Question

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

• A. $\frac{7\pi }{2}$
• B. $\frac{5\pi }{2}$
• C. $\frac{\pi }{2}$
• D. No solution

Answer 1

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

$(x-1{)}^2+3+3\sin (ax+b)=0$
$(x-1{)}^2+3=-3\sin (ax+b)$
$L.H.S\ge 3,$
$\Rightarrow \sin (ax+b)=-1$ and $x=1$
$\Rightarrow \sin (b+a)=-1$