JEE Advanced · 2006 · 09 Apr 2006 · Application of Derivatives

We have $g^{\prime }(x)=f(x)$ by the fundamental theorem of calculus.

For critical points, $g^{\prime }(x)=0$:

• In $(1,2]$: $2-e^{x-1}=0\Rightarrow e^{x-1}=2\Rightarrow x=1+\ln 2$
• In $(2,3]$: $x-e=0\Rightarrow x=e$

Now $g^{\prime \prime }(x)=f^{\prime }(x)$:


At $x=1+\ln 2$: $g^{\prime \prime }(1+\ln 2)=-e^{\ln 2}=-2<0$, so local maximum.

At $x=e$: $g^{\prime \prime }(e)=1>0$, so local minimum.

Additionally, since $f(x)$ is discontinuous at $x=1$ and $x=2$, we need to check these points:

• At $x=1$: $f(1^{-})=e>0$ and $f(1^{+})=2-1=1>0$, but $f$ changes from increasing to decreasing behavior, giving a local maximum.
• At $x=2$: $f(2^{-})=2-e>0$ (since $e\approx 2.718$, this is negative), $f(2^{+})=2-e<0$, giving a local minimum.

Therefore, options (A) and (B) are correct.