Question

If f (x) is differentiable function and f'(x) = log 1/3 (log3 (sin x + a)), then find the
interval for a so that f (x) is decreasing for all real values of x.​

Answer 1

For a function $f(x)$ being a decreasing function in a particular interval, $f'(x)<0$ at that interval.

Here,

$\longrightarrow f'(x)=\log_{\frac{1}{3}}\left(\log_3\left(\sin x+a\right)\right)$

If $f(x)$ is decreasing $\forall x\in\mathbb{R},$

$\longrightarrow f'(x)<0$

$\longrightarrow\log_{\frac{1}{3}}\left(\log_3\left(\sin x+a\right)\right)<0$

$\longrightarrow\dfrac{\log\left(\log_3\left(\sin x+a\right)\right)}{\log\left(\frac{1}{3}\right)}<0$

Since $\log\left(\dfrac{1}{3}\right)<0,$

$\longrightarrow\log\left(\log_3\left(\sin x+a\right)\right)>0$

$\longrightarrow\log_3\left(\sin x+a\right)>1$

$\longrightarrow\dfrac{\log\left(\sin x+a\right)}{\log3}>1$

Since $\log3>0,$

$\longrightarrow\log\left(\sin x+a\right)>\log3$

$\longrightarrow\sin x+a>3$

$\longrightarrow a>3-\sin x$

We know that,

$\longrightarrow \sin x\in[-1,\ 1]$

$\longrightarrow -\sin x\in[-1,\ 1]$

$\longrightarrow 3-\sin x\in[2,\ 4]$

So the maximum value of $3-\sin x$ is 4.

Value of $a$ should be greater than this 4.

Therefore,

$\longrightarrow\underline{\underline{a\in(4,\ \infty)}}$