Question

Thenumber of integral values of k, for which the equation

$7 \cos x + 5 \sin x = 2 k + 1$ has a solution is

• A. 4
• B. 8
• C. 10
• D. 12

Answer 1

Answer: (B)

8

Answer 2

$- \sqrt { 7 ^ { 2 } + 5 ^ { 2 } } \leq ( 7 \cos x + 5 \sin x ) \leq \sqrt { 7 ^ { 2 } + 5 ^ { 2 } }$ So, for solution

$- \sqrt { 74 } \leq ( 2 k + 1 ) \leq \sqrt { 74 }$ or $- 8.6 \leq 2 k + 1 \leq 8.6$ or

$- 9.6 \leq 2 k \leq 7.6$ or $- 4.8 \leq k \leq 3.8$.

So, integral values of k are $- 4 , - 3 , - 2 , - 1,0,1,2,3$

(eight values)