TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

Given function.
f(x)=x−log(‌

1+x

x

),x>0
Differentiating w.r.t. x, we get
f′(x)=‌

d

dx

(x)−‌

d

dx

‌log(‌

1+x

x

)
=1−‌

1

1+x x

⋅‌

d

dx

(‌

1+x

x

)‌‌ [from chain rule]
f′(x)=1+‌

1

x(x+1)

Since x>0, therefore there is no possibility of f′(x) to be zero at any value of x.
So, for the given function f′(x)≠0
Also, for ∀x∈R‌‌f′(x)>0
i.e., function f(x) is strictly increasing its defined domain.
So, reason correctly explains given assertion.