Question

Consider the curve $\sin x+\sin y=1$, lying in the first quadrant , then
$\begin{array}{cccc}& \text{List- I}& & \text{List-II}\\ \text{(I)}& \underset{x\to \pi /2}{lim}\frac{d^{2}y}{d{x}^2}=& \text{(P)}& 0\\ \text{(II)}& \underset{x\to 0^{+}}{lim}{x}^{3/2}\frac{d^{2}y}{d{x}^2}=& \text{(Q)}& 1\\ \text{(III)}& \underset{x\to 0^{+}}{lim}{x}^2\frac{d^{2}y}{d{x}^2}=& \text{(R)}& \frac{1}{\sqrt{2}}\\ \text{(IV)}& \underset{x\to \pi /2}{lim}\frac{dy}{dx}=& \text{(S)}& \frac{1}{2\sqrt{2}}\\ & & \text{(T)}& \sqrt{2}\\ & & \text{(U)}& 3\end{array}$

Which of the following is the only CORRECT combination?

• A. $I\to T$
• B. $II\to Q$
• C. $III\to R$
• D. $IV\to P$

Answer 1

The correct option is D $IV\to P$

$\cos x+\cos y\frac{dy}{dx}=0\cdots \left(1\right)\Rightarrow -\sin x+\cos y\frac{d^{2}y}{d{x}^2}-{\left(\frac{dy}{dx}\right)}^2\sin y=0\therefore \frac{d^{2}y}{d{x}^2}=\frac{\sin x+\sin y{\left(\frac{dy}{dx}\right)}^2}{\cos y}$

From$\left(1\right)$, we get
$\frac{d^{2}y}{d{x}^2}=\frac{\sin x{\cos }^{2}y+\sin y{\cos }^{2}x}{{\cos }^{3}y}=\frac{\sin x(1-{\sin }^{2}y)+\sin y(1-{\sin }^{2}x)}{{\cos }^{3}y}$
$\sin y=1-\sin x\Rightarrow {\sin }^{2}y=1+{\sin }^{2}x-2\sin x\Rightarrow 1-{\cos }^{2}y=1+{\sin }^{2}x-2\sin x\Rightarrow {\cos }^{2}y=\sin x(2-\sin x)$

Now,
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{\sin x-\sin x{\sin }^{2}y+\sin y-{\sin }^{2}x\sin y}{\left(\sin x{)}^{3/2}\right(2-\sin x{)}^{3/2}}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{1-\sin x+{\sin }^{2}x}{\left(\sin x{)}^{3/2}\right(2-\sin x{)}^{3/2}}$

$\left(I\right)$
$\underset{x\to \pi /2}{lim}\frac{d^{2}y}{d{x}^2}=\underset{x\to \pi /2}{lim}\frac{1-\sin x+{\sin }^{2}x}{\left(\sin x{)}^{3/2}\right(2-\sin x{)}^{3/2}}=1$
$\left(I\right)\to \left(Q\right)$

$\left(II\right)$
$\underset{x\to 0^{+}}{lim}{x}^{3/2}\frac{dy}{dx}=\underset{x\to 0^{+}}{lim}\frac{x^{3/2}(1-\sin x+{\sin }^{2}x)}{\left(\sin x{)}^{3/2}\right(2-\sin x{)}^{3/2}}=\frac{1}{2^{3/2}}=\frac{1}{2\sqrt{2}}$
$\left(II\right)\to \left(S\right)$

$\left(III\right)$
$\underset{x\to 0^{+}}{lim}{x}^2\frac{dy}{dx}=\underset{x\to 0^{+}}{lim}\frac{x^{3/2}(1-\sin x+{\sin }^{2}x)x^{1/2}}{{\sin }^{3/2}x(2-\sin x{)}^{3/2}}=\frac{1}{2^{3/2}}\times 0=0$
$\left(III\right)\to \left(P\right)$

$\left(IV\right)$
$\underset{x\to \pi /2}{lim}\frac{dy}{dx}=\underset{x\to \pi /2}{lim}\frac{-\cos x}{\cos y}=0[\because \cos y=\sqrt{\sin x(2-\sin x)}]$
$\left(IV\right)\to \left(P\right)$