The number of integral values of k for which | vedclass

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(C) The given equation is $7 \cos x + 5 \sin x = 2k + 1$.We know that the range of the expression $a \cos x + b \sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{

Vedclass · Accepted Answer

$8$

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(C) The given equation is $7 \cos x + 5 \sin x = 2k + 1$.We know that the range of the expression $a \cos x + b \sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.Here,$a = 7$ and $b = 5$,so the range is $[-\sqrt{7^2 + 5^2}, \sqrt{7^2 + 5^2}] = [-\sqrt{74}, \sqrt{74}]$.Since $\sqrt{64} For the equation to have a solution,we must have $-\sqrt{74} \leq 2k + 1 \leq \sqrt{74}$.Substituting the approximate value: $-8.6 \leq 2k + 1 \leq 8.6$.Subtracting $1$ from all parts: $-9.6 \leq 2k \leq 7.6$.Dividing by $2$: $-4.8 \leq k \leq 3.8$.Since $k$ is an integer,the possible values for $k$ are $\{-4, -3, -2, -1, 0, 1, 2, 3\}$.The total number of such integral values is $8$.

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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