Question

Let $F\left(x\right)={\int }_0^x(t-1)(t-2{)}^{2}dt,$then

• A. $(1,-17/12)$ is a point of minimum
• B. $(2,-4/3)$ is a point of inflexion
• C. $(4/3,-112/81)$ is a point of inflexion
• D. $(1,-3)$ is a point of minimum

Answer 1

Answer: (A), (B), (C)

The correct options are
A $(1,-17/12)$ is a point of minimum
B $(2,-4/3)$ is a point of inflexion
C $(4/3,-112/81)$ is a point of inflexion

$F^{\prime }\left(x\right)=(x-1)(x-2{)}^2$ so $F^{\prime }\left(x\right)=0$ implies that $x=1$ and $x=2$
Around $x=1$ $F^{\prime }\left(x\right)$ changes sign from negative to positive.
Hence $F\left(x\right)$ is minimum at $x=1$ and $F\left(1\right)={\int }_0^1(t^3-5t^2+8t-4)dt=\frac{1}{4}-\frac{5}{3}+4-4=-\frac{17}{12}$
Hence,$(1,-17/12)$ is a point of minimum.
Moreover, $F^{\prime \prime }\left(x\right)=3x^2-10x+8=(3x-4)(x-2)$ and $F^{\prime \prime }\left(x\right)=0$ implies that $x=2,4/3.$
As $F^{\prime \prime }\left(x\right)$ changes sign around $x=2$ and $4/3$.

So $x=2$ and $x=4/3$ are points of inflexion.
Also $F\left(2\right)=-4/3$ and $F\left(4/3\right)=-112/81$.

So $(2,-4/3)$ and $(4/3,-112/81)$ are points of inflexion.