Question

If the equation $x^2+4+3\sin (ax+b)-2x=0$ has at least one real solution, where $a,b\in [0,2\pi ],$ then a possible value of $(a+b)$ can be

• A. $\frac{5\pi }{2}$
• B. $\frac{7\pi }{2}$
• C. $\frac{\pi }{2}$
• D. None of these

Answer 1

The correct option is B $\frac{7\pi }{2}$

$x^2+4+3\sin (ax+b)-2x=0$
$\Rightarrow (x-1{)}^2+3+3\sin (ax+b)=0$
$\Rightarrow (x-1{)}^2+3=-3\sin (ax+b)$
L.H.S. $\ge 3$ and R.H.S. $\in [-3,3]$
$\Rightarrow$ Only possibility is $\sin (ax+b)=-1$ and $x=1$
$\Rightarrow \sin (a+b)=-1$
$\Rightarrow a+b=(4n+3)\frac{\pi }{2},n\in Z$