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JEE Main 24-Jan-2025 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 24-Jan-2025 Paper - Shift 24 Marks
MCQ
Let $(2,3)$ be the largest open interval in which the function $f(x)=2 \log _{c}(x-2)-x^{2}+a x+1$ 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:
  • A
    280
  • ✓
    360
  • C
    420
  • D
    160
✓

Answer

Correct option: B.
360
(B)
Sol. $\quad f^{\prime}(x)=\frac{2}{x-2}-2 x+a \geq 0$
$\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{-2}{(\mathrm{x}-2)^{2}}-2<0$
$\mathrm{f}^{\prime}(\mathrm{x}) \downarrow$
$\mathrm{f}^{\prime}(3) \geq 0$
$2-6+a \geq 0$
$a \geq 4$
$\mathrm{a}_{\text {min }}=4$
$g(x)=(x-1)^{3}(x+2-a)^{2}$
$\mathrm{g}(\mathrm{x})=(\mathrm{x}-1)^{3}(\mathrm{x}-2)^{2}$
$\mathrm{g}^{\prime}(\mathrm{x})=(\mathrm{x}-1)^{3} 2(\mathrm{x}-2)+(\mathrm{x}-2)^{2} 3(\mathrm{x}-1)^{2}$
$=(x-1)^{2}(x-2)(2 x-2+3 x-6)$
$=(x-1)^{2}(x-2)(5 x-8)<0$
$x \in\left(\frac{8}{5}, 2\right)$
$100(a+b-c)=100\left(4+\frac{8}{5}-2\right)=360$

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