JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Application of Derivatives

For $f(x)=2\ln (x-2)-x^2+ax+1$ on $x>2$:


Since $f^{\prime \prime }(x)<0$ throughout, $f^{\prime }$ is strictly decreasing on the whole domain. For $(2,3)$ to be the *largest* interval on which $f$ is increasing, $f^{\prime }$ must cross zero exactly at $x=3$ (positive just before, negative just after):


With $a=4$, $g(x)=(x-1{)}^3(x+2-a{)}^2=(x-1{)}^3(x-2{)}^2$. Differentiating:


Since $(x-1{)}^2\ge 0$, $g^{\prime }(x)<0$ exactly when $(x-2)(5x-8)<0$, i.e. between the roots $x=\frac{8}{5}$ and $x=2$:


So $b=\frac{8}{5},c=2$, and