Let f x =   lim n → ∞ ⁡ n n x + n x + n 2 … . . x + n n n ! x 2 + n 2   x…

Let fx= limn→∞⁡nnx+nx+n2…..x+nnn!x2+n2 x2+n24…..x2+n2n2xn , for all x>0 . Then
  1. f12≥f1
  2. f13 ≤f23
  3. f ′ 2 ≤0
  4. f ′ 3f3 ≥ f ′ 2f2

Solution

Taking log both side
ln fx= limn→∞⁡xn ln ∏r=1nx+1rn∏r=1nx2+1rn2  1∏r=1nrn
=x limn→∞⁡1n ∑r=1nlnxrn+1xrn2+1
Express summation in definite integration using r n =t & 1 n =dt
=x ∫01ln1+tx1+t2x2dt
Put tx=z
ln fx= ∫0xln1+z1+z2dz
⇒ f′xfx= ln 1+x1+x2 ( ∵f( x )  is  always  positive )
∀ x∈( 0,1 )
ln 1+x 1+ x 2 >0
⇒ f'( x ) f( x ) >0  ∀  x∈ ( 0,1 )
as  f( x )>0⇒f'( x )>0  ∀  x∈( 0,1 )
So   f( x ) is increasing ( 0,1 ) and f ′ ( 1 )=0    ( x>1⇒ f ′ ( x )<0 )
⇒ f12<f1, f13<f23,  f ′ 2<0
Also f ′ 3f 3- f ′  2f 2=ln 410- ln35 
= ln 46<0 ⇒ f ′ 3f 3< f ′  2f 2

Asked in: JEE Advanced 2016 (Paper 2)

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