The number of integral values of k,for which | vedclass

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(B) The range of the expression $a\cos x + b\sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.For the equation $7\cos x + 5\sin x = 2k + 1$,the range

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$8$

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(B) The range of the expression $a\cos x + b\sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.For the equation $7\cos x + 5\sin x = 2k + 1$,the range of $7\cos x + 5\sin x$ is $[-\sqrt{7^2 + 5^2}, \sqrt{7^2 + 5^2}] = [-\sqrt{74}, \sqrt{74}]$.Since $\sqrt{74} \approx 8.6$,the condition for a solution is $-\sqrt{74} \le 2k + 1 \le \sqrt{74}$.Substituting the approximate value: $-8.6 \le 2k + 1 \le 8.6$.Subtracting $1$ from all parts: $-9.6 \le 2k \le 7.6$.Dividing by $2$: $-4.8 \le k \le 3.8$.The integral values of $k$ are $\{-4, -3, -2, -1, 0, 1, 2, 3\}$.Counting these,we find there are $8$ such values.

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

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