TG EAMCET 4 May 2025 Shift 2 Paper

‌cosh(x)=‌

ex+e−x

2

‌tanh(x)=‌

ex−e−x

ex+e−x

‌x=‌

ecosh−1(x)+e−cosh−1(x)

2

‌x=‌

y+ 1 y

2

‌y2−2xy+1=0
‌y=‌

2x±√4x2−4

2

‌ecosh−1(x)=x+√x2−1
‌cosh−1(x)=ln(x+√x2−1)
Again,x=‌

etanh−1(x)−e−tanh−1(x)

etanh−1(x)+e−tanh−1(x)

Lety=etanh−1(x)
‌⇒‌‌x=‌

y2−1

y2+1

‌⇒‌‌xy2+x=y2−1
‌⇒‌‌‌

x+1

1−x

=y2
‌⇒‌‌etanh−1(x)=√‌

1+x

1−x

‌⇒‌‌tanh−1(x)=‌

1

2

‌ln(‌

1+x

1−x

)
Now,cosh−1(x)=tanh−1(x)
ln(x+√x2−1)=‌

1

2

‌ln(‌

1+x

1−x

)
No solution
Aasx≥1forln(x+√x2−1)to be real. And for ln(‌

1+x

1−x

) to be defined
Now, cosh−1(x)=coth−1(x)
ln(x+√x2−1)=‌

1

2

‌ln(‌

x+1

x−1

)
Now, ln(x+√x2−1) is strictly increasing function, ∀x≥1 and ‌

1

2

‌ln(‌

x+1

x−1

) is strictly decreasing function for x>1.
⇒ Both the graph will cut and one point exactly.