Relations and Functions

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New answer posted•a year ago
S
a year ago

 f (x)= [2nn=2, 4, 6....n−1n=3, 7, 11, 15....n+12n=1, 5, 9, 13....

for n = 2, 4, 6 ……

f (n) = 4, 8, 12, ….4 (n) form

for n = 3, 7, 11, 15, ….

f (n) = 1, 3, 5, 7, …. (4n + 1) or, (4n + 3) from

∴ f is one and onto.

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New answer posted•a year ago
S
a year ago

 1(20−a)(40−a)+1(40−a)(60−a)+1(60−a)(80−a)+.....+1(180−a)(200−a)=1256

LHS = 120(120−a−140−a)+120(140−a−160−a)+...+120(1180−a−1200−a)

=120(120−a−1200−a)=120.180(20−a)(200−a)

⇒a2−220a+4000−2304=0⇒a2220a+1696=0

a=220±2048=2122

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New answer posted•a year ago
S
a year ago

R = { (a, b): b = pq, where p, q ≥3 are prime}

⇒60×11=660

p, q ∈  {3, 5, 7, 11, 13, 17, 19, 23, 29, 3, 37, 41, 43, 47, 53, 59} total 16

p, q ∈  {3, 5, 7, 11, 13, 17, 19}

{3, 5, 7, 11, 13, 17, 19}

7 + 3 + 1 = 11

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New answer posted•a year ago
S
a year ago

. ?f(n)={2n, n=1,2,3,4,5

                    2n−11,n=6,7,8,9,10

∴f(1)=2,f(2)=4,....,f(5)=10

And f(6) = 1, f(7) = 3, f(8) = 5, …., f(10) = 9

Now,

f(g(n))={n+1if  n  is  oddn−1,if  n  is  even

∴f(g(10))=9⇒g(1)=1

∴g(10)(g(1)+g(2)+g(3)+g(4)+g(5))=190

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New answer posted•a year ago
S
a year ago

This is a True or False Type Question as classified in NCERT Exemplar

Sol:

    ' + '     i s     a     b i n a r y     o p e r a t i o n     o n     t h e     s e t     N     b u t     i t     h a s     n o     i d e n t i t y     e l e m e n t .     H e n c e ,     t h e     s t a t e m e n t     i s     ' F a l s e ' .

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New answer posted•a year ago
S
a year ago

This is a True or False Type Question as classified in NCERT Exemplar

Sol:

  Only  bijective  functions  are  invertible.  Hence,   the  statement  is  'False'.

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New answer posted•a year ago
S
a year ago

This is a True or False Type Question as classified in NCERT Exemplar

Sol:

          L e t   f ( x )   =   2 x     , g ( x )   =   x − 1     a n d     h ( x ) = 2 x + 3                           f o { g o h ( x ) }     =     f o { g ( 2 x + 3 ) }                                                                             =     f ( 2 x + 3 − 1 )   =   f ( 2 x + 2 )   =   2 ( 2 x + 2 )   = 4 x + 4     a n d               ( f o g ) o h ( x ) = ( f o g ) { h ( x ) }                                                                                     = f o g ( 2 x + 3 )                                                                                   = f ( 2 x + 3 − 1 )   =   f ( 2 x + 2 )   =   2 ( 2 x + 2 )   = 4 x + 4               ∴       f o { g o h ( x ) }   =   ( f o g ) o h ( x ) = 4 x + 4 H e n c e , t h e     s t a t e m e n t     i s   T r u e .

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New answer posted•a year ago
S
a year ago

This is a True or False Type Question as classified in NCERT Exemplar

Sol:

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New answer posted•a year ago
New answer posted•a year ago