Q34 / 54 · JEE Advanced 2017 (Paper 1) · Mathematics

Let fx=x+loge⁡x-xloge⁡x,x∈0, ∞
•    Column 1 contains information about zeros of f( x ) , f'( x ) and f''( x )
•    Column 2 contains information about the limiting behaviour of f( x ) , f'( x ) and f''( x ) at infinity.
•    Column 3 contains information about increasing-decreasing nature of f( x ) and f'( x )
Column 1 Column 2 Column 3
(I) fx=0 for some x∈1, e2 (i) limx→∞⁡fx=0 (P) f is increasing in (0, 1)
(II) f′x=0 for some x in 1, e (ii) limx→∞⁡fx=-∞ (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

  • A
    (III) (iv) (P)
  • B
    (I) (ii) (R)
  • Correct answer
    C
    (II) (iii) (S)
  • D
    (IV) (i) (S)

Solution

fx=x+ℓnx-xℓnx, x>0

f'x=1+1x-ℓnx-1

f''x= -1x2-1x=-x+1x2

In Column 1,

(I)  f1 fe2<0 so true

(II)   f'1 f'e<0 so true

(III)   f'x is always positive for (0,1) so it cannot be zero. So (III) is false.

(IV) Is false

In Column 2

As   limx→∞⁡fx= limx→∞⁡x1+ℓn xx- ℓn x= -∞

∴ (i)  is false (ii) is true

limx→∞⁡f'x= -∞ ,so (iii) is true.

limx→∞⁡f''x=0, so (iv) is true.

(P)  f'x  is positive in (0, 1),so true.

(Q)  f'x<0 ,for in e, e2 ,so true.

As  f''x<0 ∀ x>0 therefore, R is false, S is true.

Alternate Solution:

fx=x+ ℓn x-xℓn x

f'x=1x-ℓn x=0 at x=x0 where x0∈1, e

f''x= -1x2-1x<0 ∀ x>0 ⇒  fx  concave down

Figure

Hence, the correct answer is option: 'C'.

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