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Is the given relation a function? Give reasons for your answer.

(i) h = { ( 4 , 6 ) , ( 3 , 9 ) , ( − 1 1 , 6 ) , ( 3 , 1 1 ) }   

(ii) f = { (x,x)?x is a real number }     

(iii) g = { (n, 1n,n)?n is a positive integer }  

(iv) s = { (n,n2)?n is a positive integer }  

(v) t = { (x,3)?x is a real number }  

0 5 Views | Posted 4 months ago
Asked by Shiksha User

  • 1 Answer

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    Answered by

    Payal Gupta | Contributor-Level 10

    4 months ago

    This is a short answer type question as classified in NCERT Exemplar

    ( i ) G i v e n     t h a t :     h = { ( 4 , 6 ) , ( 3 , 9 ) , ( − 1 1 , 6 ) , ( 3 , 1 1 ) }       Since  in  the  given  relation  3  has  two  images  9  and  11.  So,  h  is  not  a  function. ( i i ) f = { ( x , x ) | x     i s     a     r e a l     n u m b e r } .     H e r e ,     w e     o b s e r v e     t h a t     f o r     e v e r y     e l e m e n t     o f     d o m a i n     h a s     a               u n i q u e     i m a g e .     S o ,     f     i s     a     f u n c t i o n . ( i i i ) G i v e n     t h a t :                       g = { (n,1n)|n  is  a  positive  integer } . H e r e ,     w e     o b s e r v e     t h a t     n       i s     a     p o s i t i v e          integer  so,  for  every  element  of  domain,  there  is  a  unique  1n  image.  Hence,  g  is  a  function. ( i v ) G i v e n     t h a t :                       S = { (n,n2)|n  is  a  positive  integer } . H e r e ,     w e     o b s e r v e     t h a t     t h e     s q u a r e     o f     a n y          integer  is  a  unique  number.  So,  every  element  in  the  domain  there  is  a  unique  image.                     H e n c e ,     S     i s     a     f u n c t i o n . ( v ) G i v e n     t h a t :                       t = { ( x , 3 ) | x     i s     a     r e a l     n u m b e r } .     H e r e ,     w e     o b s e r v e     t h a t     f o r     e v e r y     r e a l     e l e m e n t        in  the  domain,  there  is  a  constant  number  3.   Hence,  t  is  a  constant  function.

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