Question

Range of values of the function $y={\log }_3\left[\frac{3\sin x+4\cos x+10}{3\sin x+4\cos x}\right]$ is equal to

• A. $(-\infty ,1]$
• B. $[1,\infty )$
• C. $[3,\infty )$
• D. $[0,1]$

Answer 1

Answer: (B)

$[1,\infty )$

Let, $y={\log }_3\left[\frac{3\sin x+4\cos x+10}{3\sin x+4\cos x}\right]$

or, $3^y=$ $\left[\frac{3\sin x+4\cos x+10}{3\sin x+4\cos x}\right]$

or, $3^y=$ $1+\left[\frac{10}{3\sin x+4\cos x}\right]$

or, $\frac{3}{10}$ $\sin x+\frac{4}{10}$ $\cos x=\frac{1}{3^y-1}$.

Let, $r\cos \theta =\frac{3}{10}$ and $r\sin \theta =\frac{4}{10}$ where $r=\frac{1}{2}$ and $\theta ={\tan }^{-1}\frac{4}{3}$.

Then $r\sin (\theta +x)=\frac{1}{3^y-1}$

or, $\sin (\theta +x)=\frac{2}{3^y-1}$.

As, $|\sin (\theta +x)|\le 1$

$\Rightarrow |\frac{2}{3^y-1}|$ $\le 1$

$\Rightarrow \frac{2}{3^y-1}$$\le 1$ and $\frac{2}{3^y-1}$ $\ge -1$

$\Rightarrow 3^y\ge 3$ and $3^y\ge -1$(Which is not possible)

So, $3^y\ge 3\Rightarrow y\ge 1$.

So, range of $y$ is $[1,\infty )$.