Question

The function
$f\left(x\right)={\int }_1^x\left(2\right(t-1\left)\right(t-2{)}^3+3(t-1{)}^2(t-2{)}^2)$ dt attains its maximum at $x=$

• A. 1
• B. 2
• C. 7/5
• D. 5/7

Answer 1

The correct option is A 1

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

$\Rightarrow f^{\prime }\left(x\right)=2(x-1)(x-2{)}^3=0$ for extrema. $\Rightarrow x=1,2$
Also

$f^{\prime }\left(x\right)=2(x-2{)}^2+6(x-1\left)\right(x-2{)}^2$

Clearly $f^{\prime \prime }\left(1\right)<0$ and $f^{\prime \prime }\left(2\right)=0$

Hence $x=1$ is the point of maxima.