If {y-{y-{y- ∞}}}={x+{x+{x+ ∞}}} then d y/d x=

Correct Answer is : y−x+1y−x−1y-x+1/y-x-1y−x−1y−x+1

Correct Answer is : y−x+1y−x−1y-x+1/y-x-1y−x−1y−x+1

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MHTCET 25 Apr 2025 Shift 2 Paper

Question : 143 of 150

Marks: +1, -0

\sqrt{y-\sqrt{y-\sqrt{y-\ldots \ldots \ldots \infty}}}=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \ldots \ldots \infty}}}

\frac{d y}{d x}=

Solution:

Concept:

Let the common value of both infinite radicals be

p

, then differentiate the resulting algebraic equations.

Explanation:

Let

\sqrt{y-\sqrt{y-\cdots \infty}}=\sqrt{x+\sqrt{x+\cdots \infty}}=p

From the first radical:

\sqrt{y-p}=p \Rightarrow y-p=p^2

Differentiating with respect to

y

1-\frac{dp}{dy}=2p\frac{dp}{dy}

\Rightarrow (2p+1)\frac{dp}{dy}=1

\Rightarrow \frac{dy}{dp}=2p+1

From the second radical:

\sqrt{x+p}=p \Rightarrow x+p=p^2

Differentiating with respect to

x

1+\frac{dp}{dx}=2p\frac{dp}{dx}

\Rightarrow (2p-1)\frac{dp}{dx}=1

\Rightarrow \frac{dx}{dp}=2p-1

Therefore,

\frac{dy}{dx}=\frac{\frac{dy}{dp}}{\frac{dx}{dp}}=\frac{2p+1}{2p-1}

Also, from

x+p=y-p

, we get

y-x=2p

, so

p=\frac{y-x}{2}

Substituting:

\frac{dy}{dx}=\frac{2\left(\frac{y-x}{2}\right)+1}{2\left(\frac{y-x}{2}\right)-1} =\frac{y-x+1}{y-x-1}

Answer:

\frac{dy}{dx}=\frac{y-x+1}{y-x-1}

, i.e. option C.

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