Question

The domain of the function $f(x)=\text{Sec}^{-1}(3 x-4)+\text{Tan} h^{-1}\left(\frac{x+3}{5}\right)$ 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,
$f(x)=\sec ^{-1}(3 x-4)+\tanh ^{-1}\left(\frac{x+3}{5}\right)$
$=\sec ^{-1}(3 x-4)+\frac{1}{2} \ln \left(\frac{1+\frac{x+3}{5}}{1-\frac{x+3}{5}}\right) $
$=\sec ^{-1}(3 x-4)+\frac{1}{2} \ln \left(\frac{8+x}{2-x}\right)$
Now, for $\sec ^{-1}(3 x-4)$
$3 x-4 \leq-1 \cup 3 x-4 \geq 1$
$\Rightarrow 3 x \leq 3 \cup 3 x \geq 5 $
$\Rightarrow x \leq 1 \cup x \geq \frac{5}{3}$
$\therefore x \in(-\infty, 1] \cup x \in\left[\frac{5}{3}, \infty\right)$
Again, for $\ln \left(\frac{8+x}{2-x}\right)$
$\frac{8+x}{2-x}>0$

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