CUET UG 2025 — Mathematics Algebra

Given: $x = \frac{1-t}{1+t}$x=1+t1−t​ and $y = \frac{3t}{1+t}$y=1+t3t​

Both $x$x and $y$y are given in terms of $t$t, so:

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$$dxdy​=dx/dtdy/dt​

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For $x = \frac{1-t}{1+t}$x=1+t1−t​:

$$\frac{dx}{dt} = \frac{(1+t)(-1) - (1-t)(1)}{(1+t)^2}$$dtdx​=(1+t)2(1+t)(−1)−(1−t)(1)​

$$= \frac{-1-t - 1+t}{(1+t)^2}$$=(1+t)2−1−t−1+t​

$$= \frac{-2}{(1+t)^2}$$=(1+t)2−2​

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For $y = \frac{3t}{1+t}$y=1+t3t​:

$$\frac{dy}{dt} = \frac{(1+t)(3) - (3t)(1)}{(1+t)^2}$$dtdy​=(1+t)2(1+t)(3)−(3t)(1)​

$$= \frac{3+3t - 3t}{(1+t)^2}$$=(1+t)23+3t−3t​

$$= \frac{3}{(1+t)^2}$$=(1+t)23​

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$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$$dxdy​=dx/dtdy/dt​

$$= \frac{\frac{3}{(1+t)^2}}{\frac{-2}{(1+t)^2}}$$=(1+t)2−2​(1+t)23​​

$$= \frac{3}{-2}$$=−23​

$$= -\frac{3}{2}$$=−23​

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Since $\frac{dy}{dx} = -\frac{3}{2}$dxdy​=−23​ is a constant:

$$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{3}{2}\right) = 0$$dx2d2y​=dxd​(−23​)=0

Therefore, $\frac{d^2y}{dx^2} = 0$dx2d2y​=0