Question

If $0\le a\le 3,0\le b\le 3$ and the equation $x^2+4+3cos(ax+b)=2x$ has at least
one solution then the value of $a+b$ is

• A. 0
• B. $\frac{\pi }{2}$
• C. $\pi$
• D. none of these

Answer 1

Answer: (C)

$\pi$

We have, $x^2+4+3\cos (ax+b)=2x$

$\Rightarrow (x-1{)}^2+3(1+\cos (ax+b))=0$

solution of the above equation is only possible if,

$x=1$ and $\cos (ax+b)=-1$

hence,
$\Rightarrow \cos (a+b)=-1\Rightarrow a+b=\pi$