \\ f (x) = f (y) \\ \Rightarrow \frac{1}{x} = \frac{1}{y} \\ \Rightarrow x = y \\ So, f : R_{*} \to R_{*} is one - one . \\ Onto : \end{array}$

Let f( x )=y where y be any element of R * ( Co−domain )

$\begin{array}{l} \Rightarrow \frac{1}{x} = y or x = \frac{1}{y} \\ Now, f (\frac{1}{y}) = \frac{1}{(\frac{1}{y})} = y \\ Since Range = Co - domain \\ ∴ f is onto . \\ Thus, the given function (f) is one - one and onto . \\ Consider function f : N \to R_{*} defined by \\ f (x) = \frac{1}{x} \\ For any x, y \in N we see that \\ f (x) = f (y) \\ \Rightarrow \frac{1}{x} = \frac{1}{y} \\ \Rightarrow x = y \\ So f : N \to R_{*} is one - one function \\ Since fractional numbers like \frac{2}{3}, \frac{2}{5} etc . in co - domain R_{*} have \\ R_{*} have no pre image in domain N . \\ So, f : N \to R_{*} is not onto . \end{array}$

Q.2 (i) f: N→ N given by f(x)= x²

$\begin{array}{l} (i i) f : Z \to Z g i v e n b y f (x) = x^{2} \\ (i i i) f : R \to R g i v e n b y f (x) = x^{2} \\ (i v) f : N \to N g i v e n b y f (x) = x^{3} \\ (v) f : Z \to Z g i v e n b y f (x) = x^{3} \end{array}$

Ans

$\begin{array}{l} (i) f:N \to N is given by, f (x) = x^{2} \\ To check for injectivity let f (x) = f (y) \\ \Rightarrow x^{2} = y^{2} \\ \Rightarrow x = y (\begin{array}{l} T h e r e are no negative \\ natural numbers . \end{array}) \\ ∴ f (x) is injective . \\ To check for onto : \\ Let f (x) =y \\ \Rightarrow x^{2} =y \\ \Rightarrow x= \sqrt{y} \\ ∴ f (\sqrt{y}) =y \\ Since range is not equal to co-domain . \\ f(x) is not onto or surjective \\ (ii) f: Z \to Z is given by, \\ f(x) = x^{2} \\ For injectivity let f(x) = f(y) \\ \Rightarrow x^{2} = y^{2} \\ \Rightarrow x = \pm y \\ ∴ f is not injective . \\ For onto: \\ Let (x) = y \\ \Rightarrow x^{2} = y \\ \Rightarrow x = \sqrt{y} \\ ∴ f (\sqrt{y}) = y \\ Hence, function f is neither injective nor surjective . \\ (i i i) f : R \to R is given by, f (x) = x^{2} \\ F o r injectivity: \\ f (x) = f (y) \\ \Rightarrow x^{2} = y^{2} \\ \Rightarrow x = \pm y (\begin{array}{l} x can not take more than one \\ value for injectivity . \end{array}) \\ ∴ f is not injective . \\ For onto: \\ Let f (x) = y \Rightarrow x^{2} = y \\ \Rightarrow x = \sqrt{y} \\ ∴ f (\sqrt{y}) = y \\ Since range is not equal to co-domain . \\ ∴ f is not surjective . \\ (i v) f : N \to N is given by, f (x) = x^{3} \\ F o r injectivity: \\ f (x) = f (y) \\ \Rightarrow x^{3} = y^{3} \\ \Rightarrow x = y \\ ∴ f is injective . \\ For onto: \\ Let f (x) = y \Rightarrow x^{3} = y \\ \Rightarrow x = \sqrt[3]{y} \\ ∴ f (\sqrt[3]{y}) = y (y \in N b u t \sqrt[3]{y} \notin N) \\ Since range is not equal to co-domain . \\ ∴ f is not surjective . \\ Thus, given function is injective but not surjective . \\ (v) f : Z \to Z is given by, f (x) = x^{3} \\ F o r injectivity: \\ f (x) = f (y) \\ \Rightarrow x^{3} = y^{3} \\ \Rightarrow x = y \\ ∴ f is injective . \\ For onto: \\ Let f (x) = y \Rightarrow x^{3} = y \\ \Rightarrow x = \sqrt[3]{y} \\ ∴ f (\sqrt[3]{y}) = y (y \in Z b u t \sqrt[3]{y} \notin Z) \\ Since range is not equal to co-domain . \\ ∴ f is not surjective . \\ Thus, given function is injective but not surjective . \end{array}$

