Let A = { 1 , 2 , 3 , . . . . . , 1 0 } a n d f : A A be defined as

f ( k ) = { k + 1 i f k i s o d d k i f k i s e v e n            

Then the number of possible functions g : A ® A such that gof = f is:

 

Option 1 -

A

Option 2 -

B

Option 3 -

C

Option 4 -

D

0 2 Views | Posted 2 weeks ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    2 weeks ago
    Correct Option - 3


    Detailed Solution:

    Given f (k) = { k + 1 , k i s o d d k , k i s e v e n

      ? g : A A           such that g (f (x) = f (x)

    Case I : If x is even then g (x) = x . (i)

    Case II : If x is odd then g (x + 1) = x + 1 . (ii)

    From (i) & (ii), g (x) = x, when x is even

    So total no. of functions = 105 × 1 = 105

Similar Questions for you

A
alok kumar singh

R1 = { (1, 1) (1, 2), (1, 3)., (1, 20), (2, 2), (2, 4). (2, 20), (3, 3), (3, 6), . (3, 18),
(4, 4), (4, 8), . (4, 20), (5, 5), (5, 10), (5, 15), (5, 20), (6, 6), (6, 12), (6, 18), (7. 7),
(7, 14), (8, 8), (8, 16), (9, 9), (9, 18), (10, 10), (10, 20), (11, 11), (12, 12), . (20, 20)}

n (R1) = 66

R2 = {a is integral multiple of b}

So n (R1 – R2) = 66 – 20 = 46

as R1 Ç R2 = { (a, a) : a Î s} = { (1, 1), (2, 2), ., (20, 20)}

A
alok kumar singh

g o f ( x ) = { f ( x ) , f ( x ) < 0 f ( x ) , f ( x ) > 0

= { e l n x = x ( 0 , 1 ) e x ( , 0 ) l n x ( 1 , )

Therefore, gof (x) is many one and into

A
alok kumar singh

  ?  R is symmetric relation

⇒   (y, x) R V (x, y) R

(x, y)  R 2x = 3y and (y, x) R 3x = 2y

Which holds only for (0, 0)

Which does not belongs to R.

Value of n = 0

A
alok kumar singh

f is increasing function

x < 5x < 7x

f (x) < f (5x) < f (7x)

  f ( x ) f ( x ) < f ( 5 x ) f ( x ) < f ( 7 x ) f ( x )           

l i m x f ( x ) f ( x ) < l i m x f ( 5 x ) f ( x ) < l i m x f ( 7 x ) f ( x )             

-> 1 < l i m x f ( 5 x ) f ( x ) < 1 l i m x f ( 5 x ) f ( x ) = 1

l i m x ( f ( 5 x ) f ( x ) 1 ) = 0            

R
Raj Pandey

f ( x ) = x 1 x + 1 f ( f ( x ) ) = x 1 x + 1 1 x 1 x + 1 + 1 = 1 x

f 3 ( x ) = x + 1 x 1 f 4 ( x ) = x 1 x + 1 + 1 x 1 x + 1 1 = x

S o , f 6 ( 6 ) + f 7 ( 7 ) = f 2 ( 6 ) + f 3 ( 7 )

= 1 6 7 + 1 7 1 = 9 6 = 3 2

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 688k Reviews
  • 1800k Answers

Learn more about...

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.

Need guidance on career and education? Ask our experts

Characters 0/140

The Answer must contain atleast 20 characters.

Add more details

Characters 0/300

The Answer must contain atleast 20 characters.

Keep it short & simple. Type complete word. Avoid abusive language. Next

Your Question

Edit

Add relevant tags to get quick responses. Cancel Post