Home > NCERT Solutions > Ncert Solutions class 12 maths chapter 1 exercise 1.2
Q.1
Show that the function f : R *→R* definedby f(x)=1x isone–one and onto, where R* is these t of all non–zerorealnumbers. Is there sulttrue, if the domain R * is replaced by Nwithco–domain being same as R * ?
Ans
Let x,y∈R* such f(x)=f(y)one−one:f(x)=f(y)⇒ 1x=1y⇒ x=ySo, f:R*→R* is one–one.Onto:
Let f( x )=y where y be any element of R * ( Co−domain )
⇒ 1x=y or x=1yNow, f(1y)=1(1y)=ySince Range = Co–domain∴f is onto.Thus, the given function (f) is one–one and onto.Consider function f: N→R* defined byf(x)=1xFor any x,y∈N we see thatf(x)=f(y)⇒ 1x=1y⇒ x=ySo f:N→R* is one–one functionSince fractional numbers like 23,25 etc. in co–domain R* haveR* have no pre image in domain N.So, f:N→R* is not onto.
Q.2 (i) f: N→ N given by f(x)= x2
(ii) f: Z →Z given by f(x)= x 2 (iii) f: R →R given by f(x)= x 2 (iv) f: N →N given by f(x)= x 3 (v) f: Z→Z given by f(x)= x 3 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaaykW7ieqacaWFOaGaa8xAaiaa=LgacaWFPaGaa8hiaiaa=zgacaWF6aGaa8hiaiaa=PfacaWFGaGaeyOKH4Qaa8Nwaiaa=bcacaaMc8UaaGPaVlaaykW7caWFNbGaa8xAaiaa=zhacaWFLbGaa8NBaiaa=bcacaWFIbGaa8xEaiaa=bcacaWFMbGaa8hkaiaa=HhacaWFPaGaa8xpaiaa=bcacaWF4bWaaWbaaSqabeaacaWFYaaaaaGcbaGaa8hkaiaa=LgacaWFPbGaa8xAaiaa=LcacaWFGaGaa8Nzaiaa=PdacaWFGaGaa8Nuaiaa=bcacqGHsgIRcaWFsbGaa8hiaiaaykW7caaMc8Uaa83zaiaa=LgacaWF2bGaa8xzaiaa=5gacaWFGaGaa8Nyaiaa=LhacaWFGaGaa8Nzaiaa=HcacaWF4bGaa8xkaiaa=1dacaWFGaGaa8hEamaaCaaaleqabaGaa8NmaaaaaOqaaiaa=HcacaWFPbGaa8NDaiaa=LcacaWFGaGaa8Nzaiaa=PdacaWFGaGaa8Ntaiaa=bcacqGHsgIRcaWFobGaa8hiaiaaykW7caaMc8Uaa83zaiaa=LgacaWF2bGaa8xzaiaa=5gacaWFGaGaa8Nyaiaa=LhacaWFGaGaa8Nzaiaa=HcacaWF4bGaa8xkaiaa=1dacaWFGaGaa8hEamaaCaaaleqabaGaa83maaaaaOqaaiaaykW7caWFOaGaa8NDaiaa=LcacaWFGaGaa8Nzaiaa=PdacaWFGaGaa8NwaiabgkziUkaa=PfacaWFGaGaa8hiaiaa=bcacaWFNbGaa8xAaiaa=zhacaWFLbGaa8NBaiaa=bcacaWFIbGaa8xEaiaa=bcacaWFMbGaa8hkaiaa=HhacaWFPaGaa8xpaiaa=bcacaWF4bWaaWbaaSqabeaacaWFZaaaaaaaaa@ABD5@
(i) f:N→N is given by, f( x )= x 2 To check for injectivity let f( x )=f( y ) ⇒ x 2 = y 2 ⇒ x=y ( There are no negative natural numbers. ) ∴f( x ) is injective. To check for onto: Let f( x )=y ⇒ x 2 =y ⇒ x= y ∴ f( y )=y Since range is not equal to co-domain. f(x) is not onto or surjective (ii) f: Z→Z is given by, f(x)= x 2 For injectivity let f(x)=f(y) ⇒ x 2 = y 2 ⇒ x=±y ∴f is not injective. For onto: Let (x)=y ⇒ x 2 =y ⇒ x= y ∴ f( y )=y Hence, function f is neither injective nor surjective. ( iii ) f:R→R is given by, f( x )= x 2 For injectivity: f( x )=f( y ) ⇒ x 2 = y 2 ⇒ x=±y ( x can not take more than one value for injectivity. ) ∴f is not injective. For onto: Let f( x )=y⇒ x 2 =y ⇒ x= y ∴ f( y )=y Since range is not equal to co-domain. ∴f is not surjective. ( iv ) f:N→N is given by, f( x )= x 3 For injectivity: f( x )=f( y ) ⇒ x 3 = y 3 ⇒ x=y ∴f is injective. For onto: Let f( x )=y⇒ x 3 =y ⇒ x= y 3 ∴ f( y 3 )=y ( y∈N but y 3 ∉N ) Since range is not equal to co-domain. ∴f is not surjective. Thus, given function is injective but not surjective. ( v ) f:Z→Z is given by, f( x )= x 3 For injectivity: f( x )=f( y ) ⇒ x 3 = y 3 ⇒ x=y ∴f is injective. For onto: Let f( x )=y⇒ x 3 =y ⇒ x= y 3 ∴ f( y 3 )=y ( y∈Z but y 3 ∉Z ) Since range is not equal to co-domain. ∴f is not surjective. Thus, given function is injective but not surjective. 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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.
