Question

The smallest possible integral value of $k$ for which the equation $7\cos x+5\sin x=2k+1$ has a solution is

• A. $-2$
• B. $-3$
• C. $-5$
• D. $-4$

Answer 1

The correct option is D $-4$

Given equation is,

$7\cos x+5\sin x=2k+1$

$\begin{array}{c}\Rightarrow \frac{7}{\sqrt{7^2+5^2}}\cos x+\frac{5}{\sqrt{7^2+5^2}}\sin x=\frac{2k+1}{\sqrt{7^2+5^2}}\\ \Rightarrow \sin \varphi \cos x+\cos \varphi \sin x=\frac{2k+1}{\sqrt{49+25}}\\ \Rightarrow \sin \left(x+\varphi \right)=\frac{2k+1}{\sqrt{74}}\end{array}$

Here $\tan \varphi =\frac{7}{5}$.

As we know that the range of sine function is from $[-1,1]$. So,

$\begin{array}{c}\Rightarrow -1\le \sin \left(x+\varphi \right)\le 1\\ \Rightarrow -1\le \frac{2k+1}{\sqrt{74}}\le 1\\ \Rightarrow -\sqrt{74}\le 2k+1\le \sqrt{74}\\ \Rightarrow -4.8\le k\le 3.8\end{array}$

Here we can conclude that the smallest possible integer value of $k$ is equal to $2$.