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Functions question

2021 · 26 Feb · Shift 2 · Q33
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  5. /2021 · 26 Feb · Shift 2 · Q33

Functions question

2021 · 26 Feb · Shift 2 · Q33

JEE MainMathematicsFunctionsMCQ+4 / −1
Let A={1,2,3,....,10}A = \{ 1,2,3,....,10\}A={1,2,3,....,10} and f:A→Af:A \to Af:A→A be defined as f(k)={k+1if k is oddkif k is evenf(k) = \left\{ {\begin{matrix} {k + 1} & {if\,k\,is\,odd} \\ k & {if\,k\,is\,even} \\ \end{matrix} } \right.f(k)={k+1k​ifkisoddifkiseven​ Then the number of possible functions g:A→Ag:A \to Ag:A→A such that gof=fgof = fgof=f is :
  1. A
    55
  2. B
    105
  3. C
    5!
  4. D
    10C5
View written solutionFree

Correct answer: 100000

  1. Understand the given function fff

We have A={1,2,3,…,10}A = \{1,2,3,\dots,10\}A={1,2,3,…,10} and

f(k)={k+1,k oddk,k evenf(k)= \begin{cases} k+1,& k\text{ odd}\\ k,& k\text{ even} \end{cases}f(k)={k+1,k,​k oddk even​

So let us list the values:

f(1)=2,  f(2)=2f(1)=2,\; f(2)=2f(1)=2,f(2)=2 f(3)=4,  f(4)=4f(3)=4,\; f(4)=4f(3)=4,f(4)=4 f(5)=6,  f(6)=6f(5)=6,\; f(6)=6f(5)=6,f(6)=6 f(7)=8,  f(8)=8f(7)=8,\; f(8)=8f(7)=8,f(8)=8 f(9)=10,  f(10)=10f(9)=10,\; f(10)=10f(9)=10,f(10)=10

Hence the range of fff is

Im⁡(f)={2,4,6,8,10}\operatorname{Im}(f)=\{2,4,6,8,10\}Im(f)={2,4,6,8,10}

and each even number has exactly two preimages:

{1,2}↦2,{3,4}↦4,{5,6}↦6,{7,8}↦8,{9,10}↦10.\{1,2\}\mapsto 2,\quad \{3,4\}\mapsto 4,\quad \{5,6\}\mapsto 6,\quad \{7,8\}\mapsto 8,\quad \{9,10\}\mapsto 10.{1,2}↦2,{3,4}↦4,{5,6}↦6,{7,8}↦8,{9,10}↦10.
  1. Interpret the condition g∘f=fg\circ f = fg∘f=f

The condition means

g(f(k))=f(k)for every k∈A.g(f(k))=f(k)\quad \text{for every }k\in A.g(f(k))=f(k)for every k∈A.

Now f(k)f(k)f(k) can only be one of 2,4,6,8,102,4,6,8,102,4,6,8,10. Therefore we must have

g(2)=2,g(4)=4,g(6)=6,g(8)=8,g(10)=10.g(2)=2,\quad g(4)=4,\quad g(6)=6,\quad g(8)=8,\quad g(10)=10.g(2)=2,g(4)=4,g(6)=6,g(8)=8,g(10)=10.

So ggg must fix all elements in the image of fff.

  1. What about the odd elements 1,3,5,7,91,3,5,7,91,3,5,7,9?

These elements are not in the image of fff. Since the equation g(f(k))=f(k)g(f(k))=f(k)g(f(k))=f(k) only restricts values of ggg on {2,4,6,8,10}\{2,4,6,8,10\}{2,4,6,8,10}, there is no restriction on

g(1),  g(3),  g(5),  g(7),  g(9).g(1),\; g(3),\; g(5),\; g(7),\; g(9).g(1),g(3),g(5),g(7),g(9).

Each of these 5 inputs can be sent to any of the 10 elements of AAA.

So the total number of possible functions is

105=100000.10^5 = 100000.105=100000.
  1. Check options

The computed value is

100000,100000,100000,

which is not among the listed options:

  • A: 555555
  • B: 105105105
  • C: 5!=1205! = 1205!=120
  • D: (105)=252\binom{10}{5}=252(510​)=252

So none of the options match the actual count.

  1. Conclusion

The correct number of functions g:A→Ag:A\to Ag:A→A satisfying g∘f=fg\circ f=fg∘f=f is

105=100000.\boxed{10^5 = 100000}.105=100000​.

Therefore the stored answer B\boxed{B}B​ is incorrect.

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