MAH-CET 2026 — Mathematics PYQ

Step 1: Apply Componendo and Dividendo

We are given the equation:

$$\frac{x}{1}=\frac{e^{2y}+1}{e^{2y}-1}$$

Applying the property $\frac{a}{b}=\frac{c}{d}⟹\frac{a+b}{a-b}=\frac{c+d}{c-d}$:

$$\frac{x+1}{x-1}=\frac{(e^{2y}+1)+(e^{2y}-1)}{(e^{2y}+1)-(e^{2y}-1)}$$

$$\frac{x+1}{x-1}=\frac{2e^{2y}}{2}$$

$$\frac{x+1}{x-1}=e^{2y}$$

Step 2: Take natural logarithm on both sides

Taking $\ln$ on both sides to isolate $y$:

$$\ln \left(\frac{x+1}{x-1}\right)=2y$$

$$\ln (x+1)-\ln (x-1)=2y$$

$$y=\frac{1}{2}[\ln (x+1)-\ln (x-1)]$$

Step 3: Differentiate with respect to $x$

Differentiating both sides using the chain rule:

$$\frac{dy}{dx}=\frac{1}{2}\left[\frac{1}{x+1}-\frac{1}{x-1}\right]$$

Take a common denominator inside the bracket:

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

$$\frac{dy}{dx}=\frac{1}{2}\left[\frac{-2}{x^2-1}\right]$$

$$\frac{dy}{dx}=\frac{-1}{x^2-1}$$

Multiplying the negative sign into the denominator:

$$\frac{dy}{dx}=\frac{1}{1-x^2}$$