Question

The equation $k\cos x-3\sin x=k+1$ is solvable only if $k$ belongs to the interval

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

Answer 1

Answer: (A), (C)

$[4,\infty )$
$(-\infty ,4]$

$kcosx-3sinx=k+1$

$\Rightarrow k(cosx-1)=1+3sinx$
$\Rightarrow k(\frac{1-ta{n}^2\left(x/2\right)}{1+ta{n}^2\left(x/2\right)}-1)=1+3.\frac{2tan\left(x/2\right)}{1+ta{n}^2\left(x/2\right)}$
$\Rightarrow -k.2ta{n}^2\left(x/2\right)=1+ta{n}^2\left(x/2\right)+6tan\left(x/2\right)$
$\Rightarrow -2k=1+co{t}^2\left(x/2\right)+6cot\left(x/2\right)=-8+(3+cot(x/2){)}^2$
$\Rightarrow k=4-\frac{1}{2}(1+cot(x/2){)}^2$
Thus range of k for which equation is solvable is,
$k\in [-\infty ,4]$
Hence, option 'C' is correct.