Question

If $2{\sin }^{-1}{x}_0+{\tan }^{-1}{x}_0=\frac{5\pi }{\sqrt{-x_0^2+2x_0+15}},$ then the possible value(s) of $\frac{dy}{dx}$ to the curve $y=\frac{2x}{y+x}$ at $x=x_0$ is (are)

• A. $\frac{1}{3}$
• B. $-\frac{4}{3}$
• C. $-2$
• D. $\frac{1}{2}$

Answer 1

Answer: (A), (B)

$-\frac{4}{3}$

$2{\sin }^{-1}{x}_0+{\tan }^{-1}{x}_0=\frac{5\pi }{\sqrt{-x_0^2+2x_0+15}}$
$\Rightarrow 2{\sin }^{-1}{x}_0+{\tan }^{-1}{x}_0=\frac{5\pi }{\sqrt{16-(x_0-1{)}^2}}$
Clearly, $x_0\in [-1,1]$
$R.H.S.=\frac{5\pi }{\sqrt{16-(x_0-1{)}^2}}$
$R.H.S.\ge \frac{5\pi }{4}$
$L.H.S.=2{\sin }^{-1}{x}_0+{\tan }^{-1}{x}_0$
$L.H.S.\le \frac{5\pi }{4}$

So, the equation is true only when $L.H.S.=R.H.S.=\frac{5\pi }{4}$
$\therefore x_0=1$ is the only solution.

Now, $y=\frac{2x}{y+x}\cdots \left(1\right)$
At $x=1,y=\frac{2}{y+1}$
$\Rightarrow y^2+y-2=0$
$\Rightarrow y=1,-2$
From $\left(1\right),$
$y^2+xy=2x$
Differentiating w.r.t. $x,$ we get
$2y\frac{dy}{dx}+x\frac{dy}{dx}+y=2$
$\Rightarrow \frac{dy}{dx}=\frac{2-y}{2y+x}$

At $(1,1)$
$\frac{dy}{dx}{|}_{(1,1)}=\frac{2-1}{2+1}=\frac{1}{3}$

At $(1,-2)$
$\frac{dy}{dx}{|}_{(1,-2)}=\frac{2+2}{-4+1}=-\frac{4}{3}$