Question

If $x^2$ + 5 = 2x - 4 cos (a + bx) where a, b $ϵ$ (0, 5) is satisfied for alteast one real x, then the minimum value of a + bx is

• A. 0
• B. $\frac{\pi }{2}$
• C. $\pi$
• D. $\frac{3\pi }{2}$

Answer 1

The correct option is C

$\pi$

$x^2$ + 5 = 2x - 4 cos (a + bx)
$x^2$ - 2x + 1 + 4 = - 4 cos (a + bx)
$(x-1{)}^2$ + 4 (1 + cos (a + bx)) = 0
$\therefore$ x - 1 = 0, cos (a + bx) = -1
$\therefore$ minimum of a + bx = $\pi$