Question

If $0<a<5,0<b<5$ and $\frac{x^2+5}{2}=x-2cos(a+bx)$ is satisfied for at least one real $x$, then the greatest value of $a+b$ is

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $3\pi$
• D. $4\pi$

Answer 1

The correct option is C $3\pi$

Given,$\frac{x^2+5}{2}=x-2cos(a+bx)$
$\frac{x^2-2x+5}{2}=-2cos(a+bx)$
$\frac{(x-1{)}^2+4}{2}=-2cos(a+bx)$
$2+\frac{(x-1{)}^2}{2}=-2cos(a+bx)$
Here, $2+\frac{(x-1{)}^2}{2}⩾2$ and $-2⩽2+\frac{(x-1{)}^2}{2}⩽2$
$\Rightarrow 2+\frac{(x-1{)}^2}{2}=2$
$\Rightarrow x=1$ can be the only solution. $-2cos(a+b)=2$
$cos(a+b)=-1$
$a+b=(2n+1)\pi$

$\Rightarrow a+b=3\pi$ $(\because 3\pi <10<5\pi )$ is the greatest value.
Hence, option 'C' is correct.