Question

The number of solutions $x$ of the equation $\sin\, (x+x^{2})-\sin\, (x^{2})=\sin\,x$ in the interval $[2, 3]$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

Given,
$\sin (x+x^{2})-\sin (x^{2})=\sin\,x$
$\Rightarrow \sin(x+x^{2})=\sin(x^{2})+\sin\,x$
$\Rightarrow 2\,\sin \left(\frac{x+x^{2}}{2}\right)\cos \left(\frac{x+x^{2}}{2}\right)$
$=2\sin \left(\frac{x^{2}+x}{2}\right)\cos \left(\frac{x^{2}-x}{2}\right)$
$\sin \left(\frac{x+x^{2}}{2}\right)=0 $
or $\cos\left(\frac{x^{2}+x}{2}\right)-\cos\left(\frac{x^{2}-x}{2}\right)=0$
$\frac{x^{2}+x}{2}=x\pi$ or $2\sin \frac{x^{2}}{2} \sin \frac{x}{2}=0$
$\Rightarrow \frac{x^{2}+x}{2}=0, \pi, 2\pi$, or $\frac{x^{2}}{2}=0, \pi, \frac{n}{2}$
$=0, \pi, 2\pi$
$\Rightarrow x^{2}+x=2\pi$ or $x^{2}=2\pi$
$\Rightarrow x^{2}+x-2\pi=0$ or $x=\sqrt{2\pi}$
$\Rightarrow x=\frac{-1\pm\sqrt{1+8\pi}}{2}$ or $2 <\,\sqrt{2}\pi<\,3$
$\Rightarrow \sqrt{1+8\pi}=\sqrt{25.14}$
$\therefore x=\frac{5.2-1}{2}$
$=\frac{4.2}{2}=2.1$
$\because$ Total numbers of solution lies between $\left(2,3\right)=2$