Question

If a, b are real numbers and $\alpha$ is a real root of $x^2+12+3\sin(a+bx)+6x=0$ then the value of $\cos(a+b\alpha)$ for the least positive value of $a+b\alpha$ is

• A. -1
• B. $1/\sqrt{2}$
• C. $1/2$
• D. 0

Hint:

When an equation mixes polynomial and trigonometric functions, try to find the range of each part. If the ranges only intersect at a single point, any solution must make both sides equal to that point.

Answer 1

The Correct Option is D

Step 1: Rearrange equation.
\[ x^2 + 6x + 12 + 3\sin(a+bx) = 0 \implies x^2+6x+12 = -3\sin(a+bx) \]
Step 2: Find LHS minimum.
\[ x^2+6x+12 = (x+3)^2 + 3 \implies \text{minimum} = 3 \]
Step 3: RHS range.
\[ -3 \le -3\sin(a+bx) \le 3 \]
Step 4: Intersection point.
Only possible if both sides equal 3: \[ x+3=0 \implies \alpha = -3, -3\sin(a+b\alpha)=3 \implies \sin(a-3b)=-1 \]
Step 5: Least positive angle.
\[ a+b\alpha = a-3b = 3\pi/2 \implies \cos(a+b\alpha) = \cos(3\pi/2) = 0 \]