Question

Let $f(x)=\log _2 \log _3 \log _4 \log _6\left(\sin x+a^2\right)$. Find the set of values of a for which domain of $f(x)$ is $R$.

Answer 1

SOLUTION : Given $f(x)=\log _2 \log _3 \log _4 \log _5\left(\sin x+a^2\right)$

$f(x)$ is defined only if $\log _3 \log _4 \log _5\left(\sin x+a^2\right)>0, \forall x \in R$

$\Rightarrow \quad  \log _4 \log _5\left(\sin x+a^2\right)>1, \forall x \in R \quad  \Rightarrow \quad \log _5\left(\sin x+a^2\right)>4, \forall x \in R \\$

$\Rightarrow \left(\sin x+a^2\right)>5^4, \forall x \in R \quad  \Rightarrow \quad a^2>625-\sin x, \forall x \in R \\$

$\Rightarrow \quad  a^2 \text { must be greater than maximum value of } 625-\sin x \text { which is } 626(\text { when } \sin x=-1) \\$

$\Rightarrow \quad a^2>626 \quad \Rightarrow \quad a \in(-\infty,-\sqrt{626}) \cup(\sqrt{626}, \infty)$

$\Rightarrow \quad a^2$ must be greater than maximum value of $625-\sin x$ which is 626 (when $\sin x=-1$ )

$\Rightarrow \quad a^2>626 \quad \Rightarrow \quad a \in(-\infty,-\sqrt{626}) \cup(\sqrt{626}, \infty)$