The number of integral values of k for which the equation 7 cos x + 5 sin x = 2 k + 1 has a solution is [2002]

Question

The number of integral values of $k$ for which the equation $7 \cos x + 5 \sin x = 2 k + 1$ has a solution is [2002]

Smart Achievers · Accepted Answer

(2)

$We know, - \sqrt{a^{2} + b^{2}} \leq a \cos θ + b \sin θ \leq \sqrt{a^{2} + b^{2}}$

$\Rightarrow - \sqrt{74} \leq 7 \cos x + 5 \sin x \leq \sqrt{74}$

$\Rightarrow - \sqrt{74} \leq 2 k + 1 \leq \sqrt{74} \Rightarrow - 8.6 \leq 2 k + 1 \leq 8.6$

$\Rightarrow - 4.8 \leq k \leq 3.8$

$Hence, k can take only 8 integral values.$

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