Question

Let $f\left(x\right)={\int }_1^x\{2(t-1){(t-2)}^3+3{(t-1)}^2{(t-2)}^2\}dt$ then

• A. $x=1$ is point of absolute minima
• B. $x=\frac{7}{5}$ is point of absolute maxima
• C. $x=2$ is point of inflexion
• D. None of these

Answer 1

Answer: (C)

$x=2$ is point of inflexion

$f\left(x\right)={\int }_1^x\{2(t-1){(t-2)}^3+3{(t-1)}^2{(t-2)}^2\}dt$
$\Rightarrow$ $f^{\prime }\left(x\right)=2(x-1){(x-2)}^3+3{(x-1)}^2{(x-2)}^2$
$=(x-1){(x-2)}^2[2{(x-2)}^2+3(x-1)]$
$=(x-1){(x-2)}^2(5x-7)$
for absolute values $f^{\prime }\left(x\right)=0$
$\Rightarrow$ $x=1,2,\frac{7}{5}$
and $f^{\prime \prime }\left(x\right)={(x-2)}^2(5x-7)+5(x-1){(x-2)}^2+2(x-1)(x-2)(5x-7)$
$\therefore$ $f^{\prime \prime }\left(1\right)=-2<0$
$\therefore$ $x=1$ is point of absolute maxima.
$\therefore$ choice (a) is false.
$f^{\prime \prime }\left(\frac{7}{5}\right)=5\left(\frac{2}{5}\right){(-\frac{3}{5})}^2>0$
$\therefore$ $x=\frac{7}{5}$ is point of absolute minima
$\therefore$ Choice (b) is false.
Again $f^{\prime \prime }\left(2\right)=0$ but after checking we find $f^{\prime \prime \prime }\left(2\right)\ne 0$
$\therefore$ $x=2$ is point of inflection.
Ans: C