Question

Let $f\left(x\right)=\log \left({\log }_{1/3}\right({\log }_7(\sin x+a)\left)\right)$ be defined for every real values of $x$, then the range of $a$

• A. $(2,6)$
• B. $[2,6]$
• C. $(-\infty ,2)\cup [6,\infty )$
• D. $(-\infty ,2]\cup (6,\infty )$

Answer 1

The correct option is A $(2,6)$

$f\left(x\right)=\log \left({\log }_{1/3}\right({\log }_7(\sin x+a)\left)\right)$
For the $\log$ function to be defined,
$\left(i\right){\log }_{1/3}\left({\log }_7\right(\sin x+a\left)\right)>0\Rightarrow {\log }_7(\sin x+a)<1\Rightarrow \sin x+a<7\cdots \left(1\right)$

$\left(ii\right){\log }_7(\sin x+a)>0\Rightarrow \sin x+a>1\cdots \left(2\right)$

$\left(iii\right)\sin x+a>0\cdots \left(3\right)$

From equation $\left(1\right),\left(2\right)$ and $\left(3\right)$,
$1<\sin x+a<7$
So,
$\sin x+a>1\text{and}\sin x+a<7\Rightarrow a>1-\sin x\text{and}\Rightarrow a<7-\sin x$

We know that $\sin x\in [-1,1]$, so
$1-\sin x\in [0,2]7-\sin x\in [6,8]$
Now,
$a>1-\sin x$
Largest possible value of $1-\sin x$ is $2$,
$a>2$

$a<7-\sin x$
Smallest possible value of $7-\sin x$ is $6$,
$a<6$

Hence, the required range is
$a\in (2,6)$