Question

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

• A. $\frac{7\pi }{2}$
• B. $\frac{3\pi }{2}$
• C. $\frac{9\pi }{2}$
• D. $noneofthese$

Answer 1

The correct option is B $\frac{3\pi }{2}$

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

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

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

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

Now consider $L.H.S$ $\Rightarrow (x-1{)}^2+3$

We know that $\Rightarrow 0\le (x-1{)}^2<\infty$

$\Rightarrow 3\le (x-1{)}^2+3<\infty$

Now consider $R.H.S$ $\Rightarrow -3sin(ax+b)$

We know that $\Rightarrow -1\le sin(ax+b)\le 1$

$\Rightarrow -3\le sin(ax+b)\le 3$

$\Rightarrow 3\ge sin(ax+b)\ge -3$

So, the only possible value which satisfies these condition is $L.H.S=R.H.S=3$

First, Consider $L.H.S=3$

$\Rightarrow (x-1{)}^2+3=3$

$\Rightarrow (x-1{)}^2=0$

$\Rightarrow x=1$

Now, Consider $R.H.S=3$

$\Rightarrow -3sin(ax+b)=3$

Substitute the value of $x$ obtained in the above equation, we get

$\Rightarrow -3sin(a+b)=3$

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

$\Rightarrow a+b=si{n}^{-1}(-1)$

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