Question

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

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

Hint:

Consider the ranges of the functions on both sides of the equation to find the conditions for real solutions.

Answer 1

The Correct Option is C

Step 1: Rewrite the equation.
\( (x - 1)^2 + 3 = -3\cos(ax + b) \)
Step 2: Analyze the ranges.
LHS range: \( [3, \infty) \) RHS range: \( [-3, 3] \)
Step 3: Condition for real solutions.
Equality occurs when both sides are equal to 3.
\( (x - 1)^2 + 3 = 3 \Rightarrow x = 1 \)
\( -3\cos(ax + b) = 3 \Rightarrow \cos(ax + b) = -1 \)

Step 4: Solve for \( a + b \).
Substitute \( x = 1 \): \( \cos(a + b) = -1 \)
\( a + b = (2n + 1)\pi \)

Step 5: Apply constraints \( 0 \leq a, b \leq 3 \).
\( 0 \leq a + b \leq 6 \).
The only value of \( (2n + 1)\pi \) in this range is \( \pi \) (for \( n = 0 \)).