Consider the curve sin x -sin y-1- lying in the first quadrant - then- List- I List-II- I limx-2d2ydx2- P 0- II limx- 0-x3-2d2ydx2- Q 1- -III limx-0-x2d2ydx2- R 1-2- IV limx-2dydx- S 12-2- T -2- U 3- - Which of the following is the only INCORRECT combination-

Cos x+cos ydydx=0 ⋯(1) ⇒ sin x+cos yd^2ydx^2 (dydx)^2sin y=0 ∴d^2ydx^2=sin x+sin y(dydx)^2cos y From(1), we get d^2ydx^2=sin xcos ^2y+sin ycos ^2xcos ^3 y ...

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Standard XIII

Mathematics

Implicit Differentiation

Consider the ...

Question

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

Which of the following is the only INCORRECT combination?

I \to T

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I I \to S

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I I I \to P

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I V \to P

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Solution

The correct option is A $I \to T$
$cos x + cos y \frac{d y}{d x} = 0 \dots (1) \Rightarrow - sin x + cos y \frac{d^{2} y}{d x^{2}} - {(\frac{d y}{d x})}^{2} sin y = 0 ∴ \frac{d^{2} y}{d x^{2}} = \frac{sin x + sin y {(\frac{d y}{d x})}^{2}}{cos y}$

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

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

$(I I)$
$lim x \to 0^{+} x^{3 / 2} \frac{d y}{d x} = lim x \to 0^{+} \frac{x^{3 / 2} (1 - sin x + {sin}^{2} x)}{(sin x)^{3 / 2} (2 - sin x)^{3 / 2}} = \frac{1}{2^{3 / 2}} = \frac{1}{2 \sqrt{2}}$
$(I I) \to (S)$

$(I I I)$
$lim x \to 0^{+} x^{2} \frac{d y}{d x} = lim x \to 0^{+} \frac{x^{3 / 2} (1 - sin x + {sin}^{2} x) x^{1 / 2}}{{sin}^{3 / 2} x (2 - sin x)^{3 / 2}} = \frac{1}{2^{3 / 2}} \times 0 = 0$
$(I I I) \to (P)$

$(I V)$
$lim x \to π / 2 \frac{d y}{d x} = lim x \to π / 2 \frac{- cos x}{cos y} = 0 [∵ cos y = \sqrt{sin x (2 - sin x)}]$
$(I V) \to (P)$ 1

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Consider the curve sinx+siny=1, lying in the first quadrant , then List- IList-II(I)limx→π/2d2ydx2=(P) 0(II)limx→0+x3/2d2ydx2=(Q) 1(III) limx→0+x2d2ydx2=(R) 1√2(IV)limx→π/2dydx=(S) 12√2(T) √2(U) 3 Which of the following is the only INCORRECT combination?