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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Similar questions

Q. 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 CORRECT combination?

$\begin{matrix} C o l u m n I & C o l u m n 2 & C o l u m n 3 (I) & f (x) = t a n (2 t a n^{- 1} \sqrt{\frac{\sqrt{1 + \sqrt{x}} - 1}{\sqrt{1 + \sqrt{x}} + 1}}) & (i) & \int f (x) d x = \frac{4}{3} x^{\frac{3}{4}} + C & (P) & \int_{0}^{1} f (x) d x = \frac{4}{3} (I I) & f (x) = c o t (2 t a n^{- 1} \sqrt{\frac{\sqrt{1 + \sqrt{x}} - \sqrt[4]{x}}{\sqrt{1 + \sqrt{x}} + \sqrt[4]{x}}}) & (i i) & \int f (x) d x = \frac{4}{5} x^{\frac{5}{4}} + C & (Q) & \int_{0}^{1} f (x) d x = \frac{4}{5} (I I I) & f (x) ⎛ ⎜ ⎝ \frac{1 - t a n (\frac{1}{2} s i n^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}))}{1 + t a n (\frac{1}{2} s i n^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}))} ⎞ ⎟ ⎠ & (i i i) & \int f (x) d x = \frac{2}{3} x^{\frac{3}{4}} + C & (R) & \int_{0}^{1} f (x) d x = \frac{2}{3} (I V) & f (x) = \sqrt{x} t a n (2 t a n^{- 1} (\frac{\sqrt{\sqrt{1 + \sqrt{x} + 1}} - \sqrt{\sqrt{1 + \sqrt{x} - 1}}}{\sqrt{\sqrt{1 + \sqrt{x} + 1}} + \sqrt{\sqrt{1 + \sqrt{x} - 1}}})) & (i v) & \int f (x) d x = \frac{2}{5} x^{\frac{5}{4}} + C & (S) & \int_{0}^{1} f (x) d x = \frac{2}{5} \end{matrix}$

Which of the following options is only correct combination?

\begin{matrix} List- I & List-II (I) & Number of integral solutions of & (P) & 132 x + y + z = 1, x \geq - 4, y \geq - 4, z \geq - 4 is less than (Q) & 99 (I I) & Greatest term in the expression of \frac{4}{3 \sqrt{2}} {(1 + \frac{1}{\sqrt{2}})}^{12} is (R) & 120 (I I I) & If a_{1}, a_{2}, a_{3} . . . a_{100} are in H.P. then value of \sum_{i = 1}^{99} \frac{a_{i} a_{i + 1}}{a_{1} a_{100}} is - (S) & 100 (I V) & If 8 points out of 11 are in same straight line then number of triangles formed is less then & (T) & 125 \end{matrix}

Which of the following is only INCORRECT combination?

\begin{matrix} List I & List II (I) \int_{0}^{2} \frac{x^{4} + 1}{(x^{2} + 2)^{\frac{3}{4}}} d x = & (P) \sqrt{3 \sqrt{2}} (II) \int_{1}^{3} (\sqrt{1 + (x - 1)^{3}} + \sqrt[3]{x^{2} - 1}) d x = & (Q) \sqrt{2 \sqrt{3}} (III) \int_{0}^{2} \sqrt[3]{x^{3} - 3 x^{2} + 8 x - 6} d x = & (R) 6 (IV) \int_{e}^{π} \frac{2^{\frac{x}{e}} - 2^{\frac{π}{x}}}{x} d x = & (S) 0 (T) \frac{π - e}{2} \end{matrix}

Which of the following is only CORRECT combination?

A line meets x-axis and y-axis at A and B respectively and O is the origin.

$\begin{matrix} Column I & Column 2 & Column 3 Equation of AB & Area of Δ O A B (I) & Centroid Δ O A B is (1, 2) & (i) & 2 x + y = 2 & (P) & 6 sq. units (II) & Circumcenter of Δ O A B is (1, 2) & (ii) & 3 x + 4 y = 12 & (Q) & 9 sq. units (III) & Distance of the orthocentre of Δ O A B & (iii) & 2 x + y = 6 & (R) & 1 sq. units From A and B is 1 and 2 respectively (IV) & Incenter of Δ O A B is (1, 1) & (iv) & 2 x + y = 4 & (S) & 4 sq. units \end{matrix}$

Which of the following is correct combination?

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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?