NVAMediumJEE 2023Functions

JEE Mathematics 2023 Question with Solution

Let A={1,2,3,5,8,9}A = \{ 1, 2, 3, 5, 8, 9 \}. Then the number of possible functions f:AAf : A \to A such that f(mn)=f(m)f(n)f(m \cdot n) = f(m) \cdot f(n) for every m,nAm, n \in A with mnAm \cdot n \in A is equal to _____

Answer

Correct answer:432

Step-by-step solution

Standard Method

Given: A={1,2,3,5,8,9}A = \{1,2,3,5,8,9\} and f:AAf : A \to A satisfies

f(mn)=f(m)f(n)f(mn) = f(m)f(n)

for every m,nAm,n \in A with mnAmn \in A.

Find: The number of such functions.

From the condition with 11=11 \cdot 1 = 1,

f(1)=f(11)=f(1)f(1)f(1) = f(1\cdot 1) = f(1)\cdot f(1)

So,

f(1)=1f(1) = 1

Also, since 33=9A3 \cdot 3 = 9 \in A,

f(9)=f(33)=f(3)f(3)f(9) = f(3\cdot 3) = f(3)\cdot f(3)

Hence f(3)2=f(9)f(3)^2 = f(9), and from the given solution this gives two possibilities:

f(3)=1orf(3)=3f(3) = 1 \quad \text{or} \quad f(3) = 3

corresponding to f(9)=1f(9)=1 or f(9)=9f(9)=9.

Case 1: f(3)=1f(3)=1 and f(9)=1f(9)=1.

Then f(2),f(5),f(8)f(2), f(5), f(8) can each be chosen in 66 ways. So the number of functions in this case is

63=2166^3 = 216

Case 2: f(3)=3f(3)=3 and f(9)=9f(9)=9.

Again, f(2),f(5),f(8)f(2), f(5), f(8) can each be chosen in 66 ways. So the number of functions in this case is

63=2166^3 = 216

Therefore, total number of functions is

216+216=432216 + 216 = 432

So, the required answer is 432432.

Case Split from the Given Relations

Given: The only multiplicative relations inside AA that matter here are 11=11\cdot1=1 and 33=93\cdot3=9.

Find: Count all functions satisfying the condition.

First use

f(1)=f(11)=f(1)f(1)f(1)=f(1\cdot1)=f(1)f(1)

which forces

f(1)=1f(1)=1

Next use

f(9)=f(33)=f(3)2f(9)=f(3\cdot3)=f(3)^2

Since values of the function lie in A={1,2,3,5,8,9}A = \{1,2,3,5,8,9\}, the solution identifies only two admissible outcomes:

(f(3),f(9))=(1,1)or(3,9)(f(3),f(9))=(1,1) \quad \text{or} \quad (3,9)

Once these are fixed, the elements 2,5,82,5,8 are free to map anywhere in AA, giving

6×6×6=63=2166 \times 6 \times 6 = 6^3 = 216

functions for each admissible case.

Hence the total count is

2×216=4322 \times 216 = 432

Therefore, the answer is 432432.

Common mistakes

  • Assuming every element of AA must satisfy a multiplicative relation with every other element is incorrect, because the condition is imposed only when mnAm\cdot n \in A. Check closure inside AA before applying the rule.

  • Forgetting to use 11=11\cdot1=1 leads to missing the restriction on f(1)f(1). Since f(1)=f(1)2f(1)=f(1)^2, the valid value here is f(1)=1f(1)=1.

  • Treating f(2),f(5),f(8)f(2), f(5), f(8) as constrained without justification is a conceptual error. In the extracted solution, these values are free once the relations involving 11 and 99 are handled, so each has 66 choices.

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