Question

Let $f(x)=\int_{0}^{x}(t-1)(t-2)^{2} dt$. If $f(x) \geq k$ for all $x$ and for some $k,$ then the set of exhaustive values of $k$ is

• A. $\left(0 , \in fty\right)$
• B. $\left(0 , 2\right)$
• C. $\left(1 , \in fty\right)$
• D. $\left(-\infty,-\frac{17}{12}\right]$

Answer 1

Answer: (D)

$x=1$ is a point of global minima of $f\left(x\right)$
Minimum value of
$f\left(x\right)= \, \displaystyle \int _{0}^{1} \left(t - 1\right)\left(t - 2\right)^{2}dt$
$=\displaystyle \int _{0}^{1} \left(t^{3} - 5 t^{2} + 8 t - 4\right)dt$
$=\frac{- 17}{12}$