Since, f:R→R is given by, f( x )=[ x ] We​ see that f( 0.1 )=0,f( 0.3 )=0. ∴f( 0.1 )=f( 0.3 ), but 0.1≠0.3 So, f is not injective i.e., one-one. Now, let 0.6∈R. It is given that f( x )=[ x ] is always an integer. Thus, there does not exist any element x∈R such that f( x )=0.6. Then, f is not onto. Therefore, the greatest integer function is neither one-one nor onto. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@6345@
Q.4
Show that the Modul us Function f :R→R given by f(x)=|x|,is neither one–onen or onto, where|x|is x,if x is positiveor 0 and |x| is –x, if x is negative.
f: R→ R is given by,Modulus function is defined byf(x)={x, if x≥0−x, if x<0Here, f(−1)=|−1|=1, f(1)=|1|=1, ∴f(−1)=f(1), but −1≠1So, f is not one–one.Let −1∈R.Since, f(x)=|x| is always positive. So, there does not existany element in x in domain R such that f(x)=−1.∴f(x)​ is not onto.Therefore, the modulus function is neither one–one nor onto.
Q.5
Show that the Signum Function f : R → R, given byf(x)={ 1, if x>0 0, if x=0–1, if x<0is neither one–one noronto.
f: R→ R is given by,Modulus function is defined byf(x)={1, if x>0 0, if x=0−1, if x<0Here, f(1)=1, f(3)=1, but 1≠3So, 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 thatf(x)=−2.∴f is not onto.Hence, the signum function is neither one–one nor onto.
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.
It is given that A = { 1, 2, 3 }, B={ 4, 5, 6, 7 } f:A→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. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@DD6B@
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@
(i) f: R → R is defined as f(x) = 3−4x. For one-one Let f(x)=f(y) ⇒3−4x=3−4y ⇒ x=y ∴ f is one-one. for onto Let f(x)=y ⇒3−4x=y ⇒ x= 3−y 4 ⇒f( 3−y 4 )=3−4( 3−y 4 )=y Since Range = Co-domain ∴f is onto. Hence, f is bijective. (ii) f: R→ R is defined as f(x)=1+x 2 For one-one Let f(x)=f(y ) ⇒ 1+x 2 = 1+y 2 ⇒ x 2 = y 2 ⇒ x=±y ∴ f is not one-one. For onto: Let f( x )=y ⇒ 1+x 2 =y ⇒ x=± y−1 Since Range is not equal to Co-domain. Then given function f(x) is not onto. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@EF5D@
Q.8
Let A and B be sets. Show that f : A×B→B×A such thatf(a,b)=f(b,a) is bijective function.
f: A×B→B× A is defined as f(a, b) = (b, a). Let (a1, b1),(a2, b2)∈A×B such that f(a1, b1)=f(a2, b2)⇒(b1, a1)=(b2, a2)⇒b1=b2 and a1=a2⇒(a1,b1)=(a2,b2)∴f is one–one.Now, let (b, a)∈B×A be any element.Then, there exists (a, b)∈A×B such that f(a, b) = (b, a). [By definition of f]∴ f is onto.Hence, f is bijective.
