Question

The range of $k$ for which the equation $k\cos x-3\sin x=k+1$ has a solution is

• A. $(-\infty ,4]$
• B. $(-\infty ,4)$
• C. $(4,\infty )$
• D. $[4,\infty )$

Answer 1

The correct option is A $(-\infty ,4]$

$k\cos x-3\sin x=k+1\Rightarrow \frac{k}{\sqrt{k+9}}\cos x-\frac{3}{\sqrt{k^2+9}}\sin x=\frac{k+1}{\sqrt{k^2+9}}\Rightarrow \cos (x+\varphi )=\frac{k+1}{\sqrt{k^2+9}}(\because \cos A\cos B-\sin A\sin B=\cos (A+B\left)\right)$
Where $\cos \varphi =\frac{k}{\sqrt{k+9}}$

Now,
$-1\le \cos (x+\varphi )\le 1\Rightarrow -1\le \frac{k+1}{\sqrt{k^2+9}}\le 1\Rightarrow \left|\frac{k+1}{\sqrt{k^2+9}}\right|\le 1\Rightarrow (k+1{)}^2\le k^2+9\Rightarrow k^2+2k+1\le k^2+9\Rightarrow k\le 4\therefore k\in (-\infty ,4]$