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QJEEM22Maca26: JEE Main 2022 Mathematics - Class 11 JEEM22Maca26 Solution: Let be functions defined by , where is the maximum of the…

Let f,g:N−{1}→Nf, g: \mathbb{N}-\{1\} \rightarrow \mathbb{N} be functions defined by f(a)=αf(a)=\alpha , where α\alpha is the maximum of the powers of those primes pp such that pαp^{\alpha} divides aa , and g(a)=a+1g(a)=a+1 , for all a∈N−{1}a \in \mathbb{N}-\{1\} . Then, the function f+gf+g is

Multiple Choice Options

  • A one-one but not onto
  • B onto but not one-one
  • C both one-one and onto
  • D neither one-one nor onto

Step-by-Step Verified Solution

f,g:N−{1}→Nf,g:N - \{ 1\} \to N defined as

f(a)=αf(a) = \alpha , where α\alpha is the maximum power of those primes p such that p α\alpha divides a.

g(a)=a+1g(a) = a + 1 ,

Now,

f(2)=1,g(2)=3⇒(f+g)(2)=4f(3)=1,g(3)=4⇒(f+g)(3)=5f(4)=2,g(4)=5⇒(f+g)(4)=7f(5)=1,g(5)=6⇒(f+g)(5)=7

∵\because (f+g) (5)=(f+g) (4)(f + g)\,(5) = (f + g)\,(4)

∴\therefore f+gf + g is not one-one

Now, ∵\because fmin⁡=1, gmin⁡=3{f_{\min }} = 1,\,{g_{\min }} = 3

So, there does not exist any x∈N−{1}x \in N - \{ 1\} such that (f+g)(x)=1,2,3(f + g)(x) = 1,2,3

∴\therefore f+gf + g is not onto

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