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The number of integral values of $k$ for which the equation $7 \cos x+5 \sin x=2 k+1$ has a solution, is
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Verified Answer
The correct answer is:
8
We have,
$$
7 \cos x+5 \sin x=2 k+1
$$
Maximum and minimum value of $7 \cos x+5 \sin x$ is $\sqrt{49+25},-\sqrt{49+25} \Rightarrow \sqrt{74},-\sqrt{74}$
$\therefore-\sqrt{74} \leq 2 k+1 \leq \sqrt{74}$
$\because k$ is an integer
$$
\Rightarrow-8 \leq 2 k+1 \leq 8 \Rightarrow-9 \leq 2 k \leq 7 \Rightarrow-\frac{9}{2} \leq k \leq \frac{7}{2}
$$
$\therefore \quad k=\{-4,-3,-2,-1,0,1,2,3\}$
Total number of $k=8$
$$
7 \cos x+5 \sin x=2 k+1
$$
Maximum and minimum value of $7 \cos x+5 \sin x$ is $\sqrt{49+25},-\sqrt{49+25} \Rightarrow \sqrt{74},-\sqrt{74}$
$\therefore-\sqrt{74} \leq 2 k+1 \leq \sqrt{74}$
$\because k$ is an integer
$$
\Rightarrow-8 \leq 2 k+1 \leq 8 \Rightarrow-9 \leq 2 k \leq 7 \Rightarrow-\frac{9}{2} \leq k \leq \frac{7}{2}
$$
$\therefore \quad k=\{-4,-3,-2,-1,0,1,2,3\}$
Total number of $k=8$
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