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. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

We know $\sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$

$\therefore \sin (x+x^2)-\sin \left(x^2\right)=2\cos \left(\frac{x+x^2+x^2}{2}\right)\sin \left(\frac{x+x^2-x^2}{2}\right)$

$=2\cos \left(\frac{x+2x}{2}\right)\sin \frac{x}{2}$

and $\sin x=2\cos \frac{x}{2}\sin \frac{x}{2}$

Now, $\sin (\lambda +x^2)-\sin \left(x^2\right)=\sin x$

or, $2\cos \left(\frac{x+2x^2}{2}\right)\sin \frac{x}{2}=2\cos \frac{x}{2}\sin \frac{x}{2}$

or, $\frac{x+2x^2}{2}=\frac{x}{2}\Rightarrow x+2x^2=x\Rightarrow x=0$

$\therefore$ In any given internal say $[2,3]$

the no.of soln $=0$