Question

Let $f$ be real valued function on $R$ defined as $f\left(x\right)=x^4(1-x{)}^2.$ Then which of the following statements is (are) CORRECT?

• A. $f^{\prime }\left(x\right)=0$ for some $x\in (0,1)$
• B. $f^{\prime \prime }\left(x\right)$ vanishes exactly twice in $R$
• C. $f\left(x\right)$ is an even function
• D. $f\left(x\right)$ is monotonically increasing in $(0,\frac{2}{3})\cup (1,\infty )$

Answer 1

Answer: (A), (D)

$f\left(x\right)$ is monotonically increasing in $(0,\frac{2}{3})\cup (1,\infty )$

Here, we can see that $f\left(x\right)\ne f(-x)$
So, clearly $f\left(x\right)$ is not an even function.

Consider $f\left(x\right)=x^4(1-x{)}^2,x\in [0,1]$
$\therefore f\left(x\right)$ being a polynomial, is continuous and differentiable also.
$f\left(0\right)=0=f\left(1\right)$
So, Rolle's theorem is applicable here.
$f^{\prime }\left(x\right)=4x^3+6x^5-10x^4$
So, there exists at least one $x\in (0,1)$ such that $f^{\prime }\left(x\right)=0$

$f^{\prime }\left(x\right)=4x^3+6x^5-10x^4$
$=2x^3(2+3x^2-5x)$
$=2x^3(x-1)(3x-2)$
$f^{\prime }\left(x\right)>0$ in $(0,\frac{2}{3})\cup (1,\infty )$
So, $f\left(x\right)$ is monotonically increasing in $(0,\frac{2}{3})\cup (1,\infty )$


$f^{\prime \prime }\left(x\right)=0$
$\Rightarrow 12x^2+30x^4-40x^3=0$
$\Rightarrow 2x^2(6+15x^2-20x)=0$
$\Rightarrow x=0$ or $6+15x^2-20x=0$
$15x^2-20x+6=0$
$D=400-360=40>0$
So, $15x^2-20x+6=0$ has two distinct real roots.
Hence, $f^{\prime \prime }\left(x\right)=0$ has three distinct solutions.