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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Differentiation

JEE Advanced · Mathematics · Differentiation

If y= x+ x+ x+ x+ , where x>0 , then (d y)/(d x) can be:

If y=x+x+x+x+……y=\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots \ldots}}}}, where x>0x>0, then dydx\frac{d y}{d x} can be:

Options

  1. A12y−1\frac{1}{2 y-1}
  2. Bxx+2y\frac{x}{x+2 y}
  3. C11+4x\frac{1}{\sqrt{1+4 x}}
  4. Dy2x+y\frac{y}{2 x+y}

Correct answer

D. y2x+y\frac{y}{2 x+y}

Step-by-step solution

y=x+y⇒y2=x+y∴y2−y−x=0y=\sqrt{x+y} \quad \Rightarrow \quad y^2=x+y\therefore \quad y^2-y-x=0 ...(1) (2y−1)y′=1⇒y′=12y−1⇒(2 y-1) y^{\prime}=1\Rightarrow \quad y^{\prime}=\frac{1}{2 y-1} \quad \Rightarrow(a) From equation (1), 2y=1+1+4x2 y=1+\sqrt{1+4 x} Hence, 2y−1=1+4x⇒y′=11+4x⇒2 y-1=\sqrt{1+4 x}\Rightarrow \quad y^{\prime}=\frac{1}{\sqrt{1+4 x}} \quad \Rightarrow(c) On dividing by yy, equation (1) gives y−1=xy⇒2y−2=2xyy-1=\frac{x}{y} \quad \Rightarrow \quad 2 y-2=\frac{2 x}{y} (d) : y2x+y=12xy+1=1(2y−2)+1=12y−1\frac{y}{2 x+y}=\frac{1}{\frac{2 x}{y}+1}=\frac{1}{(2 y-2)+1}=\frac{1}{2 y-1} which is same as (a).
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