CUET UG 2025 — Mathematics Calculus

Given $f(x) = x^3\log_e x = x^3 \ln x$f(x)=x3loge​x=x3lnx

Using the product rule with $u = x^3$u=x3 and $v = \ln x$v=lnx:

$$u' = 3x^2$$u′=3x2

$$v' = \frac{1}{x}$$v′=x1​

$$f'(x) = 3x^2 \cdot \ln x + x^3 \cdot \frac{1}{x}$$f′(x)=3x2⋅lnx+x3⋅x1​

$$f'(x) = 3x^2 \ln x + x^2$$f′(x)=3x2lnx+x2

$$f'(x) = x^2(3\ln x + 1)$$f′(x)=x2(3lnx+1)

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Using the product rule with $u = x^2$u=x2 and $v = (3\ln x + 1)$v=(3lnx+1):

$$u' = 2x$$u′=2x

$$v' = \frac{3}{x}$$v′=x3​

$$f''(x) = 2x \cdot (3\ln x + 1) + x^2 \cdot \frac{3}{x}$$f′′(x)=2x⋅(3lnx+1)+x2⋅x3​

$$f''(x) = 2x(3\ln x + 1) + 3x$$f′′(x)=2x(3lnx+1)+3x

$$f''(x) = 6x\ln x + 2x + 3x$$f′′(x)=6xlnx+2x+3x

$$f''(x) = 6x\ln x + 5x$$f′′(x)=6xlnx+5x

$$f''(x) = x(6\ln x + 5)$$f′′(x)=x(6lnx+5)

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At $x = e^2$x=e2:

$$f''(e^2) = e^2(6\ln(e^2) + 5)$$f′′(e2)=e2(6ln(e2)+5)

Since $\ln(e^2) = 2\ln(e) = 2$ln(e2)=2ln(e)=2:

$$f''(e^2) = e^2(6 \times 2 + 5)$$f′′(e2)=e2(6×2+5)

$$f''(e^2) = e^2(12 + 5)$$f′′(e2)=e2(12+5)

$$f''(e^2) = 17e^2$$f′′(e2)=17e2

Therefore, $f''(e^2) = 17e^2$f′′(e2)=17e2