Mathematics · Application of Derivatives

JEE Main 2025 — 24 January, Evening Shift — Question 10

Let $(2,3)$ be the largest open interval in which the function $f(x)=2 \log _{e}(x-2)-x^{2}+a x+1$

is strictly increasing and (b, c) be the largest open interval, in which the function

$g(x)=(x-1)^{3}(x+2-a)^{2}$ is strictly decreasing. Then $100(a+b-c)$ is equal to:

Answer: B

$f^{\prime}(x)=\frac{2}{x-2}-2 x+a \geq 0$

$\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{-2}{(\mathrm{x}-2)^{2}}-2<0$

$\mathrm{f}^{\prime}(\mathrm{x}) \downarrow$

$\mathrm{f}^{\prime}(3) \geq 0$

$2-6+a \geq 0$ $\mathrm{a} \geq 4$

$\mathrm{a}_{\text {min }}=4$

$\mathrm{g}(\mathrm{x})=(\mathrm{x}-1)^{3}(\mathrm{x}+2-\mathrm{a})^{2}$

$\mathrm{g}(\mathrm{x})=(\mathrm{x}-1)^{3}(\mathrm{x}-2)^{2}$

$\mathrm{g}^{\prime}(\mathrm{x})=(\mathrm{x}-1)^{3} 2(\mathrm{x}-2)+(\mathrm{x}-2)^{2} 3(\mathrm{x}-1)^{2}$

$=(\mathrm{x}-1)^{2}(\mathrm{x}-2)(2 \mathrm{x}-2+3 \mathrm{x}-6)$

$=(\mathrm{x}-1)^{2}(\mathrm{x}-2)(5 \mathrm{x}-8)<0$ $\mathrm{x} \in\left(\frac{8}{5}, 2\right)$

$100(a+b-c)=100\left(4+\frac{8}{5}-2\right)=360$

Answer key and solution verified before publishing.

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