    Question

# Let f(x)=x+lnx−xlnx, x∈(0,∞) ​​​​​​Column 1Column 2Column 3(I)f(x)=0 for some x∈(1,e2)(i)limx→∞f(x)=0(P)f is increasing in (0,1)(II)f′(x)=0 for some x∈(1,e) (ii)limx→∞f(x)=−∞ (Q)f is decreasing in (e,e2)(III)f′(x)=0 for some x∈(0,1) (iii)limx→∞f′(x)=−∞ (R)f′ is increasing in (0,1)(IV)f′′(x)=0 for some x∈(1,e) (iv)limx→∞f′′(x)=0 (S)f′ is decreasing in (e,e2) Which of the following options is the only CORRECT combination?

A
(I)(i)(P)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(II)(ii)(Q)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
(III)(iii)(R)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(IV)(iv)(S)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A (II)(ii)(Q)Column 1: (I)f(1)=1f(e2)=e2+2−2⋅e2<0⇒f(1)⋅f(e2)<0∴f(x)=0 for some x∈(1,e2) Statement is correct. (II)f′(x)=1x−lnxf′(1)=1f′(e)=1e−1<0⇒f′(1)⋅f′(e)<0∴f′(x)=0 for some x∈(1,e) Statement is correct. (III)f′(x)=1x−lnx ∴f′(x)≠0 ∀ x∈(0,1) Statement is incorrect. (IV)f′(x)=1x−lnxf′′(x)=−1x2−1x=−1x2(x+1)f′′(1)=−2f′′(e)=−e+1e2<0⇒f′′(1)⋅f′′(e)>0∴f′′(x)≠0 ∀ x∈(1,e) Statement is correct. Column 2: limx→∞f(x)=−∞ Therefore, statement (i) is incorrect and statement (ii) is correct. (iii)f′(x)=1x−lnxlimx→∞=−∞ Statement is correct. (iii)f′′(x)=−1x2−1xlimx→∞=0 Statement is correct. Column 3: (P)f′(x)=1x−lnx⇒f′(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<−1∴f′(x) is increasing in (−∞,−1)⇒f′′(x)<0 ∀x>−1∴f′(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.  Suggest Corrections  1      Explore more