Question

The function
$f\left(x\right)={\int }_1^x\left[2\right(t-1\left)\right(t-2{)}^3+3(t-1{)}^2(t-2{)}^2]dt$ has :

• A. Maximum at $x=1$
• B. Minimum at $x=\frac{7}{5}$
• C. Neither maximum nor minimum at $x=2$
• D. All of these

Answer 1

Answer: (A)

Maximum at $x=1$

We have
$f\left(x\right)={\int }_1^x\left[2\right(t-1\left)\right(t-2{)}^3+3(t-1{)}^2(t-2{)}^2]dt$
$={\int }_1^x(t-1)(t-2{)}^2[5t-7]dt$
$\Rightarrow f^{\prime }\left(x\right)=(x-1)(x-2{)}^2(5x-7)$
for max. or min. $f^{\prime }\left(x\right)=0\Rightarrow x=1,2,\frac{7}{5}$
Now $f^{\prime \prime }\left(x\right)=(x-2{)}^2(5x-7)+5(x-1\left)\right(x-2{)}^2+2(x-2)(x-1)(5x-7)$
$\Rightarrow f^{\prime \prime }\left(0\right)<0,f^{\prime \prime }\left(7/5\right)>0$ and $f^{\prime \prime }\left(2\right)=0$
Hence $f\left(x\right)$ attains its maximum at $x=1$