Question

Suppose $f\left(x\right)$ is a polynomial of degree four, having critical points at $\left(-1,0,1\right)$. If $T=\left\{x\in \mathrm{ℝ}:f\left(x\right)=f\left(0\right)\right\}$, then the sum of squares of all the elements of $T$ is:

• A. $2$
• B. $6$
• C. $8$
• D. $4$

Answer 1

The correct option is D

$4$

Explanation for the correct option :

Step-1 : Find the expression for $f\text{'}\left(x\right)$

Given that $f\left(x\right)$ is a polynomial of degree four and has critical points at $x=-1,0,1$.

So, the derivative of $f\left(x\right)$ is a polynomial of degree three and has zeros at $x=-1,0,1$i.e. $(x+1)$, $x$ and $(x-1)$ are the factors of $f\text{'}\left(x\right)$.

Hence, we must have $f\text{'}\left(x\right)=k(x+1)x(x-1)$, where $k\ne 0$is a constant.

Step-2 : Find the expression for $f\left(x\right)$

Integrating both sides of $f\text{'}\left(x\right)=k(x+1)x(x-1)$, we get

$$\begin{gathered} f\left(x\right)=k\int \left(x^3-x\right)dx \\ =k\left[\frac{x^4}{4}+\frac{x^2}{2}\right]+c \end{gathered}$$

where $c$ is the constant of integration.

Step-3 : Evaluate $f\left(0\right)$

$$\begin{gathered} f\left(x\right)=k\left[\frac{x^4}{4}+\frac{x^2}{2}\right]+c \\ \Rightarrow f\left(0\right)=c \end{gathered}$$

Step-4 : Find the elements of $T$ i.e. the points $x$ for which $f\left(x\right)=f\left(0\right)$

Now,

$$\begin{gathered} f\left(x\right)=f\left(0\right) \\ \Rightarrow k\left[\frac{x^4}{4}-\frac{x^2}{2}\right]+c=c\left[\because f\left(0\right)=0\right] \\ \Rightarrow k\left[\frac{x^4}{4}-\frac{x^2}{2}\right]=0\left[\because k\ne 0\right] \\ \Rightarrow \frac{x^4}{4}-\frac{x^2}{2}=0 \\ \Rightarrow x^4-2x^2=0 \\ \Rightarrow x^2\left(x^2-2\right)=0 \\ \Rightarrow x=0,\sqrt{2},-\sqrt{2} \end{gathered}$$

So, the elements of $T$ are $0,\sqrt{2}$and $-\sqrt{2}$.

Step-5 : Evaluate the sum of the squares of elements of $T$.

The sum of the squares of elements of $T$

$$\begin{gathered} =0^2+{\left(\sqrt{2}\right)}^2+{\left(-\sqrt{2}\right)}^2 \\ =0+2+2 \\ =4 \end{gathered}$$

Hence, option (D) is the correct answer.