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Question

Let f(x)=x+lnxxlnx, x(0,)

​​​​​​Column 1Column 2Column 3(I)f(x)=0 for some x(1,e2)(i)limxf(x)=0(P)f is increasing in (0,1)(II)f(x)=0 for some x(1,e) (ii)limxf(x)= (Q)f is decreasing in (e,e2)(III)f(x)=0 for some x(0,1) (iii)limxf(x)= (R)f is increasing in (0,1)(IV)f′′(x)=0 for some x(1,e) (iv)limxf′′(x)=0 (S)f is decreasing in (e,e2)

Which of the following options is the only CORRECT combination?

A
(I)(i)(P)
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B
(II)(ii)(Q)
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C
(III)(iii)(R)
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D
(IV)(iv)(S)
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Solution

The correct option is A (II)(ii)(Q)
Column 1:
(I)f(1)=1f(e2)=e2+22e2<0f(1)f(e2)<0f(x)=0 for some x(1,e2)
Statement is correct.
(II)f(x)=1xlnxf(1)=1f(e)=1e1<0f(1)f(e)<0f(x)=0 for some x(1,e)
Statement is correct.
(III)f(x)=1xlnx


f(x)0 x(0,1)
Statement is incorrect.
(IV)f(x)=1xlnxf′′(x)=1x21x=1x2(x+1)f′′(1)=2f′′(e)=e+1e2<0f′′(1)f′′(e)>0f′′(x)0 x(1,e)
Statement is correct.

Column 2:
limxf(x)=
Therefore, statement (i) is incorrect and statement (ii) is correct.
(iii)f(x)=1xlnxlimx=
Statement is correct.
(iii)f′′(x)=1x21xlimx=0
Statement is correct.

Column 3:
(P)f(x)=1xlnxf(x)>0 x(0,1)
Statement is correct.


(Q) Similarly from the figure
f(x)<0 x(e,e2)
Statement is correct.
(R)f′′(x)=1x2(x+1)f′′(x)>0 x<1f(x) is increasing in (,1)f′′(x)<0 x>1f(x) is decreasing in (1,)
f(x)<0 x(e,e2)
Statement is incorrect.
(S) From above, we can say that,
Statement is correct.
Now,
​​​​​​Column 1Column 2Column 3(I) Correct(i) Incorrect(P) Correct(II) Correct(ii) Correct(Q) Correct(III) Incorrect(iii) Correct(R) Incorrect(IV) Incorrect(iv) Correct(S) Correct
Using the above table we say that Option 2 is correct.

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