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QJEEM25Mafe9d: JEE Main 2025 Mathematics - Class 12 JEEM25Mafe9d Solution: Let be the largest open interval in which the function is…

Let (2,3)(2,3) be the largest open interval in which the function f(x)=2log⁡e(x−2)−x2+ax+1f(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 g(x)=(x−1)3(x+2−a)2\mathrm{g}(x)=(x-1)^3(x+2-\mathrm{a})^2 is strictly decreasing. Then 100(a+b−c)100(\mathrm{a}+\mathrm{b}-\mathrm{c}) is equal to :

Multiple Choice Options

  • A 360
  • B 420
  • C 160
  • D 280

Step-by-Step Verified Solution

amp;f′(x)=2x−2−2x+a≥0amp;f′′(x)=−2(x−2)2−2lt;0amp;f′(x)↓amp;f′(3)≥0amp;2−6+a≥0amp;a≥4amp;amin⁡=4amp; g(x)=(x−1)3(x+2−a)2amp; g(x)=(x−1)3(x−2)2amp; g′(x)=(x−1)32(x−2)+(x−2)23(x−1)2amp;=(x−1)2(x−2)(2x−2+3x−6)amp;=(x−1)2(x−2)(5x−8)lt;0amp;x∈(85,2)amp;100(a+b−c)=100(4+85−2)=360\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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