Q.3 Prove that the Greatest Integer Function f: R → R given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

Ans

$\begin{array}{l} S i n c e, \\ f : R \to R is given by, \\ f (x) = [x] \\ W e â€‹ see that f (0.1) = 0, f (0.3) = 0. \\ ∴ f (0.1) = f (0.3), b u t 0 .1 \neq 0 .3 \\ So, f is not injective i .e ., one-one . \\ Now, let 0 .6 \in R . \\ It is given that f (x) = [x] is always an integer . Thus, there does \\ not exist any element x \in R such that f (x) =0 .6 . \\ Then, f is not onto . \\ Therefore, the greatest integer function is neither one-one \\ nor onto . \end{array}$

Q.4

$\begin{array}{l} Show that the Modul us Function f : R \to R given by f (x) = | x |, \\ is neither one - onen or onto, where | x | is x, if x is positive \\ or 0 and | x | is - x, if x is negative . \end{array}$

Ans

$\begin{array}{l} f : R \to R is given by, \\ Modulus function is defined by \\ f (x) = {\begin{cases} x, if x \geq 0 \\ - x, if x <0 \end{cases} \\ Here, f (- 1) = | - 1 | = 1, f (1) = | 1 | = 1, \\ ∴ f (- 1) = f (1), but - 1 \neq 1 \\ So, f is not one - one . \\ Let - 1 \in R . \\ Since, f (x) = | x | is always positive . So, there does not exist \\ any element in x in domain R such that f (x) = - 1 . \\ ∴ f (x) â€‹ is not onto . \\ Therefore, the modulus function is neither one - one nor onto . \end{array}$

Q.5

$\begin{array}{l} Show that the Signum Function f : R \to R, given by \\ f (x) = {\begin{cases} 1, if x > 0 \\ 0, if x = 0 \\ - 1, if x < 0 \end{cases} \\ is neither one - one noronto . \end{array}$

Ans

$\begin{array}{l} f : R \to R is given by, \\ Modulus function is defined by \\ f (x) = {\begin{cases} 1, if x >0 \\ 0, if x =0 \\ - 1, if x <0 \end{cases} \\ Here, f (1) = 1, f (3) = 1, but 1 \neq 3 \\ So, f is not one - one . \\ Since, range of f is (1, 0, - 1) for element - 2 in co - domain R, \\ there does not exist any value in domain R such that \\ f (x) = - 2 . \\ ∴ f is not onto . \\ Hence, the signum function is neither one - one nor onto . \end{array}$

Q.6 Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.

Ans

$\begin{array}{l} It is given that \\ A = {1, 2, 3}, B = {4, 5, 6, 7} \\ f : A \to B is defined as f = {(1, 4), (2, 5), (3, 6)} . \\ ∴ f (1) = 4, f (2) = 5, f (3) = 6 \\ Since every element has a unique value . \\ The given function f is one-one . \end{array}$

