Question

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

then $f\left(x\right)$ has ?

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

Answer 1

The correct option is D All of these

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

$\Rightarrow f\left(x\right)={\int }_1^x(t-1)(t-2{)}^2[5t-7]dt$

By the fundamental theorem of calculus we have the result that

$\frac{d}{dx}({\int }_a^{x}f\left(x\right)dx)=f(x)$

So $f^{\prime }\left(x\right)=(x-1)(x-2{)}^2[5x-7]$

Finding the double derivative of $f\left(x\right)$ we have

$f^{\prime \prime }\left(x\right)=(x-1)(x-2{)}^2+2(x-1\left)\right(5x-7\left)\right(x-2)+(x-2{)}^2(5x-7)$

To find the minima and maxima of a continous function we need to equate the derivative of the function to $0$

So , $f^{\prime }\left(x\right)=0$

$\Rightarrow (x-1)(x-2{)}^2[5x-7]=0$

$\Rightarrow x=1,x=2,x=\frac{7}{5}$

Now using the double derivative method ,

$f^{\prime \prime }\left(1\right)=-2<0$ $\Rightarrow$ Minima at $x=1$

$f^{\prime \prime }\left(2\right)=0$ $\Rightarrow$ Neither minima nor Maxima at $x=2$

$f^{\prime \prime }\left(\frac{7}{5}\right)=\frac{18}{125}>0$ $\Rightarrow$ Maxima at $x=\frac{7}{5}$

So we have Maxima at $x=1$ , Minima at $x=\frac{7}{5}$ and Neither Minima nor Maxima at $x=2$

So the answer is correct answer is D: All Of These.