JEE PYQ: Differentiation Question 9
Question 9 - 2020 (07 Jan Shift 2)
Let $y = y(x)$ be a function of $x$ satisfying $y\sqrt{1-x^2} = k - x\sqrt{1-y^2}$ where $k$ is a constant and $y\left(\frac{1}{2}\right) = -\frac{1}{4}$. Then $\frac{dy}{dx}$ at $x = \frac{1}{2}$ is equal to:
(1) $-\frac{\sqrt{5}}{4}$ (2) $-\frac{\sqrt{5}}{2}$ (3) $\frac{2}{\sqrt{5}}$ (4) $\frac{\sqrt{5}}{2}$
Show Answer
Answer: (2) $-\frac{\sqrt{5}}{2}$
Solution
Implicit differentiation at the given point yields $y’ = -\frac{\sqrt{5}}{2}$.
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