JEE Main 2025 Jan 24 Shift 2, Mathematics Q12: Monotonocity

Step 1 — Find $a$. Differentiate $f$:

$f^{\prime }(x)=\frac{2}{x-2}-2x+a$

$f^{\prime \prime }(x)=-\frac{2}{(x-2{)}^2}-2<0$, so $f^{\prime }$ is strictly decreasing. Hence $f^{\prime }>0$ holds on an interval of the form $(2,c)$, and the largest such interval ends where $f^{\prime }=0$.

Given the largest interval is $(2,3)$, we need $f^{\prime }(3)=0$:

$f^{\prime }(3)=\frac{2}{1}-6+a=0\Rightarrow a=4$

Step 2 — Find $(b,c)$. With $a=4$, $g(x)=(x-1{)}^3(x-2{)}^2$.

$g^{\prime }(x)=3(x-1{)}^2(x-2{)}^2+2(x-1{)}^3(x-2)=(x-1{)}^2(x-2)[3(x-2)+2(x-1)]$

$=(x-1{)}^2(x-2)(5x-8)$

The factor $(x-1{)}^2\ge 0$, so the sign of $g^{\prime }$ is the sign of $(x-2)(5x-8)$. The product is negative between the roots $\frac{8}{5}$ and $2$. Hence $g$ is strictly decreasing on $\left(\frac{8}{5},2\right)$, giving $b=\frac{8}{5}$, $c=2$.

Step 3 — Combine.

$100(a+b-c)=100\left(4+\frac{8}{5}-2\right)=100\left(\frac{18}{5}\right)=360$

Correct option: (2)