Question

If the domain of the function $f(x)=\sqrt{12-3^x-3^{3-x}}+$ ${\sin }^{-1}\left(\frac{2x}{3}\right)$ is $[a,b]$, then $3a+2b$ is equal to

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

Answer: (C)

$\because 12-3^x-3^{3-x}\ge 0$
$\Rightarrow 3^x+\frac{27}{3^x}-12\le 0$
Let $3^x=t$
$\therefore t^2-12t+27\le 0$
$\Rightarrow (t-3)(t-9)\le 0$
$\Rightarrow 3\le t\le 9$
$\Rightarrow 1\le x\le 2$
For ${\sin }^{-1}\frac{2x}{3}$ to be defined
$-1\le \frac{2x}{3}\le 1\Rightarrow \frac{-3}{2}\le x\le \frac{3}{2}$
$\therefore$ Domain of $f(x)=\left[1,\frac{3}{2}\right]$
$\therefore 3a+2b=3+3=6$