Q35 / 54 · JEE Advanced 2017 (Paper 1) · Mathematics

Let fx=x+loge⁡x-xloge⁡x,x∈0, ∞. 
•    Column 1 contains information about zeros of fx, f′x and f′′x.
•    Column 2 contains information about the limiting behaviour of fx, f′x and f′′x at infinity.
•    Column 3 contains information about increasing-decreasing nature of fx 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
Which of the following options is the only Incorrect combination?

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

Solution

fx=x+ℓnx-xℓnx, x>0

f'x=1+1x-ℓnx-1

f''x= -1x2-1x=-x+1x2

Figure

(I)  f1 fe2<0, so true.

(II) f'1 f'e<0,so true.

(III) Graph of f'x , so (III) is false as the curve does not intersect X-axis i.e, f'(x)≠0 when x∈(0,1)

(IV) Is false. f''(x)≠0 as tangent to the curve of f'(x) is not parallel to X-axis.

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:

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: 'B'.

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