Question

Which of the following statement(s) is/are true for the function
$f\left(x\right)=(x-1{)}^2(x-2)+1$ defined on $[0,2]$?

• A. Range of $f$ is $\left[\frac{23}{27},1\right]$
• B. The coordinates of the turning point of the graph of $y=f\left(x\right)$ occurs at $(1,1)$ and $\left(\frac{5}{3},\frac{23}{27}\right).$
• C. The value of $p$ for which the equation $f\left(x\right)=p$ has $3$ distinct solutions lies in interval $\left(\frac{23}{27},1\right).$
• D. The area enclosed by $y=f\left(x\right),$ the lines $x=0$ and $y=1$ as $x$ varies from $0$ to $1$ is $\frac{7}{12}.$

Answer 1

Answer: (B), (C), (D)

The correct options are
B The coordinates of the turning point of the graph of $y=f\left(x\right)$ occurs at $(1,1)$ and $\left(\frac{5}{3},\frac{23}{27}\right).$
C The value of $p$ for which the equation $f\left(x\right)=p$ has $3$ distinct solutions lies in interval $\left(\frac{23}{27},1\right).$
D The area enclosed by $y=f\left(x\right),$ the lines $x=0$ and $y=1$ as $x$ varies from $0$ to $1$ is $\frac{7}{12}.$

$f\left(x\right)=(x-1{)}^2(x-2)+1$
Now, differentiating w.r.t. $x$, we get
$f^{\prime }\left(x\right)=2(x-1)(x-2)+(x-1{)}^2$
$=(x-1)\left[2\right(x-2)+x-1]$
$=(x-1)[3x-5]$

Hence,
$f^{\prime }\left(x\right)=0$ implies $x=1$ and $x=\frac{5}{3}$
Corresponding values of $y$ are $y=1$ and $y=\frac{23}{27}$ respectively.

$\therefore$ Co-ordinates of turning point of the graph of $f\left(x\right)$ occurs at $(1,1)$ and $(\frac{5}{3},\frac{23}{27})$

Now,
$f\left(x\right)=(x-1{)}^2(x-2)+1$
$=(x^2-2x+1)(x-2)+1$
$=x^3-2x^2-2x^2+4x+x-2+1$
$=x^3-4x^2+5x-1$
The value of $p$ for which the equation $f\left(x\right)=p$ has $3$ distinct solutions lies in interval $(\frac{23}{27},1)$

Area enclosed by $f\left(x\right),x=0,y=1$ and $x=1$ is
$A={\int }_0^1(1-f(x\left)\right)dx$
$\Rightarrow {\int }_0^1(4x^2-x^3-5x+2)dx$

$\Rightarrow A=\frac{7}{12}$

Hence, option B, C, and D.