TG EAMCET 2 May 2025 Shift 1 Paper

Given, f(x)=x+log(‌

x−1

x+1

)
For, log(‌

x−1

x+1

) to be well-defined, ‌

x−1

x+1

>0.
This inequality holds when
⇒ Both x−1>0 and x+1>0⇒x>1
⇒ Both x−1<0 and x+1<0⇒x<−1
So, the domain of f(x) is (−∞,−1)∪(1,∞)
Now, f′(x)=‌

d

dx

[x+log(‌

x−1

x+1

)]
‌⇒1+‌

(x+1)

(x−1)

⋅‌

(x+1)⋅1−1⋅(x−1)

(x+1)2

‌⇒1+‌

(x+1)−(x−1)

(x−1)(x+1)

‌⇒1+‌

2

(x−1)(x+1)

Since, x∈(−∞,−1)∪(1,∞), we have x2>1
⇒x2−1>0
Thus, ‌

2

x2−1

>0
So, f′(x)=1+‌

2

x2−1

>1+0=1
Since, f′(x)>0 for all x in the domain, the function f(x) is monotonically increasing.