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JEE Main 202524 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2025 · 24 January · Evening Shift · Mathematics · Application of Derivatives

Let (2,3) be the largest open interval in which the function f(x)=2...

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(a+b-c) is equal to :

Options

  1. A420
  2. B360
  3. C160
  4. D280

Correct answer

B. 360

Step-by-step solution

f′(x)=2x−2−2x+a≥0f′′(x)=−2(x−2)2−2<0f′(x)↓f′(3)≥02−6+a≥0a≥4amin⁡=4 g(x)=(x−1)3(x+2−a)2 g(x)=(x−1)3(x−2)2 g′(x)=(x−1)32(x−2)+(x−2)23(x−1)2=(x−1)2(x−2)(2x−2+3x−6)=(x−1)2(x−2)(5x−8)<0x∈(85,2)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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