Answer :

Solution :

Given : ƒ(x) = ^x∫₀  e^t(t-1)(t-2) dt.

ƒ(x) decreases, that means slope of ƒ(x) means (ƒ'(x)) will be negative.

i.e.,     ƒ'(x)<0     (Slope = d/dx ƒ(x) = ƒ'(x) )

We need to find derivative of ƒ(x)

By Leibnitz rule,

ƒ'(x)= d/dx ^x∫₀  e^t(t-1)(t-2) dt.   ƒ'(x)<0

[ e^x(x²-3x+2)(1)-e⁰(0²+0+2)(0)] < 0

(x²-3x+2)< 0

so, 1<x<2

Hence, the correct option is (B).