Question

I: lf $f\left(x\right)=\cosh x+\sinh x$ then$f(x_1+x_2+\dots ..x_n)=f\left(x_1\right).f\left(x_2\right)\dots .f\left(x_n\right)$
II. ${\tanh }^{-1}\left(x\right)=\frac{1}{2}\log \left(\frac{1+x}{1-x}\right)$ for $|x|>1$

• A. Only I is true
• B. Only II is true
• C. Both I and II are true
• D. Neither I nor II is true

Answer 1

The correct option is A Only I is true

I) $f\left(x\right)=coshx+sinhx$
$=\frac{e^x+e^{-x}+e^x-e^{-x}}{2}\Rightarrow e^x$
$f\left(x_1\right)=e^{x_1}$
$f\left(x_1\right)f\left(x_2\right)....=e^{x_1}\times e^{x_2}.....e^{x_n}$
$=e^{(x_1+x_2+.....x_n)}.....\left(1\right)$
$f(x_1+x_2+x_3.....)=e^{(x_1+x_2+.....x_n)}.....\left(2\right)$
From (1) and (2) it can be inferred that statement is true.
II) $tan{h}^{-1}x=\frac{1}{2}log\left(\frac{1+x}{1-x}\right)=y$
$tanhy=x$
$x=\frac{e^y-e^{-y}}{e^y+e^{-y}}$
$x=\frac{e^{2y}-1}{e^{2y}+1}$
$e^{2y}\times x+x=e^{2y}-1$
$1+x=e^{2y}(1-x)$
$e^{2y}=\frac{1+x}{1-x}.....\left(1\right)$
Since $e^{2y}>0\Rightarrow \frac{1+x}{1-x}>0$
Using method of intervals in number line, we get $-1<x<1$. Hence statement II is true for $\mid x\mid <1$
Statement II is false for $\mid x\mid >1$