CUET UG 2025 — Mathematics Algebra

Given $f(x) = \frac{2\log_e x}{x}$f(x)=x2loge​x​ where $x > 0$x>0, and the maximum value occurs at $x = e$x=e.

For the quotient $\frac{u}{v}$vu​ where $u = 2\ln x$u=2lnx and $v = x$v=x:

$u' = \frac{2}{x}$u′=x2​ and $v' = 1$v′=1

$$f'(x) = \frac{x \cdot \frac{2}{x} - 2\ln x \cdot 1}{x^2}$$f′(x)=x2x⋅x2​−2lnx⋅1​

$$f'(x) = \frac{2 - 2\ln x}{x^2}$$f′(x)=x22−2lnx​

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For $f'(x) = \frac{2 - 2\ln x}{x^2}$f′(x)=x22−2lnx​ where $u = 2 - 2\ln x$u=2−2lnx and $v = x^2$v=x2:

$u' = -\frac{2}{x}$u′=−x2​ and $v' = 2x$v′=2x

$$f''(x) = \frac{x^2 \cdot \left(-\frac{2}{x}\right) - (2 - 2\ln x) \cdot 2x}{(x^2)^2}$$f′′(x)=(x2)2x2⋅(−x2​)−(2−2lnx)⋅2x​

$$f''(x) = \frac{-2x - 4x + 4x\ln x}{x^4}$$f′′(x)=x4−2x−4x+4xlnx​

$$f''(x) = \frac{-6x + 4x\ln x}{x^4}$$f′′(x)=x4−6x+4xlnx​

$$f''(x) = \frac{x(-6 + 4\ln x)}{x^4}$$f′′(x)=x4x(−6+4lnx)​

$$f''(x) = \frac{-6 + 4\ln x}{x^3}$$f′′(x)=x3−6+4lnx​

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At $x = e$x=e:

$$f''(e) = \frac{-6 + 4\ln e}{e^3}$$f′′(e)=e3−6+4lne​

$$f''(e) = \frac{-6 + 4(1)}{e^3}$$f′′(e)=e3−6+4(1)​

$$f''(e) = \frac{-2}{e^3}$$f′′(e)=e3−2​

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$$e^3 f''(e) = e^3 \times \frac{-2}{e^3}$$e3f′′(e)=e3×e3−2​

$$e^3 f''(e) = -2$$e3f′′(e)=−2

Therefore, $e^3 f''(e) = -2$e3f′′(e)=−2