GATE 2022 EE Q.41: solution and answer

Official answer: ASource: Official answer key, GATE 2022 EE (IIT Kharagpur)

1. The integrand $g(t)=e^t(t-1)(t-2)$ is continuous for all $t$, so by the Fundamental Theorem of Calculus (Leibniz rule with a constant lower limit and upper limit $x$): $$f^{\prime }(x)=e^x(x-1)(x-2).$$

2. A differentiable function is decreasing on an interval where $f^{\prime }(x)<0$. Since $e^x>0$ for every real $x$, the sign of $f^{\prime }(x)$ is the sign of $(x-1)(x-2)$.

3. Sign chart of $(x-1)(x-2)$, which has roots $x=1$ and $x=2$: for $x<1$ both factors are negative, so the product is positive; for $1<x<2$ the factors have opposite signs, so the product is negative; for $x>2$ both factors are positive, so the product is positive.

4. Hence $f^{\prime }(x)<0$ exactly on $(1,2)$, so $f$ decreases on $x\in (1,2)$.

5. Check the other options: on $(2,3)$ we have $x>2$, and on $(0,1)$ and its sub-interval $(0.5,1)$ we have $x<1$; in all three $f^{\prime }(x)>0$, so $f$ increases there. The answer is (A).