JEE Advanced 2020 Concept Recapitulation Test 3 Paper 1

f(x)=x+logex−xlogex
f′(x)=1+‌

1

x

logex−x(‌

1

x

)=‌

1

x

−logex f′′(x)=−‌

1

x2

−‌

1

x

<0‌∀x∈(0,∞)
Thus, f′(x) is strictly decrea sing function for x∈(0,∞).

lim

x→∞

(‌

1

x

−logex)=

lim

x→∞

f′(x)=−∞}

lim

x→0+

(‌

1

x

−logex)=

lim

x→0+

f′(x)=∞∫
i.e., f′(x)= has only one real root in (0,∞)
f′(1)=1>0
f′(e)=‌

1

e

−1<0
so, f′(x)=0 has one root in (1,e).
Let f′(α)=0, where α∈(1,e)
Therefore, f(x) is increasing in (0,α) and decreasing in (α,∞).
f(1)=1 and f(e2)=e2+2−2e=2−e2<0
The graph of the function is as shown in the given figure.
So, f(x)=0 has one root in (1,e2)
From Column I : I and II are correct.
From Column II : ii, iii and iv are correct.
From Column III : P,Q,S are correct.