Question

If \( 0\le a,b \le 3 \) and the equation \[ x^2 + 4 + 3\cos(ax+b) = 2x \] has real solutions, then the value(s) of \( (a+b) \) is/are:

• A. \( \frac{\pi}{4} \)
• B. \( \frac{\pi}{2} \)
• C. \( \pi \)
• D. \( 2\pi \)

Hint:

For trig equations with quadratics: Compare value ranges. Match extreme values. 

Answer 1

The Correct Option is B

Concept: Rewrite equation: \[ x^2 - 2x + 4 = -3\cos(ax+b) \] LHS: \[ (x-1)^2 + 3 \ge 3 \] RHS range: \[ [-3,3] \] For equality: \[ (x-1)^2 + 3 \le 3 \Rightarrow x=1 \] Then: \[ \cos(a+b) = -1 \] So: \[ a+b = \pi \text{ or odd multiples} \] Within range: \[ \frac{\pi}{2}, \pi \]