CUET UG 2025 — Mathematics Calculus

Given $y = \sqrt{ax + b}$y=ax+b​, find $y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2$ydx2d2y​+(dxdy​)2

Rewrite as $y = (ax + b)^{1/2}$y=(ax+b)1/2

$$\frac{dy}{dx} = \frac{1}{2}(ax + b)^{-1/2} \cdot a$$dxdy​=21​(ax+b)−1/2⋅a

$$\frac{dy}{dx} = \frac{a}{2\sqrt{ax + b}}$$dxdy​=2ax+b​a​

Since $y = \sqrt{ax + b}$y=ax+b​:

$$\frac{dy}{dx} = \frac{a}{2y}$$dxdy​=2ya​

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Starting from $\frac{dy}{dx} = \frac{a}{2}(ax + b)^{-1/2}$dxdy​=2a​(ax+b)−1/2:

$$\frac{d^2y}{dx^2} = \frac{a}{2} \cdot \left(-\frac{1}{2}\right)(ax + b)^{-3/2} \cdot a$$dx2d2y​=2a​⋅(−21​)(ax+b)−3/2⋅a

$$\frac{d^2y}{dx^2} = -\frac{a^2}{4(ax + b)^{3/2}}$$dx2d2y​=−4(ax+b)3/2a2​

Since $y = \sqrt{ax + b}$y=ax+b​, we have $(ax + b)^{3/2} = y^3$(ax+b)3/2=y3:

$$\frac{d^2y}{dx^2} = -\frac{a^2}{4y^3}$$dx2d2y​=−4y3a2​

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$$y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = y \cdot \left(-\frac{a^2}{4y^3}\right) + \left(\frac{a}{2y}\right)^2$$ydx2d2y​+(dxdy​)2=y⋅(−4y3a2​)+(2ya​)2

$$= -\frac{a^2}{4y^2} + \frac{a^2}{4y^2}$$=−4y2a2​+4y2a2​

$$= 0$$=0

Therefore, $y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0$ydx2d2y​+(dxdy​)2=0