Question

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

$f\left(x\right)=(x-1{)}^2(x-2)+1$
$f^{\prime }\left(x\right)=2(x-1)(x-2)+(x-1{)}^2$
$\Rightarrow f^{\prime }\left(x\right)=(x-1)(3x-5)$
For max/min, $f^{\prime }\left(x\right)=0$
$\therefore x=1,\frac{5}{3}$
Sign scheme of $f^{\prime }\left(x\right):$


$\therefore$ Turning points are $(1,1)$ and $(\frac{5}{3},\frac{23}{27})$

$f\left(1\right)=1$ and $f\left(\frac{5}{3}\right)=\frac{23}{27}$
$f\left(0\right)=-1$ and $f\left(1\right)=1$
Hence, range of $f\left(x\right)$ is $[-1,1].$


Graph of $f\left(x\right):$



From the graph,
$f\left(x\right)=p$ has three solutions, when $p$ lies in the interval $(\frac{23}{27},1)$


Shaded area $=$ Area of rectangle $ABCD-$ Area of curve $ACD$
$=1\times 2-\underset{0}{\overset{1}{\int }}\left\{f\right(x)-(-1\left)\right\}dx$
$=2-\underset{0}{\overset{1}{\int }}\left\{\right(x-1{)}^2(x-2)+1-(-1)\}dx$
$=2-\underset{0}{\overset{1}{\int }}(x^3-4x^2+5x-1+1)dx$
$=2-{[\frac{x^4}{4}-\frac{4x^3}{3}+\frac{5x^2}{2}]}_0^1$
$=2-\frac{17}{12}=\frac{7}{12}$