Q49 / 83 · NTA Abhyas (JEE Main) 2020 · Mathematics · Trigonometric Equations

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

A
$k \in (- \infty, 4)$
Correct answer
B
$k \in (- \infty, 4]$
C
$k \in (4, \infty)$
D
$k \in [4, \infty)$

Solution

k \cos x - 3 \sin x = k + 1

For a solution

|k + 1| \leq \sqrt{k^{2} + {(- 3)}^{2}}

\Rightarrow |k + 1| \leq \sqrt{k^{2} + 9}

On squaring both sides, we get,

k^{2} + 1 + 2 k \leq k^{2} + 9

\Rightarrow 2 k \leq 8 \Rightarrow k \leq 4

\Rightarrow k \in (- \infty, 4]

Hence, the correct answer is option: 'B'.

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