Question

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

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

Answer 1

The correct option is A

$\pi$

$x^2+5=2x-4cos(a+bx)$
$x^2+5-2x=2x-4cos(a+bx)-2x$
$x^2-2x+5=-4cos(a+bx)$
$x^2-2x+1+4=-4cos(a+bx)$
$x^2-2x+1+4+4cos(a+bx)=-4cos(a+bx)+4cos(a+bx)$
$(x-1{)}^2+4(1+cos(a+bx))=0$
$\therefore x-1=0\&cos(a+bx)+1=0$
$\therefore x=1\&cos(a+bx)=-1$
$\therefore$ Minimum value of $a+bx=\pi (\because cos\pi =-1)$