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@
Let f: N→N is defined as f( n )={ n+1 2 , if n is odd n 2 , if n is even for all n∈N. It can be observed that: f( 1 )= 1+1 2 =1, f( 2 )= 2 2 =1 [ By definition of f ] ∴f( 1 )=f( 2 ), where 1≠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∈N. Then,there exists 4m + 1∈N such that f( 4m+1 )= 4m+1+1 2 =2m+1 Case II: n is even ∴ n=2m for some m∈N. Then,there exists 4m∈N such that f( 4m )= 4m 2 =2m ∴f is onto. Hence, f is not a bijective function. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaGqaaiaa=XeacaWFLbGaa8hDaiaa=bcacaWFMbGaa8Noaiaa=bcacaWFobGaeyOKH4Qaa8Ntaiaa=bcacaWFPbGaa83Caiaa=bcacaWFKbGaa8xzaiaa=zgacaWFPbGaa8NBaiaa=vgacaWFKbGaa8hiaiaa=fgacaWFZbaabaaabaGaa8NzamaabmaabaGaa8NBaaGaayjkaiaawMcaaiaa=1dadaGabaabaeqabaWaaSaaaeaacaWFUbGaa83kaiaa=fdaaeaacaWFYaaaaiaa=XcacaWLjaGaa8xAaiaa=zgacaWFGaGaa8NBaiaa=bcacaWFPbGaa83Caiaa=bcacaWFVbGaa8hzaiaa=rgaaeaadaWcaaqaaiaa=5gaaeaacaWFYaaaaiaa=XcacaWLjaGaaCzcaiaa=LgacaWFMbGaa8hiaiaa=5gacaWFGaGaa8xAaiaa=nhacaWFGaGaa8xzaiaa=zhacaWFLbGaa8NBaaaacaGL7baacaaMc8UaaGPaVlaaykW7caaMc8Uaa8Nzaiaa=9gacaWFYbGaa8hiaiaa=fgacaWFSbGaa8hBaiaa=bcacaWFUbGaeyicI4Saa8Ntaiaa=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@D481@
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@
A = R−{3} and B = R−{1} f: A→B is defined as f( x )= x−2 x−3 Let x,y∈A such that f( x )=f( y ). ⇒ x−2 x−3 = y−2 y−3 ⇒ ( x−2 )( y−3 )=( x−3 )( y−2 ) ⇒ xy −3x−2y+ 6 = xy −2x−3y+ 6 ⇒ −3x−2y=−2x−3y ⇒ x=y ∴ f if one-one. y∈B=R-{ 1 }. Then, y≠1 The function f is onto if there exists x∈A such that f(x) = y ⇒ x−2 x−3 =y ⇒ x−2=y( x−3 ) ⇒ x−2=yx−3y ⇒ x−yx=2−3y ⇒ x( 1−y )=2−3y ⇒ x= 2−3y ( 1−y ) ∈A [ y≠1 ] Thus, for any y∈B, there exists 2−3y ( 1−y ) ∈A such that f( 2−3y 1−y )= 2−3y ( 1−y ) −2 2−3y ( 1−y ) −3 = 2−3y−2( 1−y ) 1−y 2−3y−3( 1−y ) 1−y = 2−3y−2+2y 2−3y−3+3y = −y −1 f( 2−3y 1−y )=y ∴ f is onto. Hence, function f is one-one and onto. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaGqaaiaa=feacaWFGaGaa8xpaiaa=bcacaWFsbGaeyOeI0Iaa83Eaiaa=ndacaWF9bGaa8hiaiaa=fgacaWFUbGaa8hzaiaa=bcacaWFcbGaa8hiaiaa=1dacaWFGaGaa8NuaiabgkHiTiaa=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Q.11
Letf:R→R be defined as f(x)=x4. Choose the correctanswer.(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.
f: R→R is defined as f(x)=x4.Let x,y∈R such that f(x)=f(y).Then, f(x)=f(y)⇒ x4=y4⇒ x=±y∴ f(x)=f(y) but x≠ySo, 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.
Q.12
Let f:R→R be defined asf(x)=3x.Choose the correctanswer.(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.
f:R→Rbedefinedasf(x)=3x.Let x, y∈R such that f(x) = f(y).⇒3x=3y⇒ x=y∴f is one–one.Also, for any real number (y) in co–domain R, there exists y3 in R such that f(y3)=3(y3)=y∴f is onto.Therefore, function f is one–one and onto.The correct answer is A.