Question

Let the function $f\left(x\right)=3x^2-4x+8\log (1+|x|)$ be defined on the interval $[0,1]$. The even extension of $f\left(x\right)$ of the interval $[-1,1]$ is

• A. $3x^2-4|x|+8\log (1+|x|)$
• B. $3x^2-4x+8\log (1+|x|)$
• C. $3x^2+4x-8\log (1+|x|)$
• D. $3x^2-4x-8\log (1+|x|)$

Answer 1

The correct option is A $3x^2-4|x|+8\log (1+|x|)$

Given the function $f\left(x\right)=3x^2-4x+8\log (1+|x|)$ be defined on the interval $[0,1]$.

Now we know, $x^2$ is an even function $\forall x\in R$.

But $x$ in an odd function for $x\in [-1,1]$ to make it even we are to extend it to $|x|$. And $\log (1+|x|)$ is also an even function $\forall x\in R$.

So the resulting extended even function will be $3x^2-4|x|+8\log (1+|x|)$.