MAH-CET 2026 — Mathematics PYQ

1. First Differentiation:

We are given the relation:

$$y=\log (m{\cos }^{-1}x)$$

By taking the exponential on both sides, we can rewrite it to make differentiation simpler:

$$e^y=m{\cos }^{-1}x\text{— (Equation 1)}$$

Differentiating both sides with respect to $x$ using the chain rule:

$$e^y\cdot \frac{dy}{dx}=m\cdot \left(\frac{-1}{\sqrt{1-x^2}}\right)$$

Rearranging the terms to clear the fraction:

$$\sqrt{1-x^2}\cdot e^y\cdot \frac{dy}{dx}=-m\text{— (Equation 2)}$$

2. Second Differentiation:

Squaring both sides of Equation 2 to eliminate the square root:

$$(1-x^2)\cdot e^{2y}\cdot {\left(\frac{dy}{dx}\right)}^2=m^2$$

Now, differentiating both sides with respect to $x$ using the product rule:

$$\frac{d}{dx}\left(1-x^2\right)\cdot \left[e^{2y}{\left(\frac{dy}{dx}\right)}^2\right]+(1-x^2)\cdot \frac{d}{dx}\left[e^{2y}{\left(\frac{dy}{dx}\right)}^2\right]=0$$

Let's compute the derivatives step-by-step:

• $\frac{d}{dx}(1-x^2)=-2x$

• Using the product and chain rule on the second part:

  $$\frac{d}{dx}\left[e^{2y}{\left(\frac{dy}{dx}\right)}^2\right]=\left(2e^{2y}\cdot \frac{dy}{dx}\right)\cdot {\left(\frac{dy}{dx}\right)}^2+e^{2y}\cdot \left(2\frac{dy}{dx}\cdot \frac{d^{2}y}{d{x}^2}\right)$$

  $$=2e^{2y}{\left(\frac{dy}{dx}\right)}^3+2e^{2y}\left(\frac{dy}{dx}\right)\left(\frac{d^{2}y}{d{x}^2}\right)$$

Substitute these back into our main derivative equation:

$$-2x\cdot e^{2y}{\left(\frac{dy}{dx}\right)}^2+(1-x^2)\left[2e^{2y}{\left(\frac{dy}{dx}\right)}^3+2e^{2y}\left(\frac{dy}{dx}\right)\left(\frac{d^{2}y}{d{x}^2}\right)\right]=0$$

3. Simplification:

We can factor out a common term of $2e^{2y}\left(\frac{dy}{dx}\right)$ from the entire equation (since $e^{2y}\ne 0$ and $\frac{dy}{dx}\ne 0$):

$$2e^{2y}\left(\frac{dy}{dx}\right)\left[-x\left(\frac{dy}{dx}\right)+(1-x^2){\left(\frac{dy}{dx}\right)}^2+(1-x^2)\left(\frac{d^{2}y}{d{x}^2}\right)\right]=0$$

This leaves us with:

$$(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+(1-x^2){\left(\frac{dy}{dx}\right)}^2=0$$

Isolating the terms to match the form given in the problem statement:

$$(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=-(1-x^2){\left(\frac{dy}{dx}\right)}^2\text{— (Equation 3)}$$

4. Finding the value of $k$:

From our earlier squared equation, we know that:

$$(1-x^2)\cdot {\left(\frac{dy}{dx}\right)}^2=\frac{m^2}{e^{2y}}=m^2{e}^{-2y}$$

Substitute this value into the right-hand side of Equation 3:

$$(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=-\left(m^2{e}^{-2y}\right)$$

Wait, let's re-verify the sign from step 2 carefully.

From Equation 2:

$$\sqrt{1-x^2}\frac{dy}{dx}=-m{e}^{-y}$$

Differentiating directly without squaring:

$$\frac{-x}{\sqrt{1-x^2}}\frac{dy}{dx}+\sqrt{1-x^2}\frac{d^{2}y}{d{x}^2}=-m{e}^{-y}\left(-\frac{dy}{dx}\right)=m{e}^{-y}\frac{dy}{dx}$$

Multiply the entire equation by $\sqrt{1-x^2}$:

$$-x\frac{dy}{dx}+(1-x^2)\frac{d^{2}y}{d{x}^2}=m{e}^{-y}\sqrt{1-x^2}\frac{dy}{dx}$$

From Equation 2, we know that $\sqrt{1-x^2}\frac{dy}{dx}=-m{e}^{-y}$. Substitute this into the right side:

$$(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=m{e}^{-y}\left(-m{e}^{-y}\right)$$

$$(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=-m^2{e}^{-2y}$$

Let's double check if there's an issue with the standard question textbook print. In many standard versions of this problem, the right-hand side is written as $-m^2{e}^{-2y}$, making $k=-m^2$. Let's re-verify option formatting: if the question specifies $=k{e}^{-2y}$, then $k=-m^2$.

Let's carefully verify option (c):

By direct matching, the coefficient of $e^{-2y}$ is $-m^2$. Thus, option (c) is the correct choice.