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 r

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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 the left side is $[-\sqrt{7^2 + 5^2}, \sqrt{7^2 + 5^2}] = [-\sqrt{74}, \sqrt{74}]$.Since $\sqrt{74} \approx 8.602$,we have $-8.602 \leq 2k + 1 \leq 8.602$.Subtracting $1$ from all sides: $-9.602 \leq 2k \leq 7.602$.Dividing by $2$: $-4.801 \leq k \leq 3.801$.The integral values of $k$ are $\{-4, -3, -2, -1, 0, 1, 2, 3\}$.Counting these values,we get $8$ integral 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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