Question

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

• A. $x^2+x+\sin x+\cos x-\log (1+|x|)$
• B. $-x^2+x+\sin x+\cos x-\log (1+|x|)$
• C. $-x^2+x+\sin x-\cos x-\log (1+|x|)$
• D. $x^2-x-\sin x+\cos x+\log (1+|x|)$

Answer 1

The correct option is B $-x^2+x+\sin x+\cos x-\log (1+|x|)$

Odd Extension of function mean extending the function to get odd function.
To make$f\left(x\right)$ an odd function in the interval $[-1,1]$, we re-define $f\left(x\right)$ as follows
$f\left(x\right)=\{\begin{array}{c}f\left(x\right),0\le x\le 1\\ -f(-x),-1\le x\le 0\end{array}$
$=\{\begin{array}{c}{x}^2+x+sinx-cosx+\log (1+|x|),0\le x\le 1\\ -x^2+x+sinx+cosx-\log (1+|x|),-1\le x\le 0\end{array}$
Thus the odd extension of $f\left(x\right)$ to the interval [- 1, 1] is $-x^2+x+$ sinx $+\cos x-\log (1+|x|)$
Hence, option 'B' is correct.