JEE MAIN - Mathematics (2025 - 24th January Evening Shift - No. 3)
Let $(2,3)$ be the largest open interval in which the function $f(x)=2 \log _{\mathrm{e}}(x-2)-x^2+a x+1$ is strictly increasing and (b, c) be the largest open interval, in which the function $\mathrm{g}(x)=(x-1)^3(x+2-\mathrm{a})^2$ is strictly decreasing. Then $100(\mathrm{a}+\mathrm{b}-\mathrm{c})$ is equal to :
360
420
160
280
Explanation
$$\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{x})=\frac{2}{\mathrm{x}-2}-2 \mathrm{x}+\mathrm{a} \geq 0 \\
& \mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{-2}{(\mathrm{x}-2)^2}-2<0 \\
& \mathrm{f}^{\prime}(\mathrm{x}) \downarrow \\
& \mathrm{f}^{\prime}(3) \geq 0 \\
& 2-6+\mathrm{a} \geq 0 \\
& \mathrm{a} \geq 4 \\
& \mathrm{a}_{\min }=4 \\
& \mathrm{~g}(\mathrm{x})=(\mathrm{x}-1)^3(\mathrm{x}+2-\mathrm{a})^2 \\
& \mathrm{~g}(\mathrm{x})=(\mathrm{x}-1)^3(\mathrm{x}-2)^2 \\
& \mathrm{~g}^{\prime}(\mathrm{x})=(\mathrm{x}-1)^3 2(\mathrm{x}-2)+(\mathrm{x}-2)^2 3(\mathrm{x}-1)^2 \\
& =(\mathrm{x}-1)^2(\mathrm{x}-2)(2 \mathrm{x}-2+3 \mathrm{x}-6) \\
& =(\mathrm{x}-1)^2(\mathrm{x}-2)(5 \mathrm{x}-8)<0 \\
& \mathrm{x} \in\left(\frac{8}{5}, 2\right) \\
& 100(\mathrm{a}+\mathrm{b}-\mathrm{c})=100\left(4+\frac{8}{5}-2\right)=360
\end{aligned}$$
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