Question

The domain of the function $f(x)={\text{Sec}}^{-1}(3x-4)+\text{Tan}{h}^{-1}\left(\frac{x+3}{5}\right)$ is

• A. $(-8,1)\cup \left(\frac{5}{3},2\right)$
• B. $\left(1,\frac{5}{3}\right)$
• C. $[-8,1]\cup \left[\frac{5}{3},2\right]$
• D. $(-8,1]\cup \left[\frac{5}{3},2\right)$

Answer 1

Answer: (D)

We have,
$f(x)={\sec }^{-1}(3x-4)+{\tanh }^{-1}\left(\frac{x+3}{5}\right)$
$={\sec }^{-1}(3x-4)+\frac{1}{2}\ln \left(\frac{1+\frac{x+3}{5}}{1-\frac{x+3}{5}}\right)$
$={\sec }^{-1}(3x-4)+\frac{1}{2}\ln \left(\frac{8+x}{2-x}\right)$
Now, for ${\sec }^{-1}(3x-4)$
$3x-4\le -1\cup 3x-4\ge 1$
$\Rightarrow 3x\le 3\cup 3x\ge 5$
$\Rightarrow x\le 1\cup x\ge \frac{5}{3}$
$\therefore x\in (-\infty ,1]\cup x\in \left[\frac{5}{3},\infty \right)$
Again, for $\ln \left(\frac{8+x}{2-x}\right)$
$\frac{8+x}{2-x}>0$

$x\in (-8,2)$
$\therefore$ Domain of $f(x)=(-8,1]\cup \left[\frac{5}{3},2\right)$