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Question

Let f(x)=limn⎜ ⎜nn(x+n)(x+n2)(x+nn)n!(x2+n2)(x2+n24)(x2+n2n2)⎟ ⎟xn, for all x>0. Then

A
f(12)f(1)
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B
f(13)f(23)
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C
f(2)0
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D
f(3)f(3)f(2)f(2)
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Solution

The correct option is C f(2)0
f(x)=limn⎜ ⎜nn(x+n)(x+n2)(x+nn)n!(x2+n2)(x2+n24)(x2+n2n2)⎟ ⎟xn

Taking log on both sides, we get
lnf(x)=limnxn⎢ ⎢ ⎢ln(nnn!)+nr=1ln⎜ ⎜ ⎜x+nrx2+(nr)2⎟ ⎟ ⎟⎥ ⎥ ⎥

lnf(x)=limnxn⎢ ⎢ ⎢nr=1ln(nr)+nr=1ln⎜ ⎜ ⎜x+nrx2+(nr)2⎟ ⎟ ⎟⎥ ⎥ ⎥

lnf(x)=limnxn⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢nr=1ln(1r/n)+nr=1ln⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜x+1r/nx2+(1r/n)2⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥
lnf(x)=x10ln(1/t) dt+10ln(x+1/tx2+1/t2) dt

lnf(x)=x10(lnt+ln(tx+1)lntln(t2x2+1)+lnt2)dt
Let tx=z
Then lnf(x)=x0ln(1+z1+z2)dz (1)
f(x)=e x0ln1+z1+z2dz
Differentiating (1) w.r.t. x, we get
f(x)f(x)=ln(1+x1+x2)​​​​

We know that f(x)>0 x>0
For f(x)>0
ln(1+x1+x2)>01+x1+x2>1x>x20<x<1
f(x) is increasing x(0,1)
Option (1) is incorrect.
Option (2) is correct.

Option (3):
f(x)=f(x)×ln(1+x1+x2)f(2)=f(2)×ln(3/5)0
Statement is correct.

Option (4):
f(3)f(3)=ln(2/5)f(2)f(2)=ln(3/5)
Statement is incorrect.

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