Question

The number of integral values of k for which the equation $7 \cos x + 5 \sin x = 2k + 1$ has solution is

• A. 4
• B. 8
• C. 10
• D. 12

Answer 1

Answer: (B)

8

Answer 2

Given $7 \cos x + 5 \sin x = 2k + 1$
We know that $-r \le a \cos\theta +b \sin\theta \le r$ where $r=\sqrt{a^{2} +b^{2}}$
$\therefore \, \, - \sqrt{49 +25} \le 7\cos x +5 \sin x \le\sqrt{49+25}$
$\Rightarrow - \sqrt{74} \le 2k +1 \le \sqrt{74}\Rightarrow -8.6 \le2k + 1 \le 8.6$
$\Rightarrow -9.6 \le2k \le 7.6 \Rightarrow -4.8 \le k \le 3.8$
$\Rightarrow \, \, k$ can take only 8 integral values.