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
B
8
C
10
D
12

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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