Match List-I with List-II (x > 0, y > 0, and in (C), \(y \ne \displaystyle\frac{1}{2}\))
| List-I | List-II | ||
|---|---|---|---|
| Relation | \(\displaystyle\frac{dy}{dx}\) | ||
| (A) | \(\sqrt{x} + \sqrt{y} = 3\) | (I) | \(\displaystyle\frac{3x^{2}}{2y}\) |
| (B) | \(x^{2}y = 5\) | (II) | \(-\sqrt{\displaystyle\frac{y}{x}}\) |
| (C) | \(y^{2} = x + y\) | (III) | \(\displaystyle\frac{1}{2y - 1}\) |
| (D) | \(y^{2} = x^{3}\) | (IV) | \(-\displaystyle\frac{2y}{x}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
In each case both sides are differentiated with respect to x, treating y as a function of x (implicit differentiation); y′ stands for \(\frac{dy}{dx}\).
(A) \(\sqrt{x} + \sqrt{y} = 3\)
This matches (II) \(-\sqrt{\frac{y}{x}}\).
\( \displaystyle \frac{1}{2\sqrt{x}} + \frac{y'}{2\sqrt{y}} = 0 \)
\( \displaystyle y' = -\frac{\sqrt{y}}{\sqrt{x}} \)
(B) \(x^{2}y = 5\)
This matches (IV) \(-\frac{2y}{x}\). By the product rule:
\(2xy + x^{2}y' = 0\)
\( \displaystyle y' = -\frac{2xy}{x^{2}} = -\frac{2y}{x} \)
(C) \(y^{2} = x + y\)
This matches (III) \(\frac{1}{2y - 1}\).
\(2yy' = 1 + y'\)
\( \displaystyle y'(2y - 1) = 1, \quad y' = \frac{1}{2y - 1} \)
(D) \(y^{2} = x^{3}\)
This matches (I) \(\frac{3x^{2}}{2y}\).
\(2yy' = 3x^{2}\)
\( \displaystyle y' = \frac{3x^{2}}{2y} \)