Q.7 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaGqabiaa=LeacaWFUbGaa8hiaiaa=vgacaWFHbGaa83yaiaa=HgacaWFGaGaa83Baiaa=zgacaWFGaGaa8hDaiaa=HgacaWFLbGaa8hiaiaa=zgacaWFVbGaa8hBaiaa=XgacaWFVbGaa83Daiaa=LgacaWFUbGaa83zaiaa=bcacaWFJbGaa8xyaiaa=nhacaWFLbGaa83Caiaa=XcacaWFGaGaa83Caiaa=rhacaWFHbGaa8hDaiaa=vgacaWFGaGaa83Daiaa=HgacaWFLbGaa8hDaiaa=HgacaWFLbGaa8NCaiaa=bcacaWF0bGaa8hAaiaa=vgacaWFGaGaa8Nzaiaa=vhacaWFUbGaa83yaiaa=rhacaWFPbGaa83Baiaa=5gacaWFGaGaa8xAaiaa=nhacaWFGaaabaGaa83Baiaa=5gacaWFLbGaa8xlaiaa=9gacaWFUbGaa8xzaiaa=XcacaWFGaGaa83Baiaa=5gacaWF0bGaa83Baiaa=bcacaWFVbGaa8NCaiaa=bcacaWFIbGaa8xAaiaa=PgacaWFLbGaa83yaiaa=rhacaWFPbGaa8NDaiaa=vgacaWFUaGaa8hiaiaa=PeacaWF1bGaa83Caiaa=rhacaWFPbGaa8Nzaiaa=LhacaWFGaGaa8xEaiaa=9gacaWF1bGaa8NCaiaa=bcacaWFHbGaa8NBaiaa=nhacaWF3bGaa8xzaiaa=jhacaWFUaaabaGaa8hiaiaa=HcacaWFPbGaa8xkaiaa=bcacaWFMbGaa8Noaiaa=bcacaWFsbGaeyOKH4Qaa8Nuaiaa=bcacaWFKbGaa8xzaiaa=zgacaWFPbGaa8NBaiaa=vgacaWFKbGaa8hiaiaa=jgacaWF5bGaa8hiaiaa=zgacaWFOaGaa8hEaiaa=LcacaWFGaGaa8xpaiaa=bcacaWFZaGaa8xlaiaa=rdacaWF4baabaGaa8hkaiaa=LgacaWFPbGaa8xkaiaa=bcacaWFMbGaa8Noaiaa=bcacaWFsbGaeyOKH4Qaa8Nuaiaa=bcacaWFKbGaa8xzaiaa=zgacaWFPbGaa8NBaiaa=vgacaWFKbGaa8hiaiaa=jgacaWF5bGaa8hiaiaa=zgacaWFOaGaa8hEaiaa=LcacaWFGaGaa8xpaiaa=bcacaWFXaGaa8hiaiaa=TcacaWFGaGaa8hEamaaCaaaleqabaGaa8Nmaaaaaaaa@D15C@

Ans

$\begin{array}{l} (i) f: R \to R is defined as f(x) = 3 - 4x . \\ For one-one \\ Let f(x) = f(y) \\ \Rightarrow 3 - 4x = 3 - 4y \\ \Rightarrow x = y \\ ∴ f is one-one . \\ for onto \\ Let f(x)=y \\ \Rightarrow 3 - 4x=y \\ \Rightarrow x= \frac{3 - y}{4} \\ \Rightarrow f (\frac{3 - y}{4}) =3 - 4 (\frac{3 - y}{4}) =y \\ Since Range = Co-domain \\ ∴ f is onto . \\ Hence, f is bijective . \\ (ii) f: R \to {R is defined as f(x)=1+x}^{2} \\ For one-one \\ Let f(x)=f(y ) \\ \Rightarrow {1+x}^{2} = {1+y}^{2} \\ \Rightarrow x^{2} = y^{2} \\ \Rightarrow x = \pm y \\ ∴ f is not one-one . \\ For onto: \\ Let f (x) = y \\ \Rightarrow {1+x}^{2} = y \\ \Rightarrow x = \pm \sqrt{y - 1} \\ Since Range is not equal to Co-domain . \\ Then given function f(x) is not onto . \end{array}$

Q.8

$\begin{array}{l} Let A and B be sets . Show that f : A \times B \to B \times A such that \\ f (a, b) = f (b, a) is bijective function . \end{array}$

Ans

$\begin{array}{l} f : A \times B \to B \times A is defined as f (a, b) = (b, a) . \\ Let (a_{1}, b_{1}), (a_{2}, b_{2}) \in A \times B such that f (a_{1}, b_{1}) = f (a_{2}, b_{2}) \\ \Rightarrow (b_{1}, a_{1}) = (b_{2}, a_{2}) \\ \Rightarrow b_{1} = b_{2} and a_{1} = a_{2} \\ \Rightarrow (a_{1}, b_{1}) = (a_{2}, b_{2}) \\ ∴ f is one - one . \\ Now, let (b, a) \in B \times A be any element . \\ Then, there exists (a, b) \in A \times B such that f (a, b) = (b, a) . \\ [By definition of f] \\ ∴ f is onto . \\ Hence, f is bijective . \end{array}$

