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Question

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


A
4
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B
8
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C
10
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D
12
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Solution

The correct option is A $$8$$
$$|7\ \cos\ x+5\ \sin\ x|\leq \sqrt{7^2+5^2}$$

$$\Rightarrow -\sqrt {7^{2} + 5^{2}} \leq (7\cos x + 5\sin x) \leq \sqrt {7^{2} + 5^{2}}$$

$$\Rightarrow -\sqrt {74} \leq (2K + 1) \leq \sqrt {74}$$

$$\Rightarrow -8.6 \leq (2K + 1)\leq 8.6$$

$$\Rightarrow -9.6 \leq 2K \leq 7.6$$

$$\Rightarrow -4.8 \leq K\leq 3.8$$

So, integral values of $$K$$ are

$$-4, -3, -2, -1, 0, 1, 2, 3$$ (eight values)

Hence, option B is correct.

Maths

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