Question

If $f\left(x\right)=\underset{2}{\overset{x}{\int }}\left\{2\right(t-2\left)\right(t-3{)}^3+3(t-2{)}^2(t-3{)}^2\}dt$, then which of the following is/are correct?

• A. $x=2$ is point of local maximum
• B. $x=3$ is point of local minimum
• C. $x=\frac{12}{5}$ is point of local minimum
• D. $x=3$ is point of inflection

Answer 1

Answer: (A), (C), (D)

The correct options are
A $x=2$ is point of local maximum
C $x=\frac{12}{5}$ is point of local minimum
D $x=3$ is point of inflection

$f^{\prime }\left(x\right)=2(x-2)(x-3{)}^3+3(x-2{)}^2(x-3{)}^2$
$=(x-2)(x-3{)}^2[2(x-3)+3(x-2)]$
$=(x-2)(x-3{)}^2[5x-12]$
For $f^{\prime }\left(x\right)=0$, we get
$x=2,$ $x=3$ and $x=\frac{12}{5}$

$f^{\prime \prime }\left(x\right)=(x-3{)}^2(5x-12)+2(x-2\left)\right(5x-12\left)\right(x-3)+5(x-2\left)\right(x-3{)}^2$
$f^{\prime \prime }\left(2\right)=-2<0$
$x=2$ is local maximum.
$f^{\prime \prime }\left(3\right)=0$ and $f^{\prime \prime \prime }\left(3\right)\ne 0$
$\Rightarrow x=3$ is point of inflection.
$f^{\prime \prime }\left(12/5\right)=5\frac{2}{5}\frac{9}{25}>0$
$\Rightarrow x=\frac{12}{5}$ is point of local minimum.