Q.9 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaGqabiaa=XeacaWFLbGaa8hDaiaa=bcacaWFMbGaa8Noaiaa=bcacaWFobGaeyOKH4Qaa8Ntaiaa=bcacaWFIbGaa8xzaiaa=bcacaWFKbGaa8xzaiaa=zgacaWFPbGaa8NBaiaa=vgacaWFKbGaa8hiaiaa=jgacaWF5baabaGaa8NzamaabmaabaGaa8NBaaGaayjkaiaawMcaaiaa=1dadaGabaabaeqabaWaaSaaaeaacaWFUbGaa83kaiaa=fdaaeaacaWFYaaaaiaa=XcacaWLjaGaa8xAaiaa=zgacaWFGaGaa8NBaiaa=bcacaWFPbGaa83Caiaa=bcacaWFVbGaa8hzaiaa=rgaaeaadaWcaaqaaiaa=5gaaeaacaWFYaaaaiaa=XcacaWLjaGaaCzcaiaa=LgacaWFMbGaa8hiaiaa=5gacaWFGaGaa8xAaiaa=nhacaWFGaGaa8xzaiaa=zhacaWFLbGaa8NBaaaacaGL7baacaaMc8UaaGPaVlaaykW7caaMc8Uaa8Nzaiaa=9gacaWFYbGaa8hiaiaa=fgacaWFSbGaa8hBaiaa=bcacaWFUbGaeyicI4Saa8Ntaiaa=5caaeaacaWFtbGaa8hDaiaa=fgacaWF0bGaa8xzaiaa=bcacaWF3bGaa8hAaiaa=vgacaWF0bGaa8hAaiaa=vgacaWFYbGaa8hiaiaa=rhacaWFObGaa8xzaiaa=bcacaWFMbGaa8xDaiaa=5gacaWFJbGaa8hDaiaa=LgacaWFVbGaa8NBaiaa=bcacaWFMbGaa8hiaiaa=LgacaWFZbGaa8hiaiaa=jgacaWFPbGaa8NAaiaa=vgacaWFJbGaa8hDaiaa=LgacaWF2bGaa8xzaiaa=5cacaWFGaaabaGaa8Nsaiaa=vhacaWFZbGaa8hDaiaa=LgacaWFMbGaa8xEaiaa=bcacaWF5bGaa83Baiaa=vhacaWFYbGaa8hiaiaa=fgacaWFUbGaa83Caiaa=DhacaWFLbGaa8NCaiaa=5caaaaa@B686@

Ans

$\begin{array}{l} L e t f : N \to N i s d e f i n e d a s \\ f (n) = {\begin{cases} \frac{n + 1}{2}, i f n i s o d d \\ \frac{n}{2}, i f n i s e v e n \end{cases} f o r a l l n \in N . \\ It can be observed that: \\ f (1) = \frac{1 + 1}{2} = 1, f (2) = \frac{2}{2} = 1 [B y definition of f] \\ ∴ f (1) = f (2), w h e r e 1 \neq 2. \\ ∴ f is not one-one . \\ Consider a natural number (n) in co-domain N . \\ Case I: n is odd \\ ∴ n = 2m + 1 for some m \in N . Then,there exists 4m + 1 \in N \\ such that \\ f (4 m + 1) = \frac{4 m + 1 + 1}{2} = 2 m + 1 \\ C a s e I I : n is even \\ ∴ n=2 m for some m \in N . Then,there exists 4 m \in N such that \\ f (4 m) = \frac{4 m}{2} = 2 m \\ ∴ f is onto . \\ Hence, f is not a bijective function . \end{array}$

Q.10 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaGqabiaa=XeacaWFLbGaa8hDaiaa=bcacaWFbbGaa8hiaiaa=1dacaWFGaGaa8NuaiabgkHiTiaa=ThacaWFZaGaa8xFaiaa=bcacaWFHbGaa8NBaiaa=rgacaWFGaGaa8Nqaiaa=bcacaWF9aGaa8hiaiaa=jfacqGHsislcaWF7bGaa8xmaiaa=1hacaWFUaGaa8hiaiaa=neacaWFVbGaa8NBaiaa=nhacaWFPbGaa8hzaiaa=vgacaWFYbGaa8hiaiaa=rhacaWFObGaa8xzaiaa=bcacaWFMbGaa8xDaiaa=5gacaWFJbGaa8hDaiaa=LgacaWFVbGaa8NBaiaa=bcaaeaacaWFMbGaa8Noaiaa=bcacaWFbbGaeyOKH4Qaa8Nqaiaa=bcacaWFKbGaa8xzaiaa=zgacaWFPbGaa8NBaiaa=vgacaWFKbGaa8hiaiaa=jgacaWF5bGaaGPaVlaa=zgadaqadaqaaiaa=HhaaiaawIcacaGLPaaacaWF9aWaaSaaaeaacaWF4bGaa8xlaiaa=jdaaeaacaWF4bGaa8xlaiaa=ndaaaGaa8Nlaiaa=bcacaWFjbGaa83Caiaa=bcacaWFMbGaa8hiaiaa=9gacaWFUbGaa8xzaiabgkHiTiaa=9gacaWFUbGaa8xzaiaa=bcacaWFHbGaa8NBaiaa=rgacaWFGaGaa83Baiaa=5gacaWF0bGaa83Baiaa=9dacaWFGaaabaGaa8Nsaiaa=vhacaWFZbGaa8hDaiaa=LgacaWFMbGaa8xEaiaa=bcacaWF5bGaa83Baiaa=vhacaWFYbGaa8hiaiaa=fgacaWFUbGaa83Caiaa=DhacaWFLbGaa8NCaiaa=5caaaaa@A35D@

