JEE Main 2025 (Online) 24th January Evening ShiftQuestion 1
Let $(2,3)$ be the largest open interval in which the function $f(x)=2\log_e(x-2)-x^{2}+ax+1$ 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 :
Differentiation and its Application
Solution
Concept. The right end $3$ of the increasing interval is where $f'=0$; that gives $a$. Then find where $g'<0$. Formulas used. (1) $\dfrac{d}{dx}\left[(x-1)^3(x-2)^2\right]=(x-1)^2(x-2)(5x-8)$Step 1 - Find $a$. $f'(x)=\frac{2}{x-2}-2x+a$, and $f'(3)=0$ gives $2-6+a=0$, so $a=4$. (Check: $f'=\frac{2}{x-2}-2(x-2)>0\iff(x-2)^2<1$, i.e. $2<x<3$.) Step 2 - $g$. With $a=4$: $g(x)=(x-1)^3(x-2)^2$, and by (1) $$g'(x)=(x-1)^2(x-2)(5x-8).$$ Step 3 - Decreasing. $(x-1)^2\ge0$, so $g'<0$ iff $(x-2)(5x-8)<0$, i.e. $\frac85<x<2$. So $b=\frac85$, $c=2$. Step 4 - Value. $100\left(4+\frac85-2\right)=100\cdot\frac{18}{5}=360$. Answer: (A)