Question:
The number of integral values of 'k' for which the equation \(3\sin x+4\cos x=k+1\) has a solution, \(k\in\mathbb{R}\) is ___________.
The number of integral values of 'k' for which the equation \(3\sin x+4\cos x=k+1\) has a solution, \(k\in\mathbb{R}\) is ___________.
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\(3\sin x+4\cos x\) ranges over \([-5,5]\), so \(-5\le k+1\le5\).
Updated On: Nov 4, 2026
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Correct Answer: 11
Solution and Explanation
\(3\sin x+4\cos x=5\sin(x+\alpha)\) has range \([-\sqrt{3^2+4^2},\sqrt{3^2+4^2}]=[-5,5]\).
A solution exists iff \(-5\le k+1\le5\), i.e. \(-6\le k\le4\).
Integers: \(-6,-5,\dots,4\), which is \(4-(-6)+1=11\) values.
Answer: 11
A solution exists iff \(-5\le k+1\le5\), i.e. \(-6\le k\le4\).
Integers: \(-6,-5,\dots,4\), which is \(4-(-6)+1=11\) values.
Answer: 11
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