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The solution of the equation $kcos x - 3 \sin x = k + 1$ is possible if

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a

k \in (- \infty, 4]

b

k \in (- \infty, \infty)

c

k \in [4, \infty)

d

None of the above

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detailed solution

Correct option is A

Here, k cos x - 3 sin x = k +1 could be rewritten as

$\begin{array}{l} \frac{k}{\sqrt{k^{2} + 9}} \cos x - \frac{3}{\sqrt{k^{2} + 9}} \sin x = \frac{k + 1}{\sqrt{k^{2} + 9}} \\ \Rightarrow \cos (x + ϕ) = \frac{k + 1}{\sqrt{k^{2} + 9}} \end{array}$

$\begin{aligned} where \cos ϕ = \frac{k}{\sqrt{k^{2} + 9}} \\ and \sin ϕ = \frac{3}{\sqrt{k^{2} + 9}} \end{aligned}$

which posses solution only, if

$\begin{array}{l} - 1 \leq \frac{k + 1}{\sqrt{k^{2} + 9}} \leq 1 \\ \Rightarrow |\frac{k + 1}{\sqrt{k^{2} + 9}}| \leq 1 \\ \Rightarrow (k + 1)^{2} \leq k^{2} + 9 \\ \Rightarrow k^{2} + 2 k + 1 \leq k^{2} + 9 \\ \Rightarrow k \leq 4 \end{array}$

Hence, the interval in which kcos x - 3sin x = k +1 admits solution for k is $(- \infty, 4]$

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