If the domain of the function f(x)= ^-1((2x)/(5x+3)) is [ , )U( , ], then |3 +10( + )+21 |…

Question

If the domain of the function $f(x)={\sec }^{-1}(\frac{2x}{5x+3})$ is $[\alpha ,\beta )U(\gamma ,\delta ]$, then $|3\alpha +10(\beta +\gamma )+21\delta |$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 24

Accepted Answer

Given that $f(x)={\sec }^{-1}(\frac{2x}{5x+3})$

Since, the domain for ${\sec }^{-1}x$ is $|x|\ge 1$

Therefore $|\frac{2x}{5x+3}|\ge 1$

$\begin{gathered} \frac{2x}{5x+3}\le -1\frac{2x}{5x+3}\ge 1 \ \frac{2x+5x+3}{5x+3}\le 0\frac{2x-5x-3}{5x+3}\ge 0 \ \frac{7x+3}{5x+3}\le 0\frac{-3(x+1)}{5x+3}\ge 0 \end{gathered}$

Case I : $7x+3\le 0$ and $5x+3>0$

$ $$\& x\le -\frac{3}{7}x>-\frac{3}{5}  \   \  \& \Rightarrow -\frac{3}{5}Case II : $7x+3\ge 0$ and $5x+3Case III : $x+1\ge 0$ and $5x+3Case IV : $x+1\le 0$ and $5x+3\ge 0$$$

$x\le -1x\ge -\frac{3}{5}$

Which is not possible

$\therefore $ Domain is $[-1,-\frac{3}{5})\cup (-\frac{3}{5},-\frac{3}{7}]$

$\therefore \alpha =-1,\beta =-\frac{3}{5},\gamma =-\frac{3}{5},\delta =-\frac{3}{7}$

$\therefore $ $

Solutions · 1