1
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 INCORRECT combination?
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 INCORRECT combination?
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Solution
The correct option is C (III)(i)(R)
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 table option 3 has incorrect combination.
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 table option 3 has incorrect combination.
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