CUET UG 2025 — Mathematics Algebra

Given: $y = \log_e\left(\frac{e^2}{x^2}\right)$y=loge​(x2e2​)

Using the quotient rule for logarithms:

$$y = \log_e(e^2) - \log_e(x^2)$$y=loge​(e2)−loge​(x2)

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Using the power rule for logarithms:

$$y = 2\log_e(e) - 2\log_e(x)$$y=2loge​(e)−2loge​(x)

Since $\log_e(e) = 1$loge​(e)=1:

$$y = 2(1) - 2\ln(x)$$y=2(1)−2ln(x)

$$y = 2 - 2\ln(x)$$y=2−2ln(x)

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Finding the first derivative:

$$\frac{dy}{dx} = \frac{d}{dx}(2) - \frac{d}{dx}(2\ln x)$$dxdy​=dxd​(2)−dxd​(2lnx)

$$\frac{dy}{dx} = 0 - 2 \cdot \frac{1}{x}$$dxdy​=0−2⋅x1​

$$\frac{dy}{dx} = -\frac{2}{x}$$dxdy​=−x2​

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Finding the second derivative:

Rewriting: $\frac{dy}{dx} = -2x^{-1}$dxdy​=−2x−1

Using the power rule:

$$\frac{d^2y}{dx^2} = -2 \cdot (-1)x^{-2}$$dx2d2y​=−2⋅(−1)x−2

$$\frac{d^2y}{dx^2} = 2x^{-2}$$dx2d2y​=2x−2

$$\frac{d^2y}{dx^2} = \frac{2}{x^2}$$dx2d2y​=x22​

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Therefore, $\frac{d^2y}{dx^2} = \frac{2}{x^2}$dx2d2y​=x22​