CUET Preparation Today
If $y =\sqrt{x+\sqrt{x+\sqrt{x+.....}}}$, then |
$(2y + 1)\frac{dy}{dx} + 3\sqrt{x} = 0$ $3\frac{dy}{dx} + 3y^2= 0$ $3x\frac{dy}{dx} +(2y + 1)= 0$ $(2y - 1)\frac{dy}{dx} -1= 0$ |
$(2y - 1)\frac{dy}{dx} -1= 0$ |
The correct answer is Option (4) → $(2y - 1)\frac{dy}{dx} -1= 0$ Given: $y = \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}}$ Let $y = \sqrt{x + y}$ Square both sides: $y^2 = x + y \Rightarrow y^2 - y - x = 0$ Differentiating both sides w.r.t $x$: $2y \frac{dy}{dx} - \frac{dy}{dx} - 1 = 0 \Rightarrow (2y - 1)\frac{dy}{dx} = 1$ |
