$\sin^{-1}(\frac{2x}{1+x^2})+px=0$ has exactly 3 solution

Question

If the set of values of $p$ for which the equation $\sin^{-1}(\frac{2x}{1+x^2})+px=0$ has exactly 3 solution is $(l,m)$.Then what is $2m-l$?

$(A)2\hspace{1cm}(B)4\hspace{1cm}(C)5\hspace{1cm}(D)6\hspace{1cm}(E)7$

$\sin^{-1}(\frac{2x}{1+x^2})+px=0$

$\Rightarrow 2\tan^{-1}x+px=0$

$\Rightarrow \tan^{-1}x=\frac{-px}{2}$

How should i do now?Please help...

Answer 1

You can see that the graph of $\tan^{-1}x$ is concave down for positive $x$ and concave up for negative $x$ and the tangent line to the graph at $x=0$ has slope $1$. So if $0<-\frac p2<1$, then the line $y=-\frac p2x$ meets the graph $y=\tan^{-1}x$ just once in the positive part and just once in the negative part (and of course once at the origin). For $-\frac p2\geq1$ and $-\frac p2\leq0$ they only meet at the origin. So $p\in(-2,0)$.

Answer 2

Hint: Graph what $\sin ( 2x / (1+x^2))$, you'll see the only way you can have a line passing through $(0,0)$ and 2 other points of the graph is if the line has slope smaller than 0 and bigger than $-2$ Since $$ \frac{2x}{1+x^2} = \sin(-px) \implies 2x \approx -px \quad \text{around zero}$$

Answer 3

Hint: For the equation to have exactly three solutions, the first requirement would be, that there would be more that $3$ points, where the slope of the tangent at those points would be $0$. For that, take the derivative of $f(x)=\sin^{-1}(\frac{2x}{1+x^2})+px$. Equate it to zero. Now $f'(x)$ should have more than $3$ solutions. Add this constraint and solve to get the interval for $p$.