If \(y = \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}}\), then
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle (2y - 1)\frac{dy}{dx} - 1 = 0\)
The expression under the first root is x plus the same infinite expression, which is y itself:
\(y = \sqrt{x + y}\)
\(y^2 = x + y\)
Differentiating with respect to x:
\( \displaystyle 2y\frac{dy}{dx} = 1 + \frac{dy}{dx} \)
\( \displaystyle (2y - 1)\frac{dy}{dx} - 1 = 0 \)