Question

The number of solutions $x$ of the equation $\sin (x+x^2)–\sin \left(x^2\right)=\sin x$ in the interval $[2,3]$ is

• A. 0.0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is C 2

Given :
$\sin (x+x^2)–\sin \left(x^2\right)=\sin x$
The interval is $[2,3]$
Now,
$\sin (x+x^2)=\sin \left(x^2\right)+\sin x$
$\Rightarrow 2\sin \left(\frac{x+x^2}{2}\right)\cos \left(\frac{x+x^2}{2}\right)=2\sin \left(\frac{x+x^2}{2}\right)\cos \left(\frac{x-x^2}{2}\right)$
$\Rightarrow 2\sin \left(\frac{x+x^2}{2}\right)[\cos \left(\frac{x+x^2}{2}\right)-\cos \left(\frac{x-x^2}{2}\right)]=0$
$\Rightarrow -2\sin \left(\frac{x+x^2}{2}\right)2\sin \left(\frac{x}{2}\right)\sin \left(\frac{x^2}{2}\right)=0$
So
$\sin \left(\frac{x^2}{2}\right)=0$ or $\sin \left(\frac{x}{2}\right)=0$
or $\sin \left(\frac{x+x^2}{2}\right)=0$

Now in interval $[2,3]$
$\sin \left(\frac{x^2}{2}\right)=0\Rightarrow x=\sqrt{2\pi }$
$\sin \left(\frac{x}{2}\right)=0\Rightarrow x=0,2\pi ,....$ no values in the given inetrval
$\sin \left(\frac{x+x^2}{2}\right)=0\Rightarrow \frac{x+x^2}{2}=n\pi$
$\Rightarrow x^2+x-2n\pi =0$
$\Rightarrow x=\frac{-1\pm \sqrt{8n\pi +1}}{2}$
checking for (neglecting the negative values)
$n=0\Rightarrow x=0$
$n=1\Rightarrow x=\frac{-1+\sqrt{8\pi +1}}{2}$
$\left[\frac{-1+\sqrt{8\pi +1}}{2}\right]\in [2,3]$
$n=2\Rightarrow x=\frac{-1+\sqrt{8\times 2\pi +1}}{2}$
$\left[\frac{-1+\sqrt{8\times 2\pi +1}}{2}\right]>3$

Hence the given equation has two solutions
$x=\sqrt{2\pi },\frac{-1+\sqrt{8\pi +1}}{2}$