Question

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

• A. $\frac{5\pi }{2}$
• B. $\frac{7\pi }{2}$
• C. $\frac{9\pi }{2}$
• D. $\frac{13\pi }{4}$

Answer 1

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

$x^2+4+3\sin (ax+b)-2x=0\Rightarrow (x-1{)}^2+3=-3\sin (ax+b)$
We know that,
$(x-1{)}^2+3\in [3,\infty )-3\sin (ax+b)\in [-3,3]$
So, the solution exists, when
$x=1$ and $\sin (a+b)=-1$
$\Rightarrow a+b=\frac{3\pi }{2},\frac{7\pi }{2}(\because a,b\in [0,2\pi \left]\right)$

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