Question

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

• A. $\frac{7\pi }{2}$
• B. $\frac{5\pi }{2}$
• C. $\frac{9\pi }{2}$
• D. None of these

Answer 1

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

$x^2+4+3\sin (ax+b)-2x=0$

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

solution of the above equation is only possible if,

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

hence,

$\Rightarrow \sin (a+b)=-1$

therefore in given interval solutions are,

$a+b=\frac{3\pi }{2},\frac{7\pi }{2}$