Ans

$\begin{array}{l} A = R - {3} a n d B = R - {1} \\ f: A \to B is defined as f (x) = \frac{x - 2}{x - 3} \\ L e t x,y \in A such that \\ f (x) = f (y) . \\ \Rightarrow \frac{x - 2}{x - 3} = \frac{y - 2}{y - 3} \\ \Rightarrow (x - 2) (y - 3) = (x - 3) (y - 2) \\ \Rightarrow x y - 3 x - 2 y + 6 = x y - 2 x - 3 y + 6 \\ \Rightarrow - 3 x - 2 y = - 2 x - 3 y \\ \Rightarrow x = y \\ ∴ f if one-one . \\ y \in B = R- {1} . T h e n, y \neq 1 \\ The function f is onto if there exists x \in A such that \\ f(x) = y \\ \Rightarrow \frac{x - 2}{x - 3} = y \\ \Rightarrow x - 2 = y (x - 3) \\ \Rightarrow x - 2 = y x - 3 y \\ \Rightarrow x - y x = 2 - 3 y \\ \Rightarrow x (1 - y) = 2 - 3 y \\ \Rightarrow x = \frac{2 - 3 y}{(1 - y)} \in A [y \neq 1] \\ Thus, for any y \in B, there exists \frac{2 - 3 y}{(1 - y)} \in A s u c h t h a t \\ f (\frac{2 - 3 y}{1 - y}) = \frac{\frac{2 - 3 y}{(1 - y)} - 2}{\frac{2 - 3 y}{(1 - y)} - 3} \\ = \frac{\frac{2 - 3 y - 2 (1 - y)}{1 - y}}{\frac{2 - 3 y - 3 (1 - y)}{1 - y}} \\ = \frac{2 - 3 y - 2 + 2 y}{2 - 3 y - 3 + 3 y} \\ = \frac{- y}{- 1} \\ f (\frac{2 - 3 y}{1 - y}) = y \\ ∴ f is onto . \\ Hence, function f is one-one and onto . \end{array}$

Q.11

$\begin{array}{l} Letf : R \to R be defined as f (x) = x^{4} . Choose the correct \\ answer . \\ (A) f is one - one onto (B) fismany - one onto . \\ (C) f is one - one but not onto (D) f is n either one - one \\ noronto . \end{array}$

Ans

$\begin{array}{l} f : R \to R is defined as f (x) = x^{4} . \\ Let x, y \in R such that f (x) = f (y) . \\ Then, f (x) = f (y) \\ \Rightarrow x^{4} = y^{4} \\ \Rightarrow x = \pm y \\ ∴ f (x) = f (y) but x \neq y \\ So, f is not one - one . \\ Consider an element 3 in co - domain R . It is clear that there \\ does not exist any x in domain R such that f (x) = 3 . \\ ∴ f is not onto . \\ Hence, function f is neither one - one nor onto . \\ The correct answer is D . \end{array}$

Q.12

$\begin{array}{l} Let f : R \to R be defined asf (x) = 3 x . Choose the correct \\ answer . \\ (A) f isone - one onto (B) f is many - one on to \\ (C) f isone - one but not onto (D) f is neither one - one nor onto . \end{array}$

Ans

$\begin{array}{l} f : R \to Rbedefinedasf (x) = 3 x . \\ Let x, y \in R such that f (x) = f (y) . \\ \Rightarrow 3 x = 3 y \\ \Rightarrow x = y \\ ∴ f is one - one . \\ Also, for any real number (y) in co - domain R, there exists \\ \frac{y}{3} in R such that f (\frac{y}{3}) = 3 (\frac{y}{3}) = y \\ ∴ f is onto . \\ Therefore, function f is one - one and onto . \\ The correct answer is A . \end{array}$

Ncert Solutions class 12 maths chapter 1 exercise 1.2

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