Question

The number of integral values of ‘$k$’ for which the equation $3\sin x+4\cos x=k+1$ has a solution, $k\in R$ is

Answer 1

Find the integral values of $k$.

An equation $3\sin x+4\cos x=k+1$ is given.

We know that, the maximum and minimum values of the expression $a\cos x+b\sin x$ can be given by $\sqrt{a^2+b^2}$ and $-\sqrt{a^2+b^2}$ respectively.

So,

$-\sqrt{3^2+4^2}\le 3\sin x+4\cos x\le \sqrt{3^2+4^2}$

From the given equation,

$$\begin{gathered} -\sqrt{9+16}\le k+1\le \sqrt{9+16} \\ \Rightarrow -\sqrt{25}\le k+1\le \sqrt{25} \\ \Rightarrow -5\le k+1\le 5 \\ \Rightarrow -6\le k\le 4 \end{gathered}$$

Therefore, the range of $k$ is $\left[-6,4\right]$. There are $11$ integers in the given interval.

Thus, the total integral values of $k$